Research Article Technology Licensing Strategy for …downloads.hindawi.com/journals/mpe/2015/179369.pdfResearch Article Technology Licensing Strategy for Network Product in a Service

Research ArticleTechnology Licensing Strategy for Network Product ina Service Industry

Xianpei Hong1 Dan Zhao2 Haiqing Hu3 and Shuang Song4

1College of Economics and Management Huazhong Agricultural University Wuhan 430070 China2School of Management Henan University of Science and Technology Luoyang 471023 China3School of Business Shandong Yingcai University Jinan 250104 China4College of Science Tianjin University of Technology Tianjin 300384 China

Correspondence should be addressed to Dan Zhao zhaodan911126com

Received 15 August 2014 Revised 15 October 2014 Accepted 2 November 2014

Academic Editor Bin Shen

Copyright copy 2015 Xianpei Hong et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Technology licensing has gained significant attention in literature and practice as a rapid and effective way to improve firmrsquoscapability of technology innovation In this paper we investigate a duopolistic service provider competition market where serviceproviders develop and sell a kind of network product In this setting we analyze the innovating service providerrsquos four licensingstrategies no licensing fixed fee licensing royalty licensing and two-part tariff licensing The literature suggests that when thenetwork products can be completely substituted two-part tariff licensing is the optimal strategy of the innovating service providerWe find that when the network products cannot be completely substituted two-part tariff licensing is not always optimal Thedegree of the product differentiation the intensity of the network effects and the RampD cost of the potential licensee play a key rolein determining the innovating service providerrsquos optimal licensing strategies

1 Introduction

Technology licensing not only is an important form oftechnology transfer under patent system but also is regardedas a crucial way to gain innovation profits through technologymarket for patent holder With the rapid development ofeconomic globalization and new technologies technologylicensing is becoming more and more common especially insome technology-intensive industries such as IBM gaininga profit of $13 billion (10 of its pretax profits) throughlicensing the technology in 2000 and Texas Instrumentswhich gets more than $40 million annually from patentlicensing [1]

Technology licensing is regarded as the fastest and mosteffective way to improve the firmsrsquo technology innovationcapability in addition to independent innovation [1] Thesignificance of technology licensing can be specified fromthe following two aspects Firstly for firms in developingcountries (such as China) confronted with overall weaknessof RampD capability a blind encouragement on innovation

may lead to higher input yet less effective output withthe loss outweighs the gain However technology licensingmay shorten the gap between abovementioned firms andthose with higher RampD efficiency in developed countriesthus preparing for further ldquoleapfrogrdquo Secondly for thosefirms with stronger innovation capability and higher RampDefficiency they should ponder over how to speed up therecovery of RampD investment to obtain higher economicprofits especially after they have patented the core tech-nology Technology licensing can provide a feasible way toimprove the return on investment for those kinds of firmsA study by Grindley and Teece [2] addresses that majorfirms in high-technology industries such as ATampT IBMTexas Instruments and Hewlett-Packard regard the use oflicensing and cross-licensing as an important part of theirbusiness strategies

In information and telecommunication industries manyinnovating service providers choose to license their technol-ogy to others to establish an industry innovation standardAs an example Qualcomm owned the core technology of

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 179369 17 pageshttpdxdoiorg1011552015179369

2 Mathematical Problems in Engineering

CDMA and LTE but it only established an innovation-basedstandard in other small markets such as Hong Kong andKorea before it entered the market of Chinese MainlandQualcomm obtained tremendous development opportunityafter it licensed its CDMA technology to China UnicomQualcomm gained a profit of 249 hundred million dollarsthrough chip distribution and technology licensing in 2013among which it gained a profit of 120 hundred milliondollars (84 of its whole profits) from China HUAWEI andZTE Corporation that have adopted CDMA technology payQualcomm a per unit charge (5) to access the CDMAnetwork That is why so many innovating firms are keeningon establishing an industry standard

Our research question arises from practice Howeverin theoretical research few address the firmrsquos technologylicensing issues for network products For a network productusers derive added value from the presence of others inthe network the additional network value depends on thenumber of other users of this product [3 4] Differentfrom the normal products the market for network productscontinues to expand with the popularity of the InternetThis phenomenon is commonly seen in reality for exampleinstant communication devices such as MSN and QQ are ofno value when users are too few However considering theincrement in the number of users of these devices their net-work values and consumer evaluations continuously increaseIn this paper we define the products with network effects asnetwork products and the products without network effectsas normal products

Based on the above analysis the purpose of this paperis to explore the following two issues by taking account ofnetwork effects of two competing firms under the situationwhere the innovator serves as an insider patentee as firmsproduce and engage in Cournot competition Firstly what isthe innovating firmrsquos optimal licensing strategy in terms offixed-fee royalty or two-part tariff licensing Secondly howdo the network intensity and product heterogeneity affect thetechnology licensing strategy for the innovating firm

The main findings of this paper are as follows (1) Forthe innovator provider no matter whether the market scalesare large or small and as long as the network intensity 120573is large enough (eg 120573 ge 05) two-part tariff licensingstrategy is the same as fixed-fee licensing strategy Howeveras for the innovator provider the optimal licensing strategyis not always two-part tariff licensing and is different fromthe case of normal product In a network product market theoptimal licensing strategy of the innovator provider dependson the market conditions to be specific when the marketscale is very small the optimal licensing strategy is fixed-fee licensing When the market scale is medium royaltylicensing is the optimal strategy when licensee firmrsquos RampDcost is very high When the market scale is very largethe optimal licensing strategy depends mainly on networkintensity product substitute levels and licensee firm RampDcost (2)When network intensity is relatively small (eg 120573 lt05) and the RampD cost is very high two-part tariff licensingis the optimal licensing strategy of the innovator provider

The reminder of this paper is organized as follows InSection 2 we briefly review the related research Section 3

describes the game process and market demand In Section 4we analyze three licensing strategies (ie no licensing fixed-fee licensing and royalty licensing) and report the resultsSection 5 extends to investigate two-part tariff licensingstrategy and Section 6 concludes this paper with a discussionof the results To simplify our exposition all proofs areprovided in the Appendix

2 Literature Review

An extensive theoretical literature has been developed toexplore the optimal licensing strategy These studies can bedivided into three categories as follows (1) optimal licensingstrategy considering the cost reduction innovation (eg [5ndash8] etc) (2) vertical product innovation to improve theproduct quality and the optimal licensing strategy (eg [9ndash14] etc) and (3) the cost reduction innovation and productinnovation For example Li andGeng [15] studied technologylicensing to the craft and product innovation of durablesThey pointed out that the optimal licensing contract isdetermined by innovation categories and innovation scaleroyalty licensing will be prioritized when it is cheaper thanthe reduction craft innovation or quality-improving productinnovation two-part tariff licensing will be prioritized whenthe scale involved is greater than that of craft innovation orvertical product innovation and fixed-fee licensing will beprioritized when the products are average

In this paper we mainly study the optimal licensingstrategies considering the cost reduction process innovationLicensing strategy has received large amount of attentionfor instance Kamien and Tauman [16] studied the optimaltechnology licensing strategy to an external innovator usingthe noncooperative game theory They pointed out thatusing fixed-fee licensing licensee firms gain lesser benefitsthan those with no licensing because of the fixed-fee andissuing number of licenses whereas royalty licensing doesnot affect licensee firmrsquos profits Thus they addressed thefixed-fee licensing being significantly better than royaltylicensing Muto [17] first introduced the product substituteparameter to technology licensing considering cost reductioncraft innovation and found that royalty licensing is signif-icantly better than fixed-fee licensing when the innovationscale is comparatively small Crama et al [18] studied theoptimal contract of RampD licensing Their results reveal thatthe optimal licensing contract should be three-part tarifflicensing namely payment in advance milestone paymentand royalty because of disparities between the estimated valuelicensor and licensee and the limited control of the licensorover the RampD efforts of the licensee Mukherjee [19] figuredout that as long as the Union has complete control overthe bargain licensee firms would pay lesser salaries underthe royalty licensing situation than under auction or fixed-fee licensing regardless of whether the Union structure is acentralized control mode or a decentralized control modeRoyalty licensing is better thanfixed-fee licensing and auctionlicensing for external innovators because the positive effectsof low salary outweigh the negative effects of firm margincosts posed by the royalty licensing rate

Mathematical Problems in Engineering 3

More recently a stream of research has emerged thatexplores the optimal strategy of network products Lin andKulatilaka [20] studied the choice of the optimal licensingstrategy for duopoly markets The results show that with theenhancement of the intensity of the network effects the opti-mal licensing strategy became fixed-fee licensing instead ofroyalty licensing However all network products are assumedreplaceable that is products are completely homogeneousHowever in reality many products have the qualities ofheterogeneity Thus the study of optimal licensing strategyfor heterogeneous products is meaningful and is a usefuladdition to the literature on network product technologylicensing

In addition this paper is also related to service contractdesign andmarket competitionThere are a growing numberof research papers on these themes we refer the reader toChoi et al [21] Chiu et al [22] Chan and Gao [23] Li et al[24] Xu et al [25] and so forth for reference

The main contributions of this paper are summarized asfollows First different with most previous research whichdid not consider the network effects and substitution coef-ficient between different products or services in the marketdemand function our model illustrates that network effectsand substitution coefficient strongly influence the choice ofthe licensing strategy Second this paper indicates that theamount of RampD investment of the competing firm to a largeextent determines the options of the innovatorrsquos technologylicensing strategies which has not been considered in otherrelated literature

3 Problem Descriptions

31 Description of the Game Process We consider a duopolymarket with two competing service providers (hereafterSP for short) and both SPs can conduct new productsdevelopmentwhich experience a network effect Suppose thatone SP has developed a new technology and the other SPlagged behind temporarily he can either develop a substitutetechnology by investing RampD cost 119896 or buy the technologyfrom the other SP As a reminder we call the SP whodevelop the new technology first the innovator SP and theSP who follows the innovator the follower SP For the sake ofsimplicity we use SP1 and SP2 to represent the innovator SPand the follower SP respectively SP1 can either monopolizethe technology or license it to SP2 SP2 can decide whetherto accept the quotation or develop a new technology by itselfTo be specific there are two cases on both SPsrsquo competitionas followsCase 1 SP1 chooses to monopolize the technology and SP2can develop a new technology by itself Both of themwill thenexperience Cournot competition in the market It is assumedthat two products cannot be substituted and are also notcompatible because of the different technological standardsHowever if the market scale is not large enough SP2 will notinvest 119896 to develop a substitute innovation In this case SP1becomes a monopolist and enjoys monopoly profitCase 2 SP1 chooses to license its technology to SP2 andproposes a licensing quotation SP2 can choose to accept it or

to refuse it If SP2 accepts the licensing it will experience theCournot competition with SP1 and only one technologicalstandard exists in the market If SP2 refuses this licensing itwill develop new technology and two technological standardswill exist in the market

32 Network Product Inverse Demand Function and NetworkIntensity The inverse demand function of formal productscan be defined as 119901(119876 120579) = 120579 minus 119876 where 119876 is the demandquantity of the product 120579 is the market scale or the potentialmaximum market demand and all SPs know the value of120579 Different from the normal product the user number canincrease the additional values for the network products andthe additional values depend on the base number of usersthe inverse function of the network products is 119901(119876 119902119890 120579) =120579 + V(119902119890) minus 119876 [20] where 119902119890 is the expectation of the networkby the users and V(119902119890) is the estimated value of the network bythe users and is an increasing function of 119902119890 120579 is the greatestmarket demand when the network products do not possessthe network additional values

In the different licensing strategies the inverse function ofthe network productions may be displayed in different waysIf SP1 does not license technology to SP2 for example the nolicensing strategy SP2makes investment 119896 then both SPs willhave two incompatible standards Thus the users of each SPcan form their own network The products of the two SPs areincompatible so we can define the inverse function as119901

119894= 120579+

V(119902119890119894)minus119902119894minus1198891199023minus119894

119894 = 1 2 and 119889 isin [0 1] is the substituting ratebetween different products A larger substituting ratemeans atougher market competition between products in particularwhen 119889 = 1 the products can be substituted perfectly

When SP1 licenses its technology to SP2 by means offixed-fee licensing royalty licensing or two-part tariff licens-ing both SPs compete under the same technology standardexperiencing a Cournot competition at the production stageIn this case the users of both SPs form a larger networkresulting in higher network value The inverse demandfunction is therefore given by 119901 = 120579 + V(119902119890

1+ 1199021198902) minus 1199021minus 1199022

According to Metcalfersquos law the total values of networkproducts are proportional to the base number of the squareof network users [26] For a certain user the value of networkproducts can be defined as V(119902) = 120573119902 and 120573 isin [0 1) Notethat we assume 120573 lt 1 to ensure that the demand function hasdownward sloping or related properties Therefore the validvalue of 120573 falls into [0 1] This value is the network intensitythat reflects the intensity of the network effects Consideringthat the base numbers of the users are the same the greater120573 is the greater the value to users brought by the networkproducts is However when 120573 and V(119902) = 0 the networkproducts will be degenerated as normal products

The network effects may affect the licensing decisions ofSP1 and the adoption of SP2 When SP1 does not license itstechnology or when both parties cannot reach an agreementon the licensing the investment of SP2 in new productsdevelopment leads to two different standards Given thiscondition only the users who buy products from the sameSP will form a network and they can simultaneously enjoythe additional network value Thus if these two standardsare compatible then users will not be willing to pay for


the network products When SPs conduct their licensingstrategy they obviously need to consider this effect

If SP1 licenses its technology to SP2 both of them willform a large network to market products instead of twoincompatible small markets In this case users have a greaterdesire to pay for the additional network value SP1 can obtainhigher price and greater benefits Using technology licensingSP1 can set the standard in its industry and gain licensingbenefits

4 Three Technology Licensing Strategies

In this section we investigate three technology licensingstrategies namely no licensing fixed-fee licensing androyalty licensing By comparing SP1rsquos profits of these threetechnology licensing strategies with product heterogeneityand network effects we explore the optimal licensing deci-sions

41 No Licensing Strategy In this case SP1 does not licensetechnology to SP2 SP2 makes investment 119896 and then bothSPs will have two incompatible standards Thus the usersof each SP can form their own network With the abovediscussion the profit functions of both SPs are given asfollows

120587119873

119894= [120579 + V (119902119890

119894) minus 119902119894minus 119902119895] 119902119894 (1)

where 119894 119895 = 1 2 and 119894 = 119895 and the superscript 119873 representsthe situation of no licensing

In order to derive both SPsrsquo optimal production decisionswe solve the first-order condition of (1) and impose a fulfilledexpectation equilibrium (FEE) condition [27] Then we canobtain the equilibrium quantities and profits as follows

119902lowast

119894=

120579

2 + 119889 minus 120573 120587

lowast

119894= [

120579

2 + 119889 minus 120573]

2

(2)

where 119894 119895 = 1 2 and 119894 = 119895Using the above function we can obtain 120587119873

2= 120587lowast2minus 119896

where 1205871198732denotes SP2rsquos net profit from investing 119896 to develop

its own standard SP2 produces the new products only when1205871198732gt 0 that is 120579 gt 120579 = (2 + 119889 minus 120573)radic119896 When 120579 le 120579 SP2

does not produce the products or withdraw from the marketIn this case SP1 becomes themonopoly in themarket and theproducts of SP1 are the only standard for the industry and SP1gains monopoly profit 120587119873

1= 1205871198721= [120579(2 minus 120573)]2 where the

superscript119872 represents the situation that SP1 monopolizesthe market Thus when SP1 does not execute a license bothSPsrsquo profits can be summarized as follows respectively

(120587119873

1 120587119873

2) =

([120579

2 minus 120573]

2

0) 120579 le 120579

([120579

2 + 119889 minus 120573]

2

[120579

2 + 119889 minus 120573]

2

minus 119896) 120579 gt 120579

(3)

42 Fixed-Fee Licensing Strategy In this case we consider asituation where SP1 licenses its technology to SP2 with fixed-fee licensing and SP2 pays a fixed-fee to SP1 for the technol-ogy Both SPs compete under the same technology standardand experience a Cournot competition at the productionstage In this licensing strategy the users of both SPs forma larger network resulting in higher network value The SPrsquosprofit is 120587119865

119894= [120579 + V(119902119890

1+ 1199021198902) minus 1199021minus 1199022]119902119894 119894 = 1 2 denoting

SP1 and SP2 respectivelyThus according to the conditions ofFEE and the optimal production decisions of the two SPs wecan obtain the following equilibriumquantity 119902lowast

119894= 120579(3minus2120573)

119894 = 1 2In the fixed-fee licensing strategy the profit functions of

both SPs are given by

(120587119865

1 120587119865

2) = ([

120579

2 minus 3120573]

2

+ 119865 [120579

2 minus 3120573]

2

minus 119865) (4)

where the superscript 119865 represents the fixed-fee licensingcase

At the licensing stage SP1 decides to implement licensingand the fixed-fee is 119865 SP1 offers a ldquotake-it-or-leave-itrdquocontract to SP2Note thatwhen there is no difference betweenrefusing and accepting licensing SP2 chooses to acceptlicensing Thus the optimal fixed-fee 119865 can be obtained by

Max119865

120587119865

1=

1205792

(3 minus 2120573)2+ 119865

st 120587119865

2=

1205792

(3 minus 2120573)2minus 119865 ge 120587

119873

2

(5)

According to (4) we can obtain the optimal prepaid fixed-fee as follows

119865lowast

=1205792

(3 minus 2120573)2minus 120587119873

2

=

1205792

(3 minus 2120573)2 120579 le 120579

(5 + 119889 minus 3120573) (119889 minus 1 + 120573) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(6)

Then we can obtain the profit of SP1 as follows

120587119865

1

=

21205792

(3 minus 2120573)2 120579 le 120579

[(2 + 119889 minus 120573)2

+ (5 + 119889 minus 3120573) (119889 minus 1 + 120573)] 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(7)

The profit of SP2 can be derived from (3) when SP1charges the maximum fixed-fee


Theorem 1 (1) When the market is small (120579 le 120579) if thenetwork intensity is large (120573 ge 120573lowast

1= 0293) SP1 makes

more profit under fixed-fee licensing than no licensing and thenfixed-fee licensing occurs Otherwise for SP1 no licensing isbetter than fixed-fee licensing (2) When the market scale islarge (120579 gt 120579) if the network intensity is large (120573 ge 120573

lowast

2=

1 minus 119889) fixed-fee licensing will occur Otherwise whether fixed-fee licensing is implemented for SP1 depends on the substituterate of network products network intensity and RampD cost ofSP2 (If the network intensity is small (120573 lt 120573lowast

2= 1minus119889) then we

have the following (1) When the product substitute rate is verysmall (119889 lt 119889

1= 0051) then regardless of RampD cost of SP2

SP1 will choose not to implement fixed-fee licensing (2) Whenthe product substitute rate is very small (119889

1le 119889 le 119889

2= 0121)

then fixed-fee licensing will occur only if the network intensityis medium (120573lowast

3lt 120573 lt 120573lowast

4) and RampD cost of SP2 is very high

(1198962le 119896 lt 119896

1) (3) When the product substitute rate is medium

(1198892lt 119889 lt 119889

3= 0196) then fixed-fee licensing will occur

only when the network intensity is very small (120573 lt 120573lowast4) and

RampD cost of SP2 is very high (1198962le 119896 lt 119896

1) (4) When the

product substitute rate is large enough (119889 ge 1198893) then as long

as RampD cost of SP2 is very high (1198962le 119896 lt 119896

1) SP1 will choose

to implement fixed-fee licensing)

Theorem 1 shows that for network products whether SP1licenses its technology to SP2 depends on the market scalenetwork intensity product substitute rate and SP2rsquos RampDcostinvestment to develop the new technology When the marketscale is very small (120579 le 120579) products of SP1 can cover the entiremarket because of the limited market volume In this case ifthe network intensity of the products is very low (120573 lt 120573

lowast

1)

then the additional network value brought by the increasein the base number of the users will not compensate for thenegative effect such as the decrease in the price as a resultof overproduction Thus the holder of the patent shouldimplement no licensing or monopolize the market Unlessthe network is strong enough the holder of the patent willsacrifice the monopolizing profits and choose to implementfixed-fee licensing

When the market scale is very large (120579 gt 120579) and thenetwork intensity is strong enough (120573 ge 120573

lowast

2) on the one

hand the products of the holder of the patent (ie SP1)may not cover the whole market if the holder does notimplement the licensing SP2 can develop its own productsand if it competes with SP1 the SP1rsquos profit will decline Onthe other hand network intensity may be strong enoughThus if SP1 chooses to implement licensing then the basenumber of the users and network intensity will increase toimprove the estimated values of users and the profit of SP1will improve Thus given this condition SP1 will choosefixed-fee licensing However when network intensity is notvery strong then whether SP1 chooses to implement fixed-fee licensing depends on the following aspects among others(1) Is the RampD cost of SP2 high or not (2) If SP2 successfullydevelops the new products what will characterize the compe-tition between the new products and the existing products(3) Does a replacement exist for the network intensity andproduct substitute rate and the RampD cost of SP2 Theseconsiderations lead to four conclusions (see the proof of

Theorem 1 in the Appendix) In particular when 119889 = 1

and the market scale is large enough fixed-fee licensing candefinitely occurThis conclusion agrees with thatmade by Linand Kulatilaka [20]

43 Royalty Licensing Strategy In this case SP1 implementstechnology licensing to SP2 with royalty licensing SP2 paysa royalty rate 119903 to SP1 Both SPs follow the same technologystandard and experience a Cournot competition in themarket The profit functions of both SPs are as follows

120587119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022

120587119877

2= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199022minus 1199031199022

(8)

where the superscript 119877 represents royalty licensingWe can obtain the following equilibrium quantities

according to the conditions of FEE and the optimal produc-tion decisions of the two SPs

119902lowast

1=120579 + (1 minus 120573) 119903

3 minus 2120573 119902

lowast

2=120579 minus (2 minus 120573) 119903

3 minus 2120573 (9)

The equilibrium profits of both SPs are as follows respec-tively

120587119877

1=1205792 + (5 minus 4120573) 120579119903 minus (1205732 minus 5120573 + 5) 1199032

(3 minus 2120573)2

120587119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(10)

And the profit function of SP2 is a decreasing function of theroyalty rate because 119889120587119877

2119889119903 = minus2(2 minus 120573)[120579 minus (2 minus 120573)119903](3 minus

2120573)2

le 0 At the licensing stage SP1 as the licensor choosesthe optimal royalty rate 119903lowast

1to maximize profit while the

optimal royalty rate cannot be higher than the royalty rate1199032 which is the highest royalty rate acceptable to SP2 and we

define 119903lowast = min(119903lowast1 1199032)Thus the highest royalty rate that SP2

will accept depends on 1205871198772= 1205871198731198712

because SP1 has completepower to bargainWe can obtain the following optimal royaltyrate by (3) and (10) Consider

1199032=

120579

2 minus 120573 120579 le 120579

120579

2 minus 120573minus3 minus 2120573

2 minus 120573radic

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(11)

According to the first-order condition the optimal roy-alty rate of SP1 can be obtained from (10) We can obtain thefollowing result

119903lowast

1=

(5 minus 4120573) 120579

2 (1205732 minus 5120573 + 5) (12)

We can test whether 119903lowast1and 119903lowast2are smaller than 120579(2 minus 120573)

the result of which is in accordance with the upper limit


of the royalty rate We need to compare the values of 119903lowast1

and 1199032to obtain the optimal royalty rate 119903lowast = min(119903lowast

1 1199032)

The following theorem yields the optimal royalty rate and itsconditions

Theorem 2 In royalty licensing mode the optimal royalty rateof SP1 depends on the critical value 120579 of the market scaleWhen120579 le 120579 the optimal royalty rate is 119903lowast = 119903lowast

1 when 120579 gt 120579 the

optimal royalty rate is 119903lowast = 1199032 where

120579 =2 (2 + 119889 minus 120573) (5 minus 5120573 + 1205732)radic119896

radic[10 + (119889 minus 8) 120573 + 1205732] [10 minus (12 + 119889) 120573 + 31205732]

(13)

Thus we can obtain both SPsrsquo profits at the licensing stageas follows

120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)

Under royalty licensing SP2 is always prepared to acceptlicensing because of Δ120587

2= 1205871198772minus 1205871198732ge 0 The question is

whether or not SP1 will implement licensing The followingresults provide the conditions under which SP1 chooses toimplement licensing

Theorem 3 When the market scale is comparatively small(120579 le 120579) then for any network intensity 120573 isin [0 1) royaltylicensing will always occur When the market scale is compar-atively large (120579 gt 120579) then the occurrence of royalty licensingdepends on the product substitution rate network intensityand the RampD cost of SP2

Theorem 3 shows that the increase in themarket scale aidsthe increase in the profit of SP1 When the market scale islarge enough the product substitute rate network intensityand RampD cost of the potential licensee determine the finaldecision of SP1 When the market scale is very small (120579 le

120579) and when the network product is regarded as a normalproduct (120573 = 0) SP1 will have no profits when it implementsroyalty licensing because the market is not large enough formore firms to enter If more firms participate in the compe-tition the overall profits will be close to zero Thus SP1 as

the one with the monopoly potential should not implementthe licensing In terms of network products (120573 gt 0) stiffercompetition among firms will decrease the industry profitsbut users give higher comments on the increase in networkscale indicating higher additional network value and moreindustry profits Given this condition the enlargement of themarket scale resulting from the positive effect (ie networkeffects) and the licensing benefits resulting from licensingoutweigh the negative effect which is the increase in thecompetition resulting from licensing

Considering the enlargement of themarket scale (120579 lt 120579 le120579) the network effect will decrease because if the networkproducts are regarded as normal products SP1 will alsochoose to implement royalty licensing Thus firms need toconsider the values of the positive effect caused by licensingand the negative effect caused by competition Howeverusers of different companies after licensing will form a largernetwork and produce greater additional network value thusleading to the positive effect caused by licensing that isgreater than the negative effect caused by competition Thepotential licensee can develop its own products and thusto some extent cause the opportunity cost to be higherwithout licensingAfter overall considerationwe find that thepositive effect is higher Thus royalty licensing is better to beimplemented when the market scale is larger

Once the market scale is large enough (120579 gt 120579) on theone hand SP1 will not be concerned about whether themarket is large enough for more companies but aboutwhether the market for network products can compensatefor the losses after licensing under the larger and the samestandard network product market The larger the networkintensity 120573 the stronger the motivation of SP1 to implementroyalty licensing On the other hand the stiffness of thecompetition between SP1 and SP2 under no licensing andwhen the standard is not the same should be consideredThe larger is the product substitute rate 119889 (which reflectsthe stiffness of the competition between the products) thestronger the motivation to implement licensing for SP1SP1 will also consider the motivation that SP2 accepts thelicensing The higher the RampD cost of SP2 the stronger themotivation to accept the licensing The lower the RampD costof SP2 the stronger its motivation to develop new productsUnder royalty licensing the lower the RampD cost of SP2the stronger the motivation of SP1 to implement the licens-ing

5 Optimal Licensing Strategies

In this section we discuss the optimal licensing strategyby comparing SP1rsquos profits under no licensing fixed-feelicensing and royalty licensing strategy This helps us betterunderstand the conditions under which SPs choose to imple-ment and accept the technology licensing

When themarket scale is small (120579 le 120579) fixed-fee licensingstrategy can be used only for SP1 when the network intensityis large enough (120573 gt 120573

lowast

1= 0293) and royalty licensing

is not influenced by network intensity 120573 Then using thecomparison of profits between the two licensing strategies we


can obtain 1205871198651= 1205871198771when 120573lowast

8= (5minusradic10)6 asymp 0306 120587119865

1lt 1205871198771

when 120573 lt 120573lowast8 and 120587119865

1gt 1205871198771when 120573 gt 120573lowast

8

When the market scale is medium (120579 lt 120579 le 120579) fixed-fee licensing can be used under four conditions and royaltylicensing always occurs Thus we need to compare SP1rsquosprofits given fixed-fee licensing and royalty licensing underthe four kinds of conditions After the comparison we candetermine that RampD cost of SP2 is 119896

5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2

(5minus5120573+1205732))1205792 andwe obtain1205871198651= 1205871198771

when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can

also obtain 1198965= 1198961when 120573lowast

9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9

and 1198965lt 1198961when 120573 gt 120573lowast

9 We can determine the following

(1)when 120573 lt 120573lowast9 for any 119896 lt 119896

1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771

However when the market scale is large enough (120579 gt 120579)the licensing strategy becomes very complicated Fixed-feelicensing can occur under four conditions and royalty licens-ing can occur under seven conditions Fifteen conditions canensure the occurrence of both licensing methods Thus theoptimal strategy can be obtained only by comparing the 15conditions However the profit functions of the fifteen con-ditions are the same considering the same licensing strategyFor convenience we first compare the profits and then inputthe limited conditions to test themAfter comparing120587119865

1under

fixed-fee licensing and 1205871198771under royalty licensing for SP1 we

obtain 1205871198771= 1205871198651when RampD cost of SP2 is

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1

Finally we compare 1198966with the 15 critical values and then

we obtain the eight conditions that enable the simultaneousoccurrence of both licensing strategies All the conclusionsare summarized in Table 1

Analyzing all the above results we can obtain Proposi-tion 4

Proposition 4 (1) When the market scale is small (120579 le 120579) if120573lowast

1le 120573 lt 120573lowast

8 royalty licensing is better than fixed-fee licensing

and if120573 gt 120573lowast8 fixed-fee licensing is better than royalty licensing

(2) When the market scale is medium (120579 lt 120579 le 120579) or largeenough (120579 gt 120579) the degree of the differentiation of the productsthe intensity of the network effects and the RampD cost of SP2play a key role in determining the optimal licensing strategy ofSP1

The result of Proposition 4 has the following implicationsfor SPsrsquo licensing strategies Firstly when the market scaleis small (120579 le 120579) for any positive network intensity

implementing technology licensing is better for SP1 Thisimplication also indicates that obtaining monopoly profits isnot the optimal choice

Secondly when the market scale is medium (120579 lt 120579 le 120579)developing new technologies and challenging the monopolyposition of SP1 are possible for SP2 When SP2 chooses itsown technology and competes with SP1 in the market theincomplete substitute of the products reflects the intensityof the competition between the companies Moreover theprediction of fierce competition between two SPs and theconsideration of SP2rsquos profitabilitywill influence SP1rsquos attitudetoward technology licensing and the SPrsquos approach to itsimplementation Thus predictably whether SP1 will chooseto implement the technology licensing depends on theproduct substitution rate 119889 and SP2rsquos RampD cost 119896 except forthe network intensity 120573

Finally when the market scale is large enough (120579 gt

120579) there is no certainty that the two strategies (ie fixed-fee licensing and royalty licensing) will always be betterthan no licensing This situation shows that except for theaforementioned key parameters (ie network intensity 120573product substitution rate119889 and SP2rsquos RampDcost 119896) technologylicensing is also influenced by market scale 120579 Based on thecritical values of the various kinds of functions of SP2rsquos RampDcost the market scale 120579 exerts influence on the essenceThusimplementing technology licensing is not always optimal forSP1 By determining that two licensing methods can occurat the same time we can reach the conclusion of 120579 gt 120579 inProposition 4

6 Two-Part Tariff Licensing

In this section we consider a two-part tariff licensing strategywhere SP1 charges a prepaid fixed-fee and a royalty per unitof SP2rsquos output In normal product market two-part tarifflicensing mechanism with its complete and flawless infor-mation is superior to both fixed-fee licensing and royaltylicensing mechanisms Is it still the optimal mechanism orstrategy for SP1 in the network product market To answerthis question we propose a model where SP1 licenses itstechnology to SP2 by means of two-part tariff licensing in thenetwork product market

Under two-part tariff licensing the profit functions ofboth SPs are as follows

120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199022minus 1199031199022minus 119865

(18)

where the superscript 119865119877 represents two-part tariff licensingSolving the above problems yields the following profits

120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)


Table 1 Conditions that enable the simultaneous occurrence of both licensing strategies

Product substitution rate and network intensity condition RampD cost condition Profits relationship

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889

5and 120573 lt 120573lowast

71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

The optimal fixed-fee and royalty rate should maximizeSP1rsquos profit as long as SP2 agrees to license In other wordsthe optimal rate fixed-fee 119865lowast and royalty 119903lowast should solve

max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)

When the market scale is small (120579 le 120579) solving the aboveproblem yields SP1rsquos optimal fixed-fee

119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)

Integrating (23) into (21) we obtain the optimal royaltyrate

119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)

Thus we have the optimal fixed-fee

119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)

In the following we will compare SP1rsquos profits underdifferent licensing strategies When 120573 ge 05 the optimalroyalty rate is 0 that is two-part tariff licensing degrades tofixed-fee licensingThenwe obtain120587119865119877

1= 1205871198651= 21205792(3minus2120573)

2According to the conclusion in Case (1) of Proposition 4we have 120587119865

1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771

with1205871198771 Substituting (24) and (25) into (20) and

combining (6) and (14) we have

120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)

where1198675= (minus5+20120573minus121205732)(1minus120573)+ (1minus2120573)

2

(5minus5120573+1205732)As shown in Figure 1119867

5gt 0Then we have120587119865119877

1minus1205871198771gt 0

Based on the above problem we have the followingproposition

Proposition 5 In a small market (120579 le 120579) if all three licensingstrategies are valid then two-part tariff licensing degrades tofixed-fee licensing when 120573 ge 05 For SP1 the optimal licensingstrategy is fixed-fee licensing When 120573 lt 05 the optimallicensing strategy is two-part tariff licensing

When the market demand is in the medium range (120579 lt120579 le 120579) it is possible for SP2 to develop new technology ifit does not implement licensing Based on (3) we can derivethat

119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5

Figure 1 Changing of1198675on 120573 isin [0 05)

Based on (19) the optimal royalty rate is the same as (24)Thus the optimal fixed-fee is

119865lowast

2=

1205792

1205732

4 (1 minus 120573)2minus

1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)

The samemethods are applied in the following parts as wecompare 120587119865119877

1with 120587119877

1and 120587119865

1under different rates of network

intensity When 120573 ge 05 the royalty rate is 0 under two-part tariff licensing it simplifies to be a fixed-fee licensingThus 120587119865119877

1= 1205871198651 Therefore the profit comparison under two-

part tariff licensing and royalty licensing essentially becomesthe profit comparison under fixed-fee licensing and royaltylicensing According to the case of 120579 lt 120579 le 120579 in Proposition 4and in conjunction with the critical value under networkintensity when 120573 ge max(120573lowast

2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771

Next we explore SP1rsquos optimal licensing strategy when120573 lt 05 Based on (24) and (28) and the profit of SP1 underfixed-fee licensing we have

120587119865119877

1minus 120587119865

1=[(1 minus 2120573) 120579 minus (1 minus 120573) 119903lowast

3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2

minus 4 (1 minus 120573) (5 minus 5120573 + 1205732

)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)

Figure 2 Changing of1198676(119889) regarding 120573 when 119889 = 1

0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7

Figure 3 Changing of the critical value 1198968minus 1198967regarding 120573 when

119889 = 0

Clearly 1198676(119889) is the increasing function of 119889 isin [0 1]

Figure 2 indicates that 1198676(1) lt 0 so we can easily obtain

0 lt 1198967le 1198961

The result shows that 1205871198651198771gt 1205871198771 when 119896

7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896

7 In the medium market scale the

validity of the two-part system depends on the comparisonbetween 120587119865119877

1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896

8 only when 119896 gt

1198968does two-part tariff licensing become valid where 119896

8=

((8(5+119889minus3120573)(1minus119889minus120573)(1minus120573)minus (1minus2120573)2

(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2

(1minus120573))1205792 As Figures 3 and 4 show that 1198968minus1198967lt

0 Therefore the aforementioned conclusions can guaranteethe validity of the two-part tariff licensing strategy

Based on the conclusions under different rates of networkintensity when the market scale is in the medium range (120579 lt120579 le 120579) we obtain the following two propositions


0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1

Figure 4 Changing of the critical value 1198968minus 1198967under different 119889

regarding 120573

Proposition 6 When 120573 ge 05 two-part tariff licensingdegrades to fixed-fee licensing As for SP1 three licensingstrategies will not occur only when the network intensity isrelatively high (eg 120573 ge max120573lowast

2 120573lowast10) If SP2rsquos RampD cost is

relatively high (eg 1198965le 119896 lt 119896

1) then fixed-fee licensing will

be the optimal strategy If SP2rsquos RampD cost is relatively low (eg119896 lt 1198965) then royalty licensing will be the optimal strategy

Proposition 7 When 120573 lt 05 as far as SP1 is concerned itsoptimal licensing strategy relies on the RampD cost of SP2 If SP2rsquosRampD cost is relatively high (eg 119896

5le 119896 lt 119896

1) then two-part

tariff licensing will be the optimal strategy If SP2rsquos RampD cost isrelatively low (eg 119896 lt 119896

7) then royalty licensing will be the

optimal strategy

However when the market scale is relatively large (120579 gt120579) the optimal fixed-fee and royalty rate are the same asthose under the condition of medium market scale Whennetwork intensity is relatively high (eg 120573 ge 05) the optimalroyalty rate is zero In this case two-part tariff licensing isequal to fixed-fee licensing By comparing the profits betweenfixed-fee licensing and royalty licensing we can determinethe optimal licensing strategy The comparison of fixed-feelicensing and royalty licensing is provided in the case of 120579 gt 120579of Proposition 4 Therefore we only have to consider whatwill happen under low network intensity (eg 120573 lt 05)Similar to the situation under a medium market scale two-part tariff licensing is always superior to fixed-fee licensingThen we only have to compare the value of two-part tarifflicensing with that of royalty licensing

As 1205871198651198771minus 1205871198771= (1 minus 120573)[radic1205792(2 + 119889 minus 120573)

2

minus 119896 minus 1205731205792(1 minus

120573)]2

(2 minus 120573)2

gt 0 when network intensity is relatively low(eg 120573 lt 05) two-part tariff licensing is always superior

to fixed-fee licensing and royalty licensing Subsequently wehave Proposition 7 at a higher market scale

Proposition 8 When themarket scale is high (120579 gt 120579) If all thethree licensing strategies are valid then (1) when 120573 ge 05 two-part tariff licensing strategy is the same as fixed-fee licensingstrategy In this case the optimal licensing strategy is given byProposition 4 (2)When 120573 lt 05 as for SP1 the two-part tarifflicensing strategy is always superior to fixed-fee licensing androyalty licensing strategy

Propositions 5 to 8 suggest that under two-part tarifflicensing royalty licensing and fixed-fee licensing are pos-sible as long as the network intensity is not extremely high(eg 120573 lt 05) and SP2rsquos RampD cost is high enough (eg1198967le 119896 lt 119896

1) Then the optimal strategy of SP1 is invariably

two-part tariff licensing which contains the merits of bothfixed-fee licensing and royalty licensing

However because of the particularity of the networkproduct market and the existence of a market scale productsubstitution level RampD cost of SP2 and especially networkeffects two-part tariff licensing does not always turn outto be the optimal strategy This situation is especially truewhen there are extreme parameters in the RampD cost of SP2and network intensity Consider the following examples (1)When the market scale is relatively small and the networkintensity is relatively high the network effect is better than thesupply side effect Fixed-fee licensing develops its advantageunder stronger network effects and a smaller market scaleThus fixed-fee licensing will be better than royalty licensingand two-part tariff licensing in themiddle position (2)Whenthe market scale is medium and the network intensity isextremely high because of the increasingly larger marketscale fixed-fee licensing becomes less influential on themarket Therefore we have to consider SP2rsquos RampD costwhen the network intensity reaches a certain critical levelAs stated in Proposition 4 high RampD cost is beneficial tothe implementation of fixed-fee licensing and low RampD costforces SP1 to implement royalty licensing to keep its leadingposition in terms of output (3) When the market scale islarge and the network intensity is relatively high the influenceof fixed-fee licensing and royalty licensing on the marketis remarkably weakened and royalty licensing has an evenweaker impactThen SP1 has to consider SP2rsquos RampD cost andalso anticipate its product substitution level

7 Conclusions

In this paper we have examined the technology licensingproblems of network product in a duopolistic SP compe-tition market We also compared the SPsrsquo profits in thefour licensing strategies no licensing fixed-fee licensingroyalty licensing and two-part tariff licensing strategy Wehave shown that the licensing strategy choices depend onthe combination of the market scale the network productsubstitution rate the network effect and SPrsquos RampD cost

When the network products can be completely substi-tuted two-part tariff licensing is always the optimal strategyfor the incumbent SP in the normal and network product


markets This finding is in accordance with the conclusionmade by Lin and Kulatilaka [20] However when the networkproducts cannot be completely substituted two-part tarifflicensing is not always optimal This condition shows that noideal strategy constantly works and that the conclusion thattwo-part tariff licensing is the optimal choice in the normalproduct market or in the network product market in whichall products can be substituted completely is only a specialcase Thus if we want to know which strategy is optimal wehave to consider the combinations of all the aforementionedparameters the details of which are in the conclusions ofTheorems 1 to 3 and Propositions 4 to 8

We now discuss some limitations in this paper andpossible avenues for future research First our analysisfocused on the assumption of information symmetry (ie allparameters are common knowledge) In reality SP2rsquos RampDcost is often private information If SP1rsquos decision is basedupon the asymmetric information then our results may nothold Second we assume that SP1 has a perfect capacity forbargaining he can give SP2 an ldquoaccept-it-or-refuse-itrdquo offerOur results might make a big difference if we assume SP1could not have the perfect capacity for bargaining Third weconsider there are only two competing SPs in the industryneglecting the superior supplier of the supply chain Whenthere exist upstream enterprises SP1rsquos licensing strategy andthe outputs of the two SPs might change a lot All of theselimitations leave us a fertile area for future research

Appendix

Relative Parameters in a Duopoly Competition

Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896

2=2 (5 + 119889 minus 3120573) (1 minus 119889 minus 120573) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732

minus (5 minus 2119889) 120573 minus 1198892

minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198968= (8 (5 + 119889 minus 3120573) (1 minus 119889 minus 120573) (1 minus 120573)

minus (1 minus 2120573)2

(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2

(1 minus 120573))minus1

) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast

3=(1 minus 119889) minus radic171198892 + 62119889 minus 7

4

120573lowast

4=(1 minus 119889) + radic171198892 + 62119889 minus 7

4

120573lowast

5=5 minus 2119889 minus radic5 minus 4119889 + 81198892

2

120573lowast

6=5 minus 2119889 + radic5 minus 4119889 + 81198892

2

120573lowast

7=minus21198893 minus 161198892 + radicΔ

2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)

Proof of Theorem 1 If fixed-fee licensing is to be imple-mented then there has to be Δ120587

1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712

ge 0 As the SP2rsquos profits remain unchangedbefore and after licensing (because SP1 is completely capableof bargaining) it is automatically qualified to accept licens-ing Accordingly the implementation of fixed-fee licensingdepends on the condition Δ120587

1= 1205871198651minus 1205871198731198711

ge 0 From (3)and (7) we know that

Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)

where 1198671(120573) = minus21205732 + 4120573 minus 1 and 119896

2= 2(5 + 119889 minus 3120573)(1 minus 119889 minus

120573)1205792(3minus2120573)2(2+119889minus120573)2Then if 120579 le 120579 makingΔ1205871ge 0 only

requires1198671(120573) = minus21205732 + 4120573 minus 1 ge 0 Substituting 120573lowast

1isin [0 1)

into1198671(120573) = minus21205732 + 4120573 minus 1 = 0 we have 120573lowast

1= (2 minus radic2)2 asymp

0293 If 120573 lt 120573lowast1 then119867

1(120573) lt 0 furthermore we haveΔ120587

1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0

What if 120579 gt 120579 From (A2) we know that 120573 gt 120573lowast2= 1 minus 119889

then we have Δ1205871ge 0 However if 120573 lt 120573lowast

2= 1 minus 119889 to make

Δ1205871ge 0 we only have to consider the values of 119896 and 119896

2

Moreover as 119896 lt 1198961= 1205792(2 + 119889 minus 120573)2 we have to compare

the values of 1198961and 1198962first As

1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)

where 1198672(120573) = minus21205732 + (1 minus 119889)120573 + 21198892 + 8119889 minus 1 if 119867

2(120573) =

minus21205732+(1minus119889)120573+21198892+8119889minus1 = 0 is solvable thenΔ = 171198892+62119889 minus 7 ge 0 As 119889 lt 119889

1= (minus32 + 6radic30)17 asymp 0051 Δ lt 0

then 1198672(120573) lt 0 moreover we have 119896

1lt 1198962 Then we have

Δ1205871lt 0


If 119889 gt 1198891= (minus32+6radic30)17 asymp 0051 then Δ ge 0 solving

1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892

+ 8119889 minus 1 = 0 and we have120573lowast3= ((1 minus 119889) minus radic171198892 + 62119889 minus 7)4 and 120573lowast

4= ((1 minus 119889) +

radic171198892 + 62119889 minus 7)4 To make 120573lowast3isin [0 1 minus 119889) we only need

21198892 +8119889minus1 le 0 Solving 21198892 +8119889minus1 = 0when 119889 isin [0 1] wehave 119889

2= (minus4+3radic2)2 asymp 0121 Accordingly when 119889

1le 119889 le

1198892 then 0 le 120573lowast

3lt 1 minus 119889 Otherwise 120573lowast

3lt 0 To make 120573lowast

4isin

[0 1minus119889) we need 1198892 +10119889minus2 le 0 We solve 1198892 +10119889minus2 = 0when 119889 isin [0 1] and thus we have 119889

3= minus5 + 3radic3 asymp 0196 If

1198891le 119889 le 119889



4ge 1 minus 119889

Then we have the following (1) If 1198891le 119889 le 119889

2 then we

have 0 le 120573lowast3lt 120573lowast4lt 1 minus 119889 In this case if 120573 le 120573lowast

3or 120573 ge 120573lowast

4

we can derive that1198672(120573) le 0 and 119896

1le 1198962 if 120573lowast3lt 120573 lt 120573lowast

4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have

120573lowast3lt 0 and 0 lt 120573lowast

4lt 1minus119889 In this case for any120573 isin [0 1minus119889) if

120573 le 120573lowast4 we can derive that 119867

2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889

3 we have 120573lowast

3lt 0

and 120573lowast4ge 1 minus 119889 In this case for any 120573 isin [0 1 minus 119889) if 120573 lt 120573lowast

4

we can derive that1198672(120573) gt 0 and 119896

1gt 1198962

To sum up the preceding proofs of 120579 gt 120579 we have thefollowing

(1) If the product substitute rate is very small (119889 lt 1198891=

0051) then regardless of RampD cost of SP2 SP1 willchoose not to implement fixed-fee licensing

(2) If the product substitute rate is very small (1198891le 119889 le

1198892= 0121) then fixed-fee licensing will occur only if

the network intensity is medium (120573lowast3lt 120573 lt 120573lowast

4) and


1)

(3) If the product substitute rate is medium (1198892lt 119889 lt

1198893= 0196) then fixed-fee licensing will occur only

when the network intensity is very small (120573 lt 120573lowast4) and


1)

(4) If the product substitute rate is large enough (119889 ge 1198893)

then as long as RampD cost of SP2 is very high (1198962le 119896 lt

1198961) SP1 will choose to implement fixed-fee licensing

Proof of Theorem 2 When 120579 le 120579 we know that 119903lowast1le 119903lowast2 The

optimal royalty rate is 119903lowast = min119903lowast1 119903lowast2 = 119903lowast

1 What about

120579 gt 120579 As 119903lowast1le 119903lowast2 we have the following

120579 le 120579 =2 (2 + 119889 minus 120573) (5 minus 5120573 + 1205732)radic119896


(A4)

if 119903lowast1gt 119903lowast2 then 120579 gt 120579 This circumstance is also caused by

120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)

where 1198673(120573) = 2(5 minus 5120573 + 1205732) minus

radic[10 + (119889 minus 8)120573 + 1205732][10 minus (12 + 119889)120573 + 31205732]

Through proof by contradiction we have 1198673(120573) gt 0 rArr

1205732[120573 minus (119889 + 2)]2 gt 0 Moreover 1205732[120573 minus (119889 + 2)]2 gt 0 is validso we have119867

3(120573) gt 0 rArr 120579 minus 120579 gt 0 Thus 120579 lt 120579 le 120579 119903lowast

1lt 119903lowast2

In other words 119903lowast = min119903lowast1 119903lowast2 = 119903lowast1 if 120579 gt 120579 119903lowast

1gt 119903lowast2Thus

119903lowast = min119903lowast1 119903lowast2 = 119903lowast

2 To sum up the conclusions under

120579 le 120579 and 120579 gt 120579 we haveTheorem 2

Proof of Theorem 3 When 120579 le 120579 from (3) and (14) we have

120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)

Therefore SP1 may choose royalty licensingWhen 120579 lt 120579 le 120579 from (3) and (14) we have

120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)

Under this circumstance SP1 will implement royalty licens-ing instead of no licensing

However if 120579 gt 120579

120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)

times ((2 minus 120573)2

(2 + 119889 minus 120573)2

)minus1

(A8)

where1198621(120573) = minus1205732+(5minus2119889)120573+1198892+4119889minus5 From this equation

we know that if1198621(120573) ge 0 rArr 120587119877

1gt 1205871198731198711

then solving1198621(120573) =

0when120573 isin [0 1) gives us120573lowast5= (5minus2119889minusradic5 minus 4119889 + 81198892)2 and

120573lowast6= (5 minus 2119889 + radic5 minus 4119889 + 81198892)2 Clearly 120573lowast

6gt 1 Moreover

0 le 120573lowast5lt 1 rArr 1198892 + 2119889 minus 1 gt 0 whereas 119889 gt 119889

4= radic2 minus 1 asymp

0414 We have 1198892 + 2119889minus 1 gt 0 119889 lt 1198894 Thus 1198892 + 2119889minus 1 le 0

Then we have that when 119889 gt 1198894= radic2 minus 1 asymp 0414120573lowast

6gt

1and 0 le 120573lowast5lt 1 If 120573 ge 120573lowast

5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5

then 1198621(120573) lt 0 However if 119889 le 119889

4 120573lowast6gt 1 and 120573lowast

5ge 1

For anywhere between 120573 isin [0 1) 1198621(120573) lt 0 When 119889 gt 119889

4

and 120573 ge 120573lowast5 we have 120587119877

1gt 1205871198731198711

Therefore we only have tocompare the values of 120587119877

1and 120587119873119871

1 when 119889 gt 119889


0414 and 0 le 120573 lt 120573lowast5or 119889 le 119889

4 indicating that 119862

1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus

120573)2radic1205792(2 + 119889 minus 120573)2 minus 119896 ge 1198622(120573) where 119862

2(120573) = [1205732 minus (5 minus

2119889)120573 minus 1198892 minus 4119889 + 5]1205792 minus (1205732 minus 5120573 + 5)(2 + 119889 minus 120573)2119896


Let

1198964=1205732 minus (5 minus 2119889) 120573 minus 119889

2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862

2(120573) gt 0We then continue to compare

the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862

5= 0 yields two results one is

smaller than 0 and the other is 1198963= (minus119862

4+radicΔ1)21198623 If 1198963gt

0 we will have 11986231198625lt 0 As 120573 ge 120573lowast

2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862

5lt 0 we have 120573 ge 120573lowast

2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711

Howeverif 120573 lt 120573lowast

2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1

(A12)The only issue left is determining the conditions in

which 1198963can be located between the lower limit of 119896

4and

the upper limit of 1198961

Then we have 1198964minus 1198963= (1205732

(2 + 119889 minus 120573)4

1205792

minusradicΔ1)21198623le

0 rArr 119889(2120573 minus 119889 minus 4) le 0 and 119889(2120573 minus 119889 minus 4) le 0 isvalid under 119889 isin [0 1] Accordingly 119896

4minus 1198963le 0 In other

words 1198964is the lower limit of 119896

3 and

1198963minus 1198961=minus1198624+ radicΔ

1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0

997904rArr radicΔ1le

21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)

We can easily derive that

120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)

Moreover 120573lowast7lt 1 rArr 31198893 minus 21198892 + 20119889 minus 25 lt 0 and 31198893 minus

21198892 + 20119889 minus 25 lt 0 is valid under 119889 isin [0 1] therefore 120573lowast7lt 1

is validWe have120573 le 120573lowast7rArr 1198963le 1198961120573 gt 120573lowast

7rArr 1198963gt 1198961Then

we compare the value of 120573lowast7and the critical value of 120573lowast

2 After

complex deduction we have 120573lowast7lt 120573lowast2rArr 1198627(119889) lt 0 where

1198627(119889) = minus21198896 + 901198895 minus 8591198894 + 30301198893 minus 31251198892 + 2250119889 minus

625 Through MATLAB we obtain 0 le 119889 le 1198895asymp 0314

After solving the equation we have 1198627(119889) le 0 rArr 120573lowast

7lt 120573lowast2

Otherwise 1198627(119889) gt 0 rArr 120573lowast

7gt 120573lowast2 We compare the values

of 120573lowast5and 120573lowast

2under 119889 gt 119889

4= radic2 minus 1 asymp 0414 and 0 le 120573 lt

120573lowast5 Consider 120573lowast

5minus 120573lowast2ge 0 rArr 21198892 minus 119889 minus 1 le 0 Moreover

21198892minus119889minus1 le 0 is invariably valid under 119889 isin [0 1] As a resultwe have 120573lowast

5ge 120573lowast2

To sum up the conclusions under 120579 gt 120579 we have thefollowing

(1) If the product substitute rate is very high (119889 gt 1198894=

radic2minus1 asymp 0414) and the network is large enough (120573 gt120573lowast5) SP1 is motivated to implement royalty licensing

(2) When the product substitute rate is very high (119889 gt

1198894) but the network intensity is not large enough

(120573 lt 120573lowast5) the following three situations may occur

(1) If the network intensity is medium (1 minus 119889 = 120573lowast2le

120573 lt 120573lowast5) regardless of the RampD cost for SP2 SP1 will

not choose to implement royalty licensing (2) If thenetwork intensity is very small (120573 lt 120573lowast

2) and if RampD

cost of SP2 is very low (1198964lt 119896 lt 119896

3) then royalty

licensing will occur If RampD cost of SP2 is very high(1198963le 119896 lt 119896

1) then SP1 will choose not to implement

royalty licensing (3)As long as RampDcost of SP2 is lowenough (119896 le 119896

4) SP1 has themotivation to implement

royalty licensing

(3) When the product substitute rate is not very high (119889 le1198894) the licensing conditions are as follows (a) If the

network intensity is very large (120573 ge 120573lowast2) regardless of

theRampDcost of SP2 SP1will not choose to implementroyalty licensing (b) If the product substitute rate ismedium (0314 asymp 119889

5lt 119889 le 119889

4) network intensity is

not very large (120573 lt 120573lowast2) and RampD cost of SP2 is very

high (1198963le 119896 lt 119896

1) SP1 will choose not to implement

royalty licensing if RampD cost of SP2 is very low (1198964lt

119896 lt 1198963) royalty licensing will occur (c) If the product

substitute rate is very low (0 le 119889 le 1198895) and network

intensity is very small (120573lowast7lt 120573 lt 120573lowast

2) royalty licensing

will occur (d) If the product substitute rate is very low(0 le 119889 le 119889

5) network intensity is very small (120573 lt

120573lowast7) and RampD cost of SP2 is very high (119896

3le 119896 lt 119896

1)

SP1 will choose not to implement royalty licensing ifRampD cost of SP2 is very small (119896

4lt 119896 lt 119896

3) royalty

licensing will occur (e) As long as RampD cost of SP2 is


low enough (119896 le 1198964) SP1 is motivated to implement

royalty licensing

Proof of Proposition 4 When 120579 le 120579 1205871198651minus 1205871198771= ((minus121205732 +

20120573 minus 5)4(3 minus 2120573)2(5 minus 5120573 + 1205732))1205792 for minus121205732 + 20120573 minus 5 = 0if one root is 120573lowast

8= (5 minus radic10)6 asymp 0306 under 120573 isin [0 1) (the

other root is (5 + radic10)6 gt 1) and if 120573 lt 120573lowast8 then 120587119877

1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865

1gt 1205871198771 Moreover 120573lowast

1le 120573 and 120573lowast

1le 120573lowast8 Case

(1) in Proposition 4 is obtainedWhen

120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)

Then we know that when 119896 gt 1198965 we have 120587119865

1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure

1198965lt 1198961to keep the conclusion valid As indicated in Figure 5

we prove the existence of 120573lowast9asymp 0319 which makes 119896

5= 1198961

throughMATLABWhen 120573 lt 120573lowast9 we have 119896

1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965

After considering all four situations under fixed-feelicensing we obtain the conclusions in the case of 120579 lt 120579 le 120579as follows

(a) If the network intensity is strong enough (120573 ge

max120573lowast2 120573lowast9) and if SP2rsquos RampD cost is extremely high (119896

5le

119896 lt 1198961) then fixed-fee licensing will be better than royalty

licensing However if SP2rsquos RampD cost is extremely low(119896 lt 119896

5) then royalty licensing will be better than fixed-fee

licensing (b) If the product substitution rate is extremely high(119889 ge 119889

3) but the network intensity is not high enough (120573 lt

max120573lowast2 120573lowast

9) or if the product substitution rate is medium

(1198892lt 119889 lt 119889

3) and the network intensity is extremely low (120573 le

max120573lowast4 120573lowast9) for any 119896 lt 119896

1 then royalty licensing will be

better than fixed-fee licensing (c) If the product substitutionrate is extremely small (119889

1le 119889 le 119889

2) the network intensity

is medium (max120573lowast3 120573lowast9 lt 120573 lt min120573lowast

4 120573lowast9) and SP2rsquos RampD

cost is extremely high then royalty licensing will be betterthan fixed-fee licensing If SP2rsquos RampD cost is extremely small(1198962le 119896 lt 119896

5) then royalty licensing will be better than

royalty licensing If SP2rsquos RampD cost is extremely high thenroyalty licensingwill be better thanfixed-fee licensing If SP2rsquosRampD cost is extremely small (119896

2le 119896 lt 119896

5) then royalty

licensing will be better than royalty licensing

0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1

Figure 5 Varying patterns of 1198961minus 1198965under different values of 119889

regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896 + (1 minus 120573) 119896))


(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)

where1198674= [minus21205732+(4minus6119889)120573+21198892+8119889minus1](2minus120573)2minus[(1minus2119889)120573+

1198892 + 4119889 minus 1](3 minus 2120573)2 As indicated in Figure 6 we invariablyhave119867

4gt 0 anywhere between intervals 120573 isin [0 1)

Let 1198966= (1198674(2 + 119889 minus 120573)2(3 minus 2120573)2(1 minus 120573))1205792 Thus we

have 1198966gt 0 As a result if 119896 gt 119896

6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896

6minus 119896) minus

120579120573radic1205792(2 + 119889 minus 120573)2 minus 119896 ge 0 As 1198675is the quadratic curve of

119896 with its open upward and its discriminant asΔ3le 0 where

Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794

as indicated in Figure 7 then 1198675ge 0 is invariably validThus

if 119896 le 1198966 we have 120587119865

1ge 1205871198771 Considering the aforementioned

15 possible combinations we have to compare the value of 1198966

and these critical values and make comparisons within thecritical values themselves

In the following we first compare 1198962with 119896

6

1198962minus 1198966= 21205734 minus (14 + 2119889)1205733 + (21198892 + 4119889 + 31)1205732 minus (41198892 +

6119889+15)120573minus1198892 +4119889+5 as indicated in Figure 8 As anywherebetween interval 120573 isin [0 1) 119896

2gt 1198966is invariably valid


0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1

Figure 6 Varying patterns of 1198674under different values of 119889

regarding 120573

We then compare the critical values of 1198962and 1198963 Consider

that 1198962minus 1198963le 0 rArr 119862

311989622+ 11986241198962+ 1198625le 0 as indicated in

Figure 9 because 119862311989622+ 11986241198962+ 1198625is invariably smaller than

0 under different values of 119889 regarding 120573 isin [0 1) Therefore1198962le 1198963

We then compare the critical values of 1198964and 119896

6under

conditions 119889 gt 1198894and 120573lowast

2le 120573 lt 120573lowast

5or 119889 le 119889

4and 120573 ge 120573lowast

2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)

where1198676= (1 minus 120573)(3 minus 2120573)2[1205732 minus (5 minus 2119889)120573 minus 1198892 minus 4119889 + 5] minus

1198674(5 minus 5120573 + 1205732) as indicated in Figure 10 then 119896

4le 1198966is

invariably valid under 119889 gt 1198894and 120573lowast

2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast

2 Thus we have 119896

3ge 1198962gt 1198966gt 1198964between intervals of

119889 isin [0 0414]The only task left is to compare the critical values of 120573lowast

4

and 120573lowast7 Moreover

120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892

minus 20119889 + 25)radic171198892 + 62119889 minus 7 minus radicΔ2

Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)

As indicated in Figure 11 1198677is a decreasing function

between 119889 isin [0 0414] regarding 119889 and 1198677lt 0 Thus we

have 120573lowast4lt 120573lowast7

0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

After considering 15 possible combinations we have eightsituations in the case of 120579 gt 120579

(a) If the product substitution rate is extremely high (119889 gt1198894) the network intensity is extremely high (120573 ge 120573lowast

5) and

SP2rsquos RampD cost is extremely high (1198966lt 119896 lt 119896

1) then royalty

licensing will be better than fixed-fee licensing However ifSP2rsquos RampD cost is extremely small (119896 lt 119896

6) then fixed-

fee licensing will be better than royalty licensing (b) If theproduct substitution rate is extremely high (119889 gt 119889

4) but the

network intensity is medium (120573lowast2le 120573 lt 120573lowast

5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1

Figure 9 Varying patterns of119862311989622+11986241198962+1198625under different values

of 119889 regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573

always be true that fixed-fee licensing is better than royaltylicensing for SP2 when it has small RampD cost (119896 le 119896

4) (c)

If the product substitution rate is not high (119889 le 1198894) and the

rate of network intensity is extremely high (120573 ge 120573lowast2) it will

always be true that fixed-fee licensing is better than royaltylicensing for any RampD cost (119896 le 119896

4) for SP2 (d) If the product

substitution rate is extremely small (1198893le 119889 le 119889

5) and the

network intensity is medium (120573lowast7lt 120573 lt 120573lowast

2) then royalty

licensing will be better than fixed-fee licensing for any RampD

004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7

Figure 11 Varying patterns of1198677regarding 119889

cost (1198962le 119896 le 119896

1) for SP2 (e) If the product substitution

rate is extremely high (119889 gt 1198895) the network intensity is

relevantly small (120573 lt 120573lowast2) and SP2rsquos RampD cost is medium

(1198962le 119896 lt 119896

3) then royalty licensing will be better than fixed-

fee licensing (f) If the product substitution rate is medium(1198893le 119889 le 119889

5) the network intensity is relevantly low (120573 lt

120573lowast7) and SP2rsquos RampDcost ismedium (119896

2le 119896 lt 119896

3) then royalty

licensing will be better than fixed-fee licensing (g) If theproduct substitution rate is extremely low (119889

2lt 119889 lt 119889

3) the

network intensity is extremely low (120573 lt 120573lowast4) and SP2rsquos RampD

cost is medium (1198962le 119896 lt 119896

3) then royalty licensing will be

better than fixed-fee licensing (h) If the product substitutionrate is extremely low (119889

1le 119889 le 119889

2) the network intensity is

extremely low (120573lowast3lt 120573 lt 120573lowast

4) and SP2rsquos RampD cost is medium

(1198962le 119896 lt 119896


fee licensing

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors would like to thank the editor and anony-mous referees for their insightful comments and suggestionsthat have significantly improved this paper Xianpei Hongrsquosresearch is partially supported by Humanity and SocialScience Youth Foundation ofMinistry of Education of ChinaGrant no 11YJC630058 the Fundamental Research Fundsfor the Central Universities Program no 2662014BQ048and Research Project of Department of Education of HubeiProvince Grant no Q20141803 Haiqing Hursquos research ispartially supported by Humanities and Social SciencesFoundation of Ministry of Education in China Grant no14YJC630049


References

[1] P Germerad ldquoThe changing role of RampDrdquo Reach-TechnologyManagement vol 44 pp 15ndash20 2001

[2] P C Grindley and D J Teece ldquoManaging intellectual capitallicensing and cross-licensing in semiconductors and electron-icsrdquoCaliforniaManagement Review vol 39 no 2 pp 8ndash41 1997

[3] M L Katz and C Shapiro ldquoSystems competition and networkeffectsrdquo Journal of Economic Perspectives vol 8 no 2 pp 93ndash1151994

[4] Y Chen and J Xie ldquoCross-market network effect with asymmet-ric customer loyalty implications for competitive advantagerdquoMarketing Science vol 26 no 1 pp 52ndash66 2007

[5] M I Kamien S S Oren and Y Tauman ldquoOptimal licensing ofcost-reducing innovationrdquo Journal of Mathematical Economicsvol 21 no 5 pp 483ndash508 1992

[6] I Macho-Stadler X Martinez-Giralt and J D Perez-CastrilloldquoThe role of information in licensing contract designrdquo ResearchPolicy vol 25 no 1 pp 43ndash57 1996

[7] A Mukherjee and E Pennings ldquoTariffs licensing and marketstructurerdquo European Economic Review vol 50 no 7 pp 1699ndash1707 2006

[8] M H Nabin X N Guyen and P M Sgro ldquoOptimal licensingpolicy under vertical product differentiationrdquo Working Paper2011

[9] A Shaked and J Sutton ldquoRelaxing price competition throughproduct differentiationrdquoTheReview of Economic Studies vol 49no 1 pp 3ndash13 1982

[10] A Shaked and J Sutton ldquoNatural oligopoliesrdquo Econometricavol 51 no 5 pp 1469ndash1483 1983

[11] A Shaked and J Sutton ldquoInvoluntary unemployment as aperfect equilibrium in a bargaining modelrdquo Econometrica vol52 no 6 pp 1351ndash1364 1984

[12] G Stamatopoulos and Y Tauman ldquoLicensing of a quality-improving innovationrdquo Mathematical Social Sciences vol 56no 3 pp 410ndash438 2008

[13] C Li and J Wang ldquoLicensing a vertical product innovationrdquoEconomic Record vol 86 no 275 pp 517ndash527 2010

[14] B Shen T-MChoi YWang andCK Y Lo ldquoThe coordinationof fashion supply chains with a risk-averse supplier under themarkdown money policyrdquo IEEE Transactions on Systems Manand Cybernetics Part A Systems and Humans vol 43 no 2 pp266ndash276 2013

[15] C Y Li and X Y Geng ldquoLicensing to a durable-goodmonopolyrdquo Economic Modelling vol 25 no 5 pp 876ndash8842008

[16] M I Kamien and Y Tauman ldquoFees versus royalties and theprivate value of a patentrdquo The Quarterly Journal of Economicsvol 101 no 3 pp 471ndash491 1986

[17] SMuto ldquoOn licensing policies in Bertrand competitionrdquoGamesand Economic Behavior vol 5 no 2 pp 257ndash267 1993

[18] P Crama B de Reyck and Z Degraeve ldquoMilestone paymentsor royalties Contract design for RampD licensingrdquo OperationsResearch vol 56 no 6 pp 1539ndash1552 2008

[19] A Mukherjee ldquoLicensing a new product fee vs royalty licens-ing with unionized labor marketrdquo Labour Economics vol 17 no4 pp 735ndash742 2010

[20] L Lin and N Kulatilaka ldquoNetwork effects and technologylicensing with fixed fee royalty and hybrid contractsrdquo Journalof Management Information Systems vol 23 no 2 pp 91ndash1182006

[21] T M Choi P S Chow and G Kannan ldquoService optimizationand controlrdquo Mathematical Problems in Engineering vol 2014Article ID 951376 3 pages 2014

[22] C H Chiu T M Choi Y Li and L Xu ldquoService competitionand service war a game-theoretic analysisrdquo Service Science vol6 no 1 pp 63ndash76 2014

[23] W K Chan and B J Gao ldquoUnfair consequence of faircompetition in service systemsmdashan agent-based and queuingapproachrdquo Service Science vol 5 no 3 pp 249ndash262 2013

[24] Y Li L Xu and D Li ldquoExamining relationships between thereturn policy product quality and pricing strategy in onlinedirect sellingrdquo International Journal of Production Economicsvol 144 no 2 pp 451ndash460 2013

[25] L Xu Z Wang J Shan and Y Wang ldquoOn advertising gamesand spillover in service systemsrdquo Mathematical Problems inEngineering vol 2013 Article ID 260823 8 pages 2013

[26] G F Gilder Telecoms How Infinite Bandwidth Will Revolution-ize Our World Free Press New York NY USA 2000

[27] M L Katz andC Shapiro ldquoOn the licensing of innovationsrdquoTheRAND Journal of Economics vol 16 no 4 pp 504ndash520 1985

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


CDMA and LTE but it only established an innovation-basedstandard in other small markets such as Hong Kong andKorea before it entered the market of Chinese MainlandQualcomm obtained tremendous development opportunityafter it licensed its CDMA technology to China UnicomQualcomm gained a profit of 249 hundred million dollarsthrough chip distribution and technology licensing in 2013among which it gained a profit of 120 hundred milliondollars (84 of its whole profits) from China HUAWEI andZTE Corporation that have adopted CDMA technology payQualcomm a per unit charge (5) to access the CDMAnetwork That is why so many innovating firms are keeningon establishing an industry standard

Our research question arises from practice Howeverin theoretical research few address the firmrsquos technologylicensing issues for network products For a network productusers derive added value from the presence of others inthe network the additional network value depends on thenumber of other users of this product [3 4] Differentfrom the normal products the market for network productscontinues to expand with the popularity of the InternetThis phenomenon is commonly seen in reality for exampleinstant communication devices such as MSN and QQ are ofno value when users are too few However considering theincrement in the number of users of these devices their net-work values and consumer evaluations continuously increaseIn this paper we define the products with network effects asnetwork products and the products without network effectsas normal products

Based on the above analysis the purpose of this paperis to explore the following two issues by taking account ofnetwork effects of two competing firms under the situationwhere the innovator serves as an insider patentee as firmsproduce and engage in Cournot competition Firstly what isthe innovating firmrsquos optimal licensing strategy in terms offixed-fee royalty or two-part tariff licensing Secondly howdo the network intensity and product heterogeneity affect thetechnology licensing strategy for the innovating firm

The main findings of this paper are as follows (1) Forthe innovator provider no matter whether the market scalesare large or small and as long as the network intensity 120573is large enough (eg 120573 ge 05) two-part tariff licensingstrategy is the same as fixed-fee licensing strategy Howeveras for the innovator provider the optimal licensing strategyis not always two-part tariff licensing and is different fromthe case of normal product In a network product market theoptimal licensing strategy of the innovator provider dependson the market conditions to be specific when the marketscale is very small the optimal licensing strategy is fixed-fee licensing When the market scale is medium royaltylicensing is the optimal strategy when licensee firmrsquos RampDcost is very high When the market scale is very largethe optimal licensing strategy depends mainly on networkintensity product substitute levels and licensee firm RampDcost (2)When network intensity is relatively small (eg 120573 lt05) and the RampD cost is very high two-part tariff licensingis the optimal licensing strategy of the innovator provider

The reminder of this paper is organized as follows InSection 2 we briefly review the related research Section 3

describes the game process and market demand In Section 4we analyze three licensing strategies (ie no licensing fixed-fee licensing and royalty licensing) and report the resultsSection 5 extends to investigate two-part tariff licensingstrategy and Section 6 concludes this paper with a discussionof the results To simplify our exposition all proofs areprovided in the Appendix

2 Literature Review

An extensive theoretical literature has been developed toexplore the optimal licensing strategy These studies can bedivided into three categories as follows (1) optimal licensingstrategy considering the cost reduction innovation (eg [5ndash8] etc) (2) vertical product innovation to improve theproduct quality and the optimal licensing strategy (eg [9ndash14] etc) and (3) the cost reduction innovation and productinnovation For example Li andGeng [15] studied technologylicensing to the craft and product innovation of durablesThey pointed out that the optimal licensing contract isdetermined by innovation categories and innovation scaleroyalty licensing will be prioritized when it is cheaper thanthe reduction craft innovation or quality-improving productinnovation two-part tariff licensing will be prioritized whenthe scale involved is greater than that of craft innovation orvertical product innovation and fixed-fee licensing will beprioritized when the products are average

In this paper we mainly study the optimal licensingstrategies considering the cost reduction process innovationLicensing strategy has received large amount of attentionfor instance Kamien and Tauman [16] studied the optimaltechnology licensing strategy to an external innovator usingthe noncooperative game theory They pointed out thatusing fixed-fee licensing licensee firms gain lesser benefitsthan those with no licensing because of the fixed-fee andissuing number of licenses whereas royalty licensing doesnot affect licensee firmrsquos profits Thus they addressed thefixed-fee licensing being significantly better than royaltylicensing Muto [17] first introduced the product substituteparameter to technology licensing considering cost reductioncraft innovation and found that royalty licensing is signif-icantly better than fixed-fee licensing when the innovationscale is comparatively small Crama et al [18] studied theoptimal contract of RampD licensing Their results reveal thatthe optimal licensing contract should be three-part tarifflicensing namely payment in advance milestone paymentand royalty because of disparities between the estimated valuelicensor and licensee and the limited control of the licensorover the RampD efforts of the licensee Mukherjee [19] figuredout that as long as the Union has complete control overthe bargain licensee firms would pay lesser salaries underthe royalty licensing situation than under auction or fixed-fee licensing regardless of whether the Union structure is acentralized control mode or a decentralized control modeRoyalty licensing is better thanfixed-fee licensing and auctionlicensing for external innovators because the positive effectsof low salary outweigh the negative effects of firm margincosts posed by the royalty licensing rate










119894= 120579+




1+ 1199021198902) minus 1199021minus 1199022









120587119873

119894= [120579 + V (119902119890

119894) minus 119902119894minus 119902119895] 119902119894 (1)



119902lowast

119894=

120579

2 + 119889 minus 120573 120587

lowast

119894= [

120579

2 + 119889 minus 120573]

2

(2)


2= 120587lowast2minus 119896




1= 1205871198721= [120579(2 minus 120573)]2 where the


(120587119873

1 120587119873

2) =

([120579

2 minus 120573]

2

0) 120579 le 120579

([120579

2 + 119889 minus 120573]

2

[120579

2 + 119889 minus 120573]

2

minus 119896) 120579 gt 120579

(3)


119894= [120579 + V(119902119890



119894= 120579(3minus2120573)



(120587119865

1 120587119865

2) = ([

120579

2 minus 3120573]

2

+ 119865 [120579

2 minus 3120573]

2

minus 119865) (4)



Max119865

120587119865

1=

1205792

(3 minus 2120573)2+ 119865

st 120587119865

2=

1205792

(3 minus 2120573)2minus 119865 ge 120587

119873

2

(5)


119865lowast

=1205792

(3 minus 2120573)2minus 120587119873

2

=

1205792

(3 minus 2120573)2 120579 le 120579

(5 + 119889 minus 3120573) (119889 minus 1 + 120573) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(6)


120587119865

1

=

21205792

(3 minus 2120573)2 120579 le 120579

[(2 + 119889 minus 120573)2

+ (5 + 119889 minus 3120573) (119889 minus 1 + 120573)] 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(7)




1= 0293) SP1 makes


lowast

2=






1le 119889 le 119889

2= 0121)


3lt 120573 lt 120573lowast


(1198962le 119896 lt 119896


(1198892lt 119889 lt 119889




1) (4) When the



1) SP1 will choose



lowast

1)



lowast

2) on the one





120587119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022

120587119877

2= [120579 + V (119902119890

1+ 119902119890


(8)



119902lowast

1=120579 + (1 minus 120573) 119903

3 minus 2120573 119902

lowast

2=120579 minus (2 minus 120573) 119903

3 minus 2120573 (9)


120587119877


(3 minus 2120573)2

120587119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(10)



2120573)2







1199032=

120579

2 minus 120573 120579 le 120579

120579


2 minus 120573radic

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(11)


119903lowast

1=

(5 minus 4120573) 120579

2 (1205732 minus 5120573 + 5) (12)






1 1199032)



1 when 120579 gt 120579 the




(13)


120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)













lowast






1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


















119894= 120579+




1+ 1199021198902) minus 1199021minus 1199022









120587119873

119894= [120579 + V (119902119890

119894) minus 119902119894minus 119902119895] 119902119894 (1)



119902lowast

119894=

120579

2 + 119889 minus 120573 120587

lowast

119894= [

120579

2 + 119889 minus 120573]

2

(2)


2= 120587lowast2minus 119896




1= 1205871198721= [120579(2 minus 120573)]2 where the


(120587119873

1 120587119873

2) =

([120579

2 minus 120573]

2

0) 120579 le 120579

([120579

2 + 119889 minus 120573]

2

[120579

2 + 119889 minus 120573]

2

minus 119896) 120579 gt 120579

(3)


119894= [120579 + V(119902119890



119894= 120579(3minus2120573)



(120587119865

1 120587119865

2) = ([

120579

2 minus 3120573]

2

+ 119865 [120579

2 minus 3120573]

2

minus 119865) (4)



Max119865

120587119865

1=

1205792

(3 minus 2120573)2+ 119865

st 120587119865

2=

1205792

(3 minus 2120573)2minus 119865 ge 120587

119873

2

(5)


119865lowast

=1205792

(3 minus 2120573)2minus 120587119873

2

=

1205792

(3 minus 2120573)2 120579 le 120579

(5 + 119889 minus 3120573) (119889 minus 1 + 120573) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(6)


120587119865

1

=

21205792

(3 minus 2120573)2 120579 le 120579

[(2 + 119889 minus 120573)2

+ (5 + 119889 minus 3120573) (119889 minus 1 + 120573)] 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(7)




1= 0293) SP1 makes


lowast

2=






1le 119889 le 119889

2= 0121)


3lt 120573 lt 120573lowast


(1198962le 119896 lt 119896


(1198892lt 119889 lt 119889




1) (4) When the



1) SP1 will choose



lowast

1)



lowast

2) on the one





120587119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022

120587119877

2= [120579 + V (119902119890

1+ 119902119890


(8)



119902lowast

1=120579 + (1 minus 120573) 119903

3 minus 2120573 119902

lowast

2=120579 minus (2 minus 120573) 119903

3 minus 2120573 (9)


120587119877


(3 minus 2120573)2

120587119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(10)



2120573)2







1199032=

120579

2 minus 120573 120579 le 120579

120579


2 minus 120573radic

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(11)


119903lowast

1=

(5 minus 4120573) 120579

2 (1205732 minus 5120573 + 5) (12)






1 1199032)



1 when 120579 gt 120579 the




(13)


120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)













lowast






1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















120587119873

119894= [120579 + V (119902119890

119894) minus 119902119894minus 119902119895] 119902119894 (1)



119902lowast

119894=

120579

2 + 119889 minus 120573 120587

lowast

119894= [

120579

2 + 119889 minus 120573]

2

(2)


2= 120587lowast2minus 119896




1= 1205871198721= [120579(2 minus 120573)]2 where the


(120587119873

1 120587119873

2) =

([120579

2 minus 120573]

2

0) 120579 le 120579

([120579

2 + 119889 minus 120573]

2

[120579

2 + 119889 minus 120573]

2

minus 119896) 120579 gt 120579

(3)


119894= [120579 + V(119902119890



119894= 120579(3minus2120573)



(120587119865

1 120587119865

2) = ([

120579

2 minus 3120573]

2

+ 119865 [120579

2 minus 3120573]

2

minus 119865) (4)



Max119865

120587119865

1=

1205792

(3 minus 2120573)2+ 119865

st 120587119865

2=

1205792

(3 minus 2120573)2minus 119865 ge 120587

119873

2

(5)


119865lowast

=1205792

(3 minus 2120573)2minus 120587119873

2

=

1205792

(3 minus 2120573)2 120579 le 120579

(5 + 119889 minus 3120573) (119889 minus 1 + 120573) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(6)


120587119865

1

=

21205792

(3 minus 2120573)2 120579 le 120579

[(2 + 119889 minus 120573)2

+ (5 + 119889 minus 3120573) (119889 minus 1 + 120573)] 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120579 gt 120579

(7)




1= 0293) SP1 makes


lowast

2=






1le 119889 le 119889

2= 0121)


3lt 120573 lt 120573lowast


(1198962le 119896 lt 119896


(1198892lt 119889 lt 119889




1) (4) When the



1) SP1 will choose



lowast

1)



lowast

2) on the one





120587119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022

120587119877

2= [120579 + V (119902119890

1+ 119902119890


(8)



119902lowast

1=120579 + (1 minus 120573) 119903

3 minus 2120573 119902

lowast

2=120579 minus (2 minus 120573) 119903

3 minus 2120573 (9)


120587119877


(3 minus 2120573)2

120587119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(10)



2120573)2







1199032=

120579

2 minus 120573 120579 le 120579

120579


2 minus 120573radic

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(11)


119903lowast

1=

(5 minus 4120573) 120579

2 (1205732 minus 5120573 + 5) (12)






1 1199032)



1 when 120579 gt 120579 the




(13)


120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)













lowast






1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1= 0293) SP1 makes


lowast

2=






1le 119889 le 119889

2= 0121)


3lt 120573 lt 120573lowast


(1198962le 119896 lt 119896


(1198892lt 119889 lt 119889




1) (4) When the



1) SP1 will choose



lowast

1)



lowast

2) on the one





120587119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022

120587119877

2= [120579 + V (119902119890

1+ 119902119890


(8)



119902lowast

1=120579 + (1 minus 120573) 119903

3 minus 2120573 119902

lowast

2=120579 minus (2 minus 120573) 119903

3 minus 2120573 (9)


120587119877


(3 minus 2120573)2

120587119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(10)



2120573)2







1199032=

120579

2 minus 120573 120579 le 120579

120579


2 minus 120573radic

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(11)


119903lowast

1=

(5 minus 4120573) 120579

2 (1205732 minus 5120573 + 5) (12)






1 1199032)



1 when 120579 gt 120579 the




(13)


120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)













lowast






1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1 1199032)



1 when 120579 gt 120579 the




(13)


120587119877

1=

51205792

4 (5 minus 5120573 + 1205732)120579 le 120579

((1 minus 2119889) 120573 + 119889

2 + 4119889 minus 1

(2 + 119889 minus 120573)2

1205792

+ 120579120573radic1205792

(2 + 119889 minus 120573)2minus 119896

+ (5 minus 5120573 + 1205732

) 119896) times ((2 minus 120573)2

)minus1

120579 gt 120579

(14)

120587119877

2=

1205792

1205732

4 (5 minus 5120573 + 1205732)2

120579 le 120579

1205792

(2 + 119889 minus 120573)2minus 119896 120579 gt 120579

(15)













lowast






1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1lt 1205871198771


1gt 1205871198771when 120573 gt 120573lowast

8


5= 1198962+ (((120573 minus 5119889)

2

+

20119889(1minus119889))4(2+119889minus120573)2


when 1198965lt 119896 lt 119896

1and 120587119865

1gt 1205871198771when 119896

2lt 119896 lt 119896

5 We can


9asymp 0319 119896

5gt 1198961when 120573 lt 120573lowast

9




1 we obtain 120587119865

1lt 1205871198771 (2)when

120573 gt 120573lowast9 if 1198962lt 119896 lt 119896

5 we obtain 120587119865

1lt 1205871198771 and if 119896

5lt 119896 lt 119896

1

we obtain 1205871198651gt 1205871198771


1under



1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(16)

where

1198674= [minus2120573

2

+ 2 (2 minus 3119889) 120573 + 21198892

+ 8119889 minus 1] (2 minus 120573)2

minus [(1 minus 2119889) 120573 + 1198892

+ 4119889 minus 1] (3 minus 2120573)2

(17)

When 1198966lt 119896 lt 119896

1 1205871198651lt 120587119877

1 when 119896 lt 119896

6 1205871198651gt 120587119877

1





1le 120573 lt 120573lowast












120587119865119877

1= [120579 + V (119902119890

1+ 119902119890

2) minus 1199021minus 1199022] 1199021+ 1199031199022+ 119865

120587119865119877

2= [120579 + V (119902119890

1+ 119902119890


(18)


120587119865119877


(3 minus 2120573)2

+ 119865 (19)

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 (20)




119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

11205871198651lt 1205871198771

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573

lowast

5119896 le min 119896

6 1198964 = 1198964

120587119865

1gt 120587119877

1

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

31205871198651lt 1205871198771

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771

119889 gt 1198894= 0414 and 120573 gt 120573lowast

5

1198966lt 119896 lt 119896

1120587119865

1lt 120587119877

1

119896 gt 1198966

1205871198651gt 1205871198771

119889 gt 1198894= 0414 and 120573lowast

2le 120573 lt 120573lowast

5119896 le min 119896

6 1198964 = 1198964

1205871198651gt 1205871198771

119889 le 1198894= 0414 and 120573 gt 120573lowast

2119896 le 119896

4120587119865

1gt 120587119877

1

01964 = 1198893le 119889 le 119889

5= 0314 and 120573lowast

7le 120573 lt 120573lowast

21198962le 119896 lt 119896

11205871198651lt 1205871198771

119889 gt 1198895and 120573 lt 120573lowast

21198962le 119896 lt 119896

31205871198651lt 1205871198771

1198893le 119889 le 119889


71198962le 119896 lt 119896

31205871198651lt 1205871198771

1198892lt 119889 lt 119889


41198962le 119896 lt 119896

3120587119865

1lt 120587119877

1

1198891le 119889 le 119889

2and 120573lowast

3lt 120573 lt 120573lowast

41198962le 119896 lt 119896

31205871198651lt 1205871198771


max119865119903

120587119865119877


(3 minus 2120573)2

+ 119865 (21)

such that

120587119865119877

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus 119865 ge 120587119873

2 (22)


119865lowast

1= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

(23)


119903lowast

3=

(1 minus 2120573) 120579

2 (1 minus 120573) 120573 lt 120573lowast

10= 05

0 120573 ge 05

(24)


119865lowast

1=

12057921205732

4 (1 minus 120573)2

120573 lt 05

1205792

(3 minus 2120573)2

120573 ge 05

(25)


1= 1205871198651= 21205792(3minus2120573)


1gt 1205871198771 Thus 120587119865119877

1gt 1205871198771 When 120573 lt 05 we

compare1205871198651198771



120587119865119877

1minus 120587119865

1=(1 minus 2120573) 120579

2 (3 minus 2120573)2119903lowast

3gt 0

120587119865119877

1minus 120587119877

1=

1198675

4 (3 minus 2120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)

(26)


2


5gt 0Then we have120587119865119877

1minus1205871198771gt 0




119865lowast

2= 120587119865119877

2minus 120587119873

2=[120579 minus (2 minus 120573) 119903]

2

(3 minus 2120573)2

minus1205792

(2 + 119889 minus 120573)2+ 119896

(27)


0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 005 01 015 02 025 03 035 04 045 050

01

02

03

04

05

06

07

08

09

1

120573

H5



119865lowast

2=

1205792

1205732


1205792

(2 + 119889 minus 120573)2+ 119896 120573 lt 05

(5 + 119889 minus 3120573) (120573 minus 1 + 119889) 1205792

(3 minus 2120573)2

(2 + 119889 minus 120573)2

+ 119896 120573 ge 05

(28)


1with 120587119877

1and 120587119865





2 120573lowast10) if 1198965le 119896 lt 119896

1 we have

1205871198651198771= 1205871198651gt 1205871198771 if 119896 lt 119896

5 then 120587119865119877

1= 1205871198651lt 1205871198771


120587119865119877

1minus 120587119865


3] 119903lowast3

(3 minus 2120573)2

gt 0

120587119865119877

1minus 120587119877

1= 119896 minus 119896

7= 119896 minus [119896

1minus

12057921205732

4 (1 minus 120573) (5 minus 5120573 + 1205732)]

(29)

where

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792

1198676= 1205732

(2 + 119889 minus 120573)2


)

(30)

0 005 01 015 02 025 03 035 04 045 05minus20

minus18

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

120573

H6(1)


0 005 01 015 02 025 03 035 04 045 05minus05

minus045

minus04

minus035

minus03

minus025

minus02

minus015

minus01

minus005

0

120573

d = 0

k8minusk7


119889 = 0



0 lt 1198967le 1198961


7lt 119896 lt 119896

1and

1205871198651198771

lt 1205871198771 when 119896 lt 119896



1and 120587119873

1 As 120587119865119877

1minus 1205871198731= 119896 minus 119896



8=


(2+119889minus120573)2

)4(3minus

2120573)2

(2+119889minus120573)2





0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 005 01 015 02 025 03 035 04 045 05minus250

minus200

minus150

minus100

minus50

0

120573

k8minusk7

d = 02

d = 08

d = 05

d = 04 d = 1


regarding 120573







5le 119896 lt 119896

1) then two-part



optimal strategy



2


120573)]2

(2 minus 120573)2








7 Conclusions






Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Appendix


Consider

1198961=

1205792

(2 + 119889 minus 120573)2 119896


(3 minus 2120573)2

(2 + 119889 minus 120573)2

1198963=minus1198624+ radicΔ

1

21198623

1198964=1205732


minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

1198966=

1198674

(2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

1198967= minus

1198676

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732) (1 minus 120573)1205792



(2 + 119889 minus 120573)2

times (4 (3 minus 2120573)2

(2 + 119889 minus 120573)2


) 1205792

120573lowast

1= 0293 120573

lowast

2= 1 minus 119889

120573lowast


4

120573lowast


4

120573lowast


2

120573lowast


2

120573lowast


2

2 (41198892 minus 40119889 + 50)

120573lowast

8= 0306 120573

lowast

9asymp 0319 120573

lowast

10= 05

1198891= 0051 119889

2= 0121 119889

3= 0196

1198894= 0414 119889

5= 0314

(A1)


1= 1205871198651minus 1205871198731198711

ge 0 Δ1205872=

1205871198652minus 1205871198731198712


1= 1205871198651minus 1205871198731198711


Δ1205871= 120587119865

1minus 120587119873119871

1=

1205792

(3 minus 2120573)2

(2 minus 120573)21198671(120573) 120579 le 120579

119896 minus 1198962

120579 gt 120579

(A2)





1isin [0 1)



0293 If 120573 lt 120573lowast1 then119867


1lt

0 If 120573 gt 120573lowast1 then119867

1(120573) ge 0 We also have Δ120587

1ge 0





2



1198961minus 1198962=

1205792

(2 + 119889 minus 120573)2

(3 minus 2120573)21198672(120573) (A3)


2(120573) =





Δ1205871lt 0



1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1198672(120573) = minus2120573

2

+ (1 minus 119889)120573 + 21198892


4= ((1 minus 119889) +




1le 119889 le




4isin



1198891le 119889 le 119889



4ge 1 minus 119889


2 then we


3or 120573 ge 120573lowast

4



4 we

have1198672(120573) gt 0 and 119896

1gt 1198962 (2) If 119889

2lt 119889 lt 119889

3 then we have




2(120573) ge 0 and 119896

1ge 1198962 if 120573 gt 120573lowast

4

we have1198672(120573) lt 0 and 119896

1lt 1198962 (3) If 119889 ge 119889


3lt 0


4


1gt 1198962







4) and


1)





1)






1 What about




(A4)


120579 minus 120579

=(2 + 119889 minus 120573)radic119896


1198673(120573)

(A5)






1lt 119903lowast2







120587119877

1minus 120587119873119871

1=

12057321205792

4 (5 minus 5120573 + 1205732) (2 minus 120573)2ge 0 (A6)


120587119877

1minus 120587119873119871

1=

1205792

4 (5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

times (1205732

minus 10119889120573 + 51198892

+ 20119889) gt 0

(A7)



120587119877

1minus 120587119873119871

1

= (1198621(120573) 1205792

+ (5 minus 5120573 + 1205732

) (2 + 119889 minus 120573)2

119896

+ 120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896)


(2 + 119889 minus 120573)2

)minus1

(A8)



1gt 1205871198731198711

then solving1198621(120573) =



6gt 1 Moreover





6gt


5 then 119862

1(120573) ge 0 if 0 le 120573 lt 120573lowast

5



5ge 1


4


1gt 1205871198731198711


1and 120587119873119871

1 when 119889 gt 119889




1(120573) lt 0

When 1198621(120573) lt 0 if 120587119877

1gt 120587119873119871

1 120579120573(2 + 119889 minus


2(120573) = [1205732 minus (5 minus



Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Let


2 minus 4119889 + 5

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

as 1198961minus 1198964=

119889 (119889 + 4 minus 2120573)

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)21205792

ge 0

(A9)

We have1198622(120573) le 0 when 119896 le 119896

4 thus 120587119877

1gt 1205871198731198711

However if1198964lt 119896 lt 119896

1 we have119862


the values of 1205871198771and 120587119873119871

1when 119862

2(120573) gt 0 and

120579120573 (2 + 119889 minus 120573)2

radic1205792

(2 + 119889 minus 120573)2minus 119896 ge 119862

2(120573)

997904rArr 11986231198962

+ 1198624119896 + 1198625le 0

(A10)

where

1198623= (5 minus 5120573 + 120573

2

)2

(2 + 119889 minus 120573)4

1198624= [1205732

(2 + 119889 minus 120573)2

minus 2 (1205732


minus 4119889 + 5) (5 minus 5120573 + 1205732

)]

times (2 + 119889 minus 120573)2

1205792

1198625= [(120573

2


minus 4119889 + 5)2

minus1205732

(2 + 119889 minus 120573)2

] 1205794

(A11)

Solving 11986231198962 + 119862

4119896 + 119862



4+radicΔ1)21198623 If 1198963gt


2= 1 minus 119889 119862

3gt 0 119862

5gt 0

120573 lt 120573lowast2 1198623gt 0 and 119862


2= 1 minus 119889 and

1198963le 0 rArr 119862

31198962 + 119862

4119896 + 119862

5gt 0 rArr 120587119877

1lt 1205871198731198711


2 1198963gt 0 we have

1198963le 1198961

997904rArr119896 lt 1198963997904rArr 119862

31198962

+ 1198624119896 + 1198625le 0 997904rArr 120587

119877

1ge 120587119873119871

1

1198963lt 119896 lt 119896

1997904rArr 119862

31198962 + 119862

4119896 + 1198625gt 0 997904rArr 120587119877

1lt 1205871198731198711

1198963gt 1198961997904rArr 119896 lt 119896

1997904rArr 119862

31198962

+ 1198624119896 + 1198625lt 0 997904rArr 120587

119877

1gt 120587119873119871

1



4and



(2 + 119889 minus 120573)4

1205792





3 and


1

21198623

minus1205792

(2 + 119889 minus 120573)2le 0


21198623

(2 + 119889 minus 120573)21205792

+ 1198624

(A13)


120573lowast


2

2 (41198892 minus 40119889 + 50) (A14)




7rArr 1198963gt 1198961Then


2 After





7lt 120573lowast2




2under 119889 gt 119889





5ge 120573lowast2










2) and if RampD


3) then royalty





royalty licensing




5lt 119889 le 119889



high (1198963le 119896 lt 119896










3le 119896 lt 119896

1)


4lt 119896 lt 119896

3) royalty




royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











royalty licensing





1gt 1205871198651 if

120573 gt 120573lowast8 then 120587119865





120579 lt 120579 le 120579

120587119865

1minus 120587119877

1= 119896 minus 119896

2minus

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

= 119896 minus 1198965

(A15)

where

1198965= 1198962+

1205732 minus 10119889120573 + 20119889 + 51198892

4 (2 + 119889 minus 120573)2

(5 minus 5120573 + 1205732)1205792

(A16)


1gt 1205871198771 when

119896 lt 1198965 we have 120587119877

1gt 1205871198651 As 119896 lt 119896

1 we have to ensure



5= 1198961


1lt 1198965and when

120573 gt 120573lowast9 we have 119896

1gt 1198965




5le








(1198892lt 119889 lt 119889





1le 119889 le 119889







2le 119896 lt 119896

5) then royalty


0 01 02 03 04 05 06 07 08 09 1minus50

minus40

minus30

minus20

minus10

0

10

20

120573

k1minusk5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1


regarding 120573

When 120579 gt 120579

120587119865

1minus 120587119877

1

= (11986741205792


(2 + 119889 minus 120573)2

times (120579120573radic1205792



(2 + 119889 minus 120573)2

(2 minus 120573)2

)minus1

(A17)






6 we have 120587119877

1gt 1205871198651 To make

1205871198771gt 1205871198651 we must have 119896 le 119896

6and 119867

5= (1 minus 120573)(119896




Δ3= [1205732minus2(1minus120573)119867

4]21205794minus4(1minus120573)2[119867

4minus1205732(2+119889minus120573)2]1205794


if 119896 le 1198966 we have 120587119865





6





0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 01 02 03 04 05 06 07 08 09 10

05

1

15

2

25

3

35

4

45

5

120573

H4

d = 02

d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573







6under


2le 120573 lt 120573lowast

5or 119889 le 119889


2

As

1198964minus 1198966=

1198676

(5 minus 5120573 + 1205732) (2 + 119889 minus 120573)2

(3 minus 2120573)2

(1 minus 120573)1205792

(A18)



4le 1198966is


2le 120573 lt 120573lowast

5or 119889 le 119889

4and

120573 ge 120573lowast




4


120573lowast

4minus 120573lowast

7=

1198677

4 (21198892 minus 20119889 + 25) (A19)

where

1198677= 38119889

2

minus 45119889 + 25

+ (21198892


Δ2= 41198892

[(119889 + 8)2

+ (8119889 + 16) (41198892

minus 40119889 + 50)]

(A20)




0 01 02 03 04 05 06 07 08 09 1minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

H5

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573

0 01 02 03 04 05 06 07 08 09 1minus6

minus4

minus2

0

2

4

6

8

10

120573

k2minusk6

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1


regarding 120573



5) and


1) then royalty


6) then fixed-


4) but the


5) then it will


0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 01 02 03 04 05 06 07 08 09 1minus900

minus800

minus700

minus600

minus500

minus400

minus300

minus200

minus100

0

100

120573

C3k2 2+C4k2+C5

d = 02

d = 08

d = 05

d = 04

d = 0

d = 01

d = 1



0 01 02 03 04 05 06 07 08 09 1minus15

minus10

minus5

0

5

10

15

20

120573

d = 02 d = 08

d = 05

d = 04

d = 0

d = 1

k4minusk6


regarding 120573


4) (c)






5) and the


2) then royalty


004 006 008 01 012 014 016 018 02minus80

minus78

minus76

minus74

minus72

minus70

minus68

minus66

d

H7


cost (1198962le 119896 le 119896




(1198962le 119896 lt 119896





2le 119896 lt 119896

3) then royalty


2lt 119889 lt 119889

3) the





1le 119889 le 119889




(1198962le 119896 lt 119896


fee licensing



Acknowledgments



References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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