Coordinating experimentation - Fernando Ochoa

Coordinating experimentationlowast

Fernando Ochoa Adagger

August 2019

PRELIMINARY AND INCOMPLETE PLEASE DO NOT CITE OR CIRCULATE

Updated Frequently Download Most Recent Version Here

Abstract

I consider a simple two-agent model of strategic experimentation with payoff

externalities and unobservable actions Each agent decides whether to invest effort

(or resources) into an independent risky project if he invests and succeeds his payoff

is greater if the other agent succeeds too I characterize the unique symmetric MPE

of the game The dynamic structure of the problem allows agents to coordinate

their experimentation in such a way that they provide strictly more effort than

they would have provided in the absence of the externality Nevertheless they

provide too little effort in a dynamically inefficient way relative to the first best

Agents start providing the socially efficient effort (maximal effort) In absence of

success agents become more pessimistic about projectsrsquo quality and begin to provide

dynamically inefficient levels of effort When they become pessimistic enough they

cease experimentation Finally I show that patience plays a non-trivial role interior

levels of patience minimize efficiency losses

lowastI am indebted to Nicolas Figueroa for unconditional advice and support on my MA thesis on whichthis paper is based I would like to thank Tibor Heumann Caio Machado and Juan Pablo Xandri fortheir invaluable feedback I am also grateful to Dirk Bergemann Marina Halac Nicolas Inostroza ErikMadsen Lucas Maestri Juan-Pablo Montero Alessandro Pavan and Juuso Toikka for helpful commentsand discussions Finally I thank the ICSI Institute (Milenio P05-004F and Basal FBO-16) for financialsupport Any error or omission is my sole responsibility

daggerDepartment of Economics New York University Email nanoochoanyuedu

1 Introduction

There are many industries where the profitability of a firmrsquos innovation depends on

other firmsrsquo innovations and vice versa For example a new and more powerful computer

hardware would be more valued by customers if new and more demanding PC software

is created but that potential software could be used only with more powerful hardware

Also decisions on how much to invest to create a new product should consider potential

innovations of intermediate goods producers that might lower their production costs as

well as the investment in RampD to reduce production costs should consider that new

products might boost intermediate goods demand in the future For example while

many tech-companies were investing in RampD to reduce the size of lithium batteries for

smartphones and other devices the mining industry was investing in RampD to reduce

lithiumrsquos production costs1

The consideration of possible future innovations by other firms in the RampD decisions

of a firm can be understood as an externality on the experimentation payoffs This

means that the success of a firmrsquos RampD increments net payoff of other firmrsquos RampD If

firms invest more when there are externalities than when there are none then they are

coordinating their experimentation The necessary conditions that lead to the coordination

of experimentation could be crucial to understanding the success of some innovation-based

industries as well as the failure (or non-existence) of others

This work provides a simple model of experimentation to understand how

externalities on experimentation payoffs affect agentsrsquo behavior I model experimentation

as a variation of the continuous-time two-armed exponential bandit model of Keller et al

(2005) In particular I follow the main variations used by Bonatti and Horner (2011)

I consider the strategic behavior of experimenting agents when externalities on the

payoffs are exogenous and actions are unobservable The goal is to determine how payoff

externalities change the dynamics of investment provision An example of this is when

firms have to decide whether to dedicate their resources to selling their usual products or

investing in RampD to try to develop a new product that will have a greater market value

(or lower production cost) if other firmsrsquo RampD is also successful I model this situation

as one where two agents have to decide whether to allocate resources to experiment on

a risky project (risky arm) of unknown quality or into a safe one (safe arm) that pays

a fixed amount each period I consider the safe arm as the experimentation cost of

opportunity The risky arm can be bad and yield zero profits or good and yield a profit

Each agentrsquos risky arm quality is independent from the others and it can be learned only

by experimentation The strategic component arises because if the risky arm is good it

1Although the motivation of this work is theoretical it is worth mentioning that identification andmeasurement of RampD spillovers have been an active line of empirical research in the last decades Seefor example Griliches (1992) Malerba et al (2013) Steurs (1995) and Park (2004)

2

yields more profit if the other agent were successful too

Most of the recent strategic experimentation literature has focused on the problem

of free riding in several settings such as teams public goods or experimental consumption

(Bonatti and Horner 2011 Georgiadis 2015 Heidhues et al 2015) My model shares

some qualitative characteristics with those models However I show that many differences

and new predictions arise when the problem of experimentation coordination is considered

To the best of my knowledge this is the first work that brings the coordination problem

to the strategic experimentation literature

I show that the mix of the learning process with the dynamic structure of the

problem allows agents to coordinate their experimentation In particular there are two

forces that shape the unique symmetric MPE of the game On one hand experimentation

is more profitable today since the expected payoff is greater because of the externality so

agents are willing to provide more effort for more time On the other hand in order to save

experimentation costs agents have incentives to stop their experimentation to wait and see

if the other agent has success and then experiment again The evolution of beliefs about

projectsrsquo quality is what changes the relative importance of both forces over time This

result is surprising since my model has the characteristics of a coordination game with

multiple symmetric equilibria (ie supermodular games) where a selection criteria (ie

global games) is necessary to obtain testable predictions Therefore my work suggests

that the dynamic structure and the learning component of strategic experimentation

games could ldquorefinerdquo the equilibria without the need to impose an extra refinement

criterion

Although agents coordinate their experimentation in equilibrium there is

under-provision of effort If the common prior over projectsrsquo quality is high enough

agents start exerting socially efficient levels of effort (maximal) for some time As

agents become more pessimistic about projectsrsquo quality effort starts to be provided in

a dynamically inefficient way (ie they provide interior levels of effort) When agents

become pessimistic enough they stop experimentation Numerical exercises show that

agents stop experimentation too early relative to the first best Finally patience plays a

non-trivial role interior levels of patience minimize efficiency losses relative to the first

best All these characteristics are different from moral hazard in teams models For

example in Bonatti and Horner (2011) effort is always provided in an inefficient way

experimentation last forever and the more impatient the agents are the more effort they

provide

3

2 Related literature and contribution

This work fits into the literature of experimentation with exponential bandits

Strategic experimentation models have been used to model the problem of

experimentation in several settings such as exogenously and self-organized teams (Bonatti

and Horner 2011 Georgiadis 2015) or experimental consumption of goods (Heidhues

et al 2015)2 This literature aims to answer questions such as how much experimentation

can be expected in equilibrium how fast it is done optimal team size and provision

dynamics A particular assumption broadly used by this literature is that the quality

(good or bad) of agentsrsquo risky arms are perfectly correlated (ie every agent face the

same risky arm) One of the main reasons why the strategic experimentation literature

share that assumption is because equilibrium with imperfectly correlated risky arms is

still an open question in the literature3 My simple two-agent model departure from the

literature because without overcoming the aforementioned limitation allows me to study

a new setting where each agent faces a different project whose quality is independent from

other agentrsquos project The externalities on the payoffs of the risky arm is what keep the

strategic component in my model of experimentation despite the non-correlation of the

risky arms

This work is also related to a vast RampD literature that studies the effect of RampD

spillovers on firmsrsquo investment decisions Nevertheless this literature usually focus on

the effects that intra-industry competition (or negative spillovers) changes the way that

inter-industry spillovers affects decisions in static settings (Steurs 1995 Bessen and

Maskin 2009) externalities in our model can be understood as inter-industy spillovers4

Chen (2012) analyze innovation frequency (ie timing of release) in an industry with two

producers of perfect complementary goods he shows that coordination failures arises in

the timing of releases (too early or too late) My model departs from this literature since I

do not assume that goods are perfect complements or that innovation is sequential Also

I can explicitly study investment dynamics

The closest related work in RampD spillovers literature is the work by Salish (2015)

She studies a dynamic model of RampD investment a la Keller et al (2005) with inter-

and intra-sectoral RampD spillovers where the profitability of an invention in one sector

(dependent) increases if there is an innovation in another sector (independent) but not

vice versa She finds that if the independent firm learns faster than the dependent there

2The insights of this literature have been incorporated by contract theory In general terms theystudy how to provide optimal incentives for teams that have to experiment See for example Moroni(2015) Shan (2017) Georgiadis (2015)

3Up to our knowledge the only exception is Klein and Rady (2011) They characterize the Markovperfect equilibrium for two agents that face negatively correlated bandits (they extend the results forthree players)

4By static setting I refer to that the investment decisions are essentially static despite the model isdynamic

4

is no effect of spillovers but if it learns slower the dependent firm will invest more relative

to a setting without spillovers The main difference between Salishrsquo model and mine is

that she assumes that one sector is independent which implies that there is no strategic

game In other words the inter-sectoral spillovers model of Salish (2015) is a particular

case of my model when just one firm benefits from the other

3 Model

There are two agents engaged in different and independent projects each project

may be good (θi = G) or bad (θi = B) Agent irsquos project is good with a probability

p isin (0 1) which is assumed to be public information

Time is continuous in t isin [0infin) Each agent continuously chooses a level of costly

effort Agent irsquos instantaneous cost of exerting effort ui isin [0 1] is ci(ui) = cui

If the project is good and agent i exerts effort ui at time t a breakthrough occurs

according to a Poisson process with intensity f(ui) = λui for some λ gt 0 If the project

is bad a breakthrough never arrives

For each player the game ends if a breakthrough occurs on his project which can

be interpreted as the successful completion of the project5 A successful project is worth

an expected net present value of π to the agent if the breakthrough occurs at some

t lt infin While a breakthrough does not occur agents reaps no benefits from their project

Nevertheless conditional on having a breakthrough each agent payoff depends on the

other agent having a breakthrough Agent i is said to be successful iff he has experienced

a breakthrough at some t lt infin A successful agent j generates an (expected) externality

γ on the payoff of agent irsquos project All agents discount the future at a common rate

r gt 0

If agent i exert effort (uit)tge0 and agent j was successful at tprime lt t and a breakthrough

arrives at some t lt infin the average discounted payoff to agent i is

r

983061eminusrt(π + γ)minus c

983133 t

0

eminusrsuisds

983062

On the other hand if agent j is not successful yet the average discounted payoff of

a breakthrough to agent i is

r

983061eminusrt(π + E[γ|pjt ujτget])minus c

983133 t

0

eminusrsuisds

983062

5As Moroni (2015) notes an alternative interpretation is that after a breakthrough there are a numberof tasks with known feasibility to be done

5

where E[γ|pjt (uj)τget] is the expected discounted externality Which is the discounted

value of the externality integrated over the probability density of a breakthrough on jrsquos

risky arm which depends on the belief at t that jrsquos arm is good pjt and on the expected

effort (uj)τget that agent j will exert thereafter

Breakthroughs are public information meaning that all agents observe the time at

which a breakthrough occurs and who obtained it6

My objective is to identify the symmetric Markov Perfect Bayesian Equilibrium

(MPBE) in pure strategies of this game That is each agent i chooses two measurable

functions ui R+ rarr [0 1] (pure strategy) to maximize his expected payoff First a

function conditional on no breakthrough having occurred Second a measurable function

that maximizes his payoff if the other agent has a breakthrough at some t isin [0infin)

In absence of a breakthrough each agent learns about the quality of his project

this is captured in the evolution of his belief that his project is good7 By Bayesrsquo rule

pit =peminusλ

983125 t0 uiτdτ

peminusλ983125 t0 uiτdτ + (1minus p)

(1)

Using (1) we can describe the evolution of an agent belief by the familiar differential

equation

pit =dpitdt

= minuspit(1minus pit)λuit (2)

Every time that an agent exert effort and a breakthrough does not arrives he becomes

more pessimistic about project quality On the equilibrium path each agent learns about

both projects

4 Analysis of the problem

In this section I solve the autarky problem to fix some basic ideas from

experimentation literature Second I solve the problem of an agent when the other

had a breakthrough Finally I characterize agentrsquos problem for a given expected payoff

externality

6As will be clear later the case of unobservable breakthroughs is trivial since a successful agent hasno reasons to hide a breakthrough

7Since the arrival of a breakthrough is a Poisson process given an effort profile uit the probability

that a breakthrough has not occurred between τ and t if the project is good is expminusλ983125 t

τuitds

6

41 Autarky

The autarky problem is the simplest problem of experimentation with exponential

bandits only one agent faces the ldquoexploitationrdquo vs ldquoexplorationrdquo trade off Obviating the

individual subscript the agentrsquos problem is

ΠA(p) = maxut

983133 infin

0

[ptλπ minus c]uteminus

983125 t0 (psλus+r)dsdt

subject to

pt = minuspt(1minus pt)λut and p0 = p

Note that the instantaneous probability assigned by the agent to a breakthrough

occurring is ptλut and the probability that no breakthrough has occurred by time t ge0 is eminus

983125 t0 psλusds then ptλute

minus983125 t0 psλusds is the probability density that agent i obtains a

breakthrough at time t Since the integrand is positive as long as ptλπ ge c its clear that

if that condition is satisfied is optimal to set ut = 1 and ut = 0 otherwise Lemma 1

summarizes the result

Lemma 1 In the autarky problem the agentrsquos optimal experimentation rule is

uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)

This is a standard result in experimentation literature qualitatively similar to the

planner solution in Bonatti and Horner (2011) and Moroni (2015)8 Note that TA is only

well defined when λpπ gt c To make exposition more clear I avoid to prove that some

sub-games are well defined by assuming the following9

Assumption 1 It is always profitable for any agent to experiment for at least some

arbitrary small amount of time λpiπi gt c foralli8This is because both models consist on a team working on the same project so the first best is

equivalent to my autarky model To be more precise Moroni (2015) has a team that works on severalmilestones of a big project but just in one milestone at a time

9Main results hold for priors p lower than pa Therefore Assumption 1 is not necessary it is onlymade for exposition purposes

7

I am going to refer to

pa =c

λπ (4)

as the threshold belief at which an agent stops experimentation in autarky Also I refer

to effort at time t as the instantaneous individual effort to total effort at t as the sum of

individual effort at that time and to aggregate effort at t as the integral of of total effort

over all time up to t

42 Agentrsquos problem after a breakthrough

In order to characterize the problem I need to solve the subgames after a

breakthrough Wlg I am going to write the problem from the perspective of agent

i I start by considering the problem of agent i when agent j had a breakthrough at some

arbitrary time t

ΠS = maxuiτ

983133 infin

t

[piτλ(π + γ)minus c]uiτeminus

983125 τt (pisλuis+r)dsdτ (5)

st piτ = minuspiτ (1minus piτ )λuiτ

Note that the problem faced by agent i is equivalent to the autarky problem starting

at time t with initial belief pit Therefore Lemma 1 implies the following

Corollary 1 If agent j has a breakthrough at some arbitrary t agent i optimally chooses

the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln

983061λ(π + γ)minus c

c

983062minus ln

9830611minus pitpit

983062983064 (6)

As before T S is not well defined for every set of parameters λ pit π γ cNevertheless is always well defined for some t at which agents were experimenting The

following Lemma formalizes this statement

Lemma 2 T S is always well defined as an outcome of a symmetric sub-game where

players were experimenting

8

Proof If in any symmetric equilibrium agent j has a breakthrough at some t then it

was rational for him to experiment at t The symmetry implies that it was also rational

for agent i to experiment at t Therefore T S is well defined because at t + dt (after jrsquos

breakthrough) agent irsquos risky arm payoff is strictly higher than at t

Note that ΠS depends only on pit (see Appendix A)

ΠS(pit) = c

983083pit

λ+ r

983077β minus βminus r

λ

983061pit

1minus pit

983062minus(1+ rλ)983078minus (1minus pit)

r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084

where β = (λ(π + γ)minus c)c

43 Agentrsquos problem before a breakthrough

When no agent has been successful yet a breakthrough for agent i at some time

t implies an instant payoff of π plus the expected discounted externality The later is

the discounted externality that he will receive given that agent j will face the problem

ΠS(pjt)

E[γ|pjt ujτget] = ΠDE(pjt) = γ

983133 infin

t

pjτλujτeminus

983125 τt (pjsλujs+r)dsdτ (7)

On the equilibrium path agent i knows that j will follow the same investment rule

when he face the problem of maximizing ΠS(pjt) that is ujt = 1 iff T Sj + t ge τ The

expected discounted externality can be rewrite as (see Appendix A)

ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078

where γ = γ(1 + rλ)

Now I can write the problem of agent i when j had no success yet10

ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c

983064uit + pjtλujtΠ

S(pit)

983070eminus

983125 t0 (pisλuis+pjsλujs+r)dsdt (8)

10As is standard in the exponential bandits literature we ignore terms of o(dt) this implies that theprobability of both agents having a breakthrough in an interval dt is zero

9

subject to

pmt = minuspmt(1minus pmt)λumtdt and pm0 = p

form = i j To understand ΠEi note that at every t there are three possibilities agent i has

a breakthrough agent j has a breakthrough or no one has breakthrough (see footnote 10)

With instantaneous probability pitλ agent i has a breakthrough on his project receiving

an expected payoff composed by the payoff of the project π plus the discounted expected

externality ΠDE(pjt) On the other hand with instantaneous probability pjtλujt agent j

has a breakthrough and agent i faces the problem ΠS(pit) If no one has a breakthrough

the agent reaps no benefits The probability that no one had a breakthrough up to time

t is expminus983125 t

0(pisλuis + pjsλujs)ds

5 Cooperative solution

The amount of effort that would be chosen cooperatively by agents (or by a planner)

is the one that maximizes the sum of individual payoffs

W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt) + ΠS(pjt)

983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus

983125 t0 (pisλuis+pjsλujs+r)dsdt

subject to


for m = i j Replacing (5) and (7) in the objective function I can rewrite the objective

function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t

[pmτλ(π + 2γ)minus c]umτeminus

983125 τt (pmsλums+r)dsdτ

Note that this specification gives a straightforward interpretation Cooperative

10

agents consider that a breakthrough on any risky arm yield a payoff π plus ΠSC the

later is the (social) expected payoff of a breakthrough on the other arm That is they

internalize that a second breakthrough implies twice the externality which is clear since

the only difference between ΠSC and ΠS is that the externality is multiplied by two But

they also consider the costs of experimentation instead of ΠDE that only considers the

expected externality The following Proposition characterizes the cooperative solution

the proof is left to Appendix B

Proposition 1 The cooperative solution while no agent is successful is to set

uCit =

983099983103

9831011 if t le TC

0 if t gt TC

where TC (at which a belief pc is reached) has no analytical solution but can be calculated

from an implicit function The cooperative solution implies maximal effort for more time

than in autarky TC gt TA for every set of parameters

If there is a breakthrough in some arm at t le TC The cooperative solution for the

unsuccessful agent is to set

uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln

983061λ(π + 2γ)minus c

c

983062minus ln

9830611minus ptpt

983062983064

After a first breakthrough the cooperative solution implies maximal effort from the

unsuccessful agent for more time than in the non-cooperative solution T SC gt T S for every

set of parameters

This is not a surprising result because of the definition of an externality In absence

of a breakthrough agents would exert maximal effort for more time than in autarky since

the expected value of their projects is greater because of the externality Also after a

first breakthrough the unsuccessful agent would exert maximal effort for more time than

in the non cooperative solution since he considers not only the increment in the expected

value of his project because of the first breakthrough but also the externality that a

breakthrough on his arm will produce on the successful agent

11

6 Non-cooperative equilibrium

In the symmetric non-cooperative Markov equilibrium each agent decides an effort

profile (function) in order to maximize (8) Since effort is not observed agents share a

common belief only on the equilibrium path This makes difficult to characterize the

equilibrium for any arbitrary history since agentrsquos best-response depends both on public

and private beliefs For this reason our proof relies on Pontryaginrsquos principle where other

agentrsquos strategies can be treated as fixed which simplifies the analysis11 The proof is

presented in Appendix C

In order to simplify the analysis I also assume that agents are sufficiently patient

Nevertheless violation of this assumption does not changes the qualitative characteristics

of the equilibrium for a large set of parameters12

Assumption 2 Agents are sufficiently patient In particular λ gt r

The following proposition characterizes the unique symmetric equilibrium of the

game Despite the simplicity of the setting it is not possible to obtain analytical solutions

to characterize the equilibrium Therefore I characterize the optimal path of our optimal

control problem by an implicit function (inverse hazard rate) of xlowastit = ln((1 minus plowastit)p

lowastit)

and obtain numerical solutions for equilibrium effort

Proposition 2 There exist a unique symmetric equilibrium in which agents exert effort

according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p

where p is the belief at which the upper bound on effort is not binding That is

agents exert maximal effort until their beliefs reach p and interior levels thereafter The

function xlowastit is characterized by the solution of the following non-linear ODE

xt =r983045eminus2xit(π + γ minus c

λ) + eminusxit

983043π minus 2c

λ

983044minus eminus(1minus r

λ)xitβminus(1+ r

λ)γ minus c

λ

983046

minuseminus2xit983043π + γ minus c

λminus cβ

λ+r


λ)xit

983059γ rλβminus(1+ r

λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)

where xit = λuit with initial condition xit = ln((1minus p)p)

Effort is weakly decreasing and strictly decreasing in the interior part of the

11In the strategic experimentation literature is common to use Dynamic Programming tools tocharacterize the equilibrium Nevertheless this tools are difficult to apply to settings in which public andprivate beliefs might differ because optimality equations are partial differential equations Therefore Ifollow the approach of Bonatti and Horner (2011) to solve the problem

12This is because the effect of patience is of second order in the equilibrium Codes are available atAppendix D to calculate the equilibria for every combination of parameters

12

equilibrium A threshold beliefmacrp lt p could exist if this threshold is reached at some

macrt

agents exert zero effort for all t gemacrt If

macrp does not exists or is never reached agents exert

positive effort forever

A numerical example of equilibrium is presented in Figure 113 The following

statements are conclusions of numerical analyses of the equilibrium for several

combinations of parameters14 All results can be replicated for any other set of parameters

using codes available at Appendix D

The symmetric equilibrium is qualitatively robust to a large set of parameters15 If

agents have a high enough prior p gt p they exert maximal effort until some t at which

pt = p For p isin (macrp p) they exert interior levels of effort until time

macrt at which pt =

macrp For

every t gtmacrt agents exert no effort In the numerical exercises I have not found a set of

parameters such thatmacrp does not exists Clearly this not mean that

macrp exists for every set

of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit

(b) Belief evolution

Figure 1 Non-cooperative equilibrium effort for parameters (π γ cλ r p) =(1 05 01 025 02 05)

A comparison between non-cooperative and cooperative solution is presented in

Figure 2 Agents provide maximal effort for more time than in autarky but for strictly

13Parameters are normalized such that π = 114The interior equilibrium effort was calculated by iterations of an integral equation (a transformation

of the ODE into an integral equation) As can be noted in all figures there is a ldquokinkrdquo in the interiorequilibrium effort That is because a lot of computational precision is needed to calculate low levels ofeffort The actual equilibrium involve a ldquofastrdquo but continuous drop of effort until zero Is possible tosolve the ODE numerically this allows us to obtain the equilibrium effort for more time but with a lot ofnoise Codes are provided in Appendix D to replicate both ways the results are qualitatively the same

15I have not found yet a combination of parameters that changes the qualitative characteristics of theequilibria

13

less time than in the cooperative solution t isin (TA TC) For beliefs pt isin (macrp p) agents

provide positive but inefficient levels of effort16 Finally agents stop experimentation

inefficiently earlymacrp gt pc

The result implies that efficiency losses are critically linked to the initial prior p

If agents share low enough priors p isin (macrp p) they internalize (qualitatively) very little of

the externality Even more there are some beliefs p isin (pcmacrp) at which in the cooperative

solution it is efficient to exert maximal effort but in the non-cooperative solution agents

do not exert any effort

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit


Figure 2 Comparison of cooperative (Red) solution and non-cooperative (Blue)equilibrium for parameters (π γ cλ r p) = (1 05 01 025 02 05)

Why has this game a unique symmetric equilibrium despite having the taste of a

usual coordination game with multiplicity of equilibria Because of its dynamic structure

Letrsquos use as an example the two usual extreme equilibria of a coordination game both

agents exert maximal effort until the autarky threshold belief pa and zero thereafter or

both exert maximal effort until the cooperative threshold belief pc and zero thereafter

If agent j follows the first strategy at pa is optimal for agent i to exert maximal effort

because he knows that a breakthrough on his arm will make optimal for agent j to start

experimenting again which has an expected value ΠDE(paj ) for agent i We refer to this

effect as incentives to experiment today On the other hand if agent j follows the later

strategy agent i becomes more pessimistic about his and agent jrsquos arm quality over time

Therefore when agent irsquos beliefs reaches some threshold belief p he considers optimal to

16I refer to inefficient experimentation to interior levels of experimentation Hypothetically imaginethat agents can commit to behave cooperatively but with a restriction over aggregate effort U =

983125infin0

(uit+ujt)dt which is less or equal than aggregate effort of the cooperative solution In that case agents willchoose to exert maximal effort until they reach U

14

wait and see if j has success This is because the value of ΠDE(pj) is not enough to

compensate the costs of experimentation so he prefers to maintain his beliefs at pi and

start experimenting again if only if j has a breakthrough The combination of this forces

is what makes agents to coordinate their experimentation ldquorefiningrdquo the equilibrium

The learning component gives the non-cooperative equilibrium its shape by changing

the incentives to experiment more today and to wait and see over time For high enough

beliefs pt the incentives to experiment today makes agents to exert maximal effort

because they know that if they have a breakthrough the other agent will continue his

experimentation Nevertheless as agents become more pessimistic about the quality of

projects the weaker the incentives to experiment today and the more tempting is to wait

and see The interior part of the equilibrium arises because both agents wants to wait

and see but they need to provide incentives to the other agent to keep experimenting

Finally I am going to refer to off the equilibrium path behavior17 Despite there

being infinite types of deviations the only thing that is relevant for a deviated agent at

time t is the aggregate effort that he exerted in [0 t) This is because if aggregate effort

of a deviated agent is lower (higher) than it would have been on the equilibrium path

he is more (less) optimistic about the quality of his arm than the other agent18 If the

aggregate effort is lower the deviated agent is more optimistic about the quality of his

arm than the other Therefore if the deviated agent is more optimistic than the other

agent he will provide maximal effort until his private belief catches up with the other

agentrsquos belief On the other hand if the agent is more pessimistic he will provide zero

effort until the other agent becomes as pessimistic as him about the quality of his arm

7 Numerical exercises

In this section I study the sensibility of cooperative and non-cooperative equilibrium

to changes in two key parameters externality γ and patience r Since I am interested in

the effects of the externality I put focus in the effort provided by agents with an initial

prior p = pa That allows me to study the changes in total and aggregate effort that would

have not been provided in absence of externalities Also since after a first breakthrough

the cooperative solution always implies more aggregate effort than the non-cooperative

equilibrium (See Proposition 1) I only consider effort provision before a first breakthrough

to make exposition more clear Finally I define efficiency loss (EL) as

EL =

983125infin0(uC

t minus uNCt )dt983125infin

0uCt dt

17This is almost direct from Bonatti and Horner (2011)18Clearly for t isin [0 t] the only possible deviation is to provide less aggregate effort Also if

macrp exists

a deviated agent that has provided too much effort until t with beliefs pt lemacrp will not provide any effort

again

15

where ut = uit + ujt and the superscripts indicate cooperative (C) and non-cooperative

(NC) solutions19

71 Effect of externality

The externality γ has a direct effect on the cooperative solution since

partΠSC(pt)partγ gt 0 is easy to note that the integrand of (9) increases and cooperative agents

are willing to experiment for more time Nevertheless the effect on the non-cooperative

equilibrium effort is not clear On one hand since partΠDE(pt)partγ gt 0 there is an incentive

to provide maximal effort for more time On the other hand partΠS(pt)partγ gt 0 so there is

also more incentives to wait and see

In Figure 3 I show changes in aggregate effort in response to changes in γ As

before parameters are normalized such that π = 1 Therefore γ must be interpreted as

a proportion of π (ie γ = 01 implies that the externality represents a 10 of π)

0 02 04 06 08 1Externality

0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)

Figure 3 Sensibility of cooperative (red) and non-cooperative (blue) equilibrium toexternality Parameters (π cλ r pa) = (1 01 025 02 04)

From Figure 3 is clear that both non-cooperative and cooperative aggregate effort

are increasing in γ while efficiency loss is almost decreasing in γ (for low levels of γ it

is not necessarily decreasing) This mean that in the non cooperative equilibrium the

incentives to experiment more today dominates the effect of wait and see Also as γ

increases agents exert interior equilibrium effort for more time (see Figure 4) That is

despite that wait and see incentives makes agents to exert inefficient levels of effort they

want to keep experimentation because of higher payoffs

19All codes to replicate results for any set of parameters are available in Appendix D

16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1

Figure 4 Examples of individual cooperative (red) and non-cooperative (blue) equilibriumeffort for different levels of externality Parameters (π cλ r pa) = (1 01 025 02 04)

72 Effect of patience

The effect of patience r on the cooperative equilibrium is direct The more impatient

the agents are the less valuable is a first breakthrough partΠSC(pt)partr lt 0 so they are

willing to experiment for less time As before effect of patience on the non-cooperative

equilibrium is not clear On one hand if agents are more impatient they value less the

expected discounted externality partΠDE(pt)partr lt 0 which reduce incentives to experiment

today On the other hand if agents are more impatient is less profitable to wait and see

partΠS(pt)partr lt 0 since they want to have a breakthrough as soon as possible In Figure 5

17

we show changes in aggregate effort in response to changes in r20

0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)

Figure 5 Sensibility of cooperative (red) and non-cooperative (blue) equilibrium topatience Parameters (π γ cλ pa) = (1 02 01 025 04)

As discussed above aggregate effort is decreasing in r in the cooperative solution

But the response of non-cooperative equilibrium aggregate effort is not trivial Figure 6

show some numerical examples that makes easier to understand the following argument

When agents are too patient r rarr 0 (see panel (a) of Figure 6) the non-cooperative

solution tends to a bang-bang solution which makes aggregate effort very low This

is because wait and see incentives becomes too strong very soon so once the expected

discounted externality is low enough they just stop experimentation Then if agents are

a little impatient (see panel (b)) they still value the discounted externality enough but

wait and see incentives becomes weaker so agents provide interior levels of effort for a

while making aggregate effort to increase Nevertheless if agents are too impatient (see

panel (c) and (d)) they have too little incentives to experiment today (ΠDE is too low) so

the aggregate effort start to decrease again

Note that efficiency losses have almost a U-shaped form That means that an

intermediate level of impatience makes the non-cooperative equilibrium more efficient

This goes against a ldquoFolk theoremrdquo intuition which is is not surprising since a well known

result in repeated games literature is that when we restrict to strong symmetric equilibria

(in this case MPBE) it is not possible to obtain a Folk Theorem This is because agents

can not condition their strategies on any history or in other words the set of possible

punishments is very coarse (See Mailath and Samuelson 2006 Ch 9) Nevertheless this

20Note that Assumption 2 is violated for some levels of r in this section As commented above violationof that assumption does not change the qualitative characteristics of the equilibrium

18

result is still interesting because it departs from the literature For example Bonatti and

Horner (2011) who studies free-riding in teams in a context of experimentation finds that

aggregate effort is increasing in r Our result could give some insights about the non

trivial role of patience when we address the coordination problem in an experimentation

setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1

Figure 6 Examples of individual cooperative (red) and non-cooperative (blue) equilibriumeffort for different levels of patience Parameters (π γ cλ r pa) = (1 02 01 025 04)

8 Discussion

My simple model provide some insights about the problem of coordination in a

dynamic strategic experimentation context Nevertheless in order to provide a simple

model I made some assumptions that could be guiding the results In this section I

19

discuss how results could vary if some key assumptions are relaxed

Uniqueness and payoff functional form I assumed that the independent payoff

(π) and the externality (γ or ΠDE) are additively separable This generates a dynamic

effect that could be crucial to obtain a unique equilibrium In absence of a breakthrough

complementarity comes from the fact that if agent i has a breakthrough agent j will

follow the investment rule described in Corollary 1 this can be understood as ldquoex-postrdquo

complementarities At the same time strategies are not perfect complements The force

that captures strategic substitutability is the wait and see incentive Actually the interior

part of the equilibrium is shaped by the interaction of this two forces agents want to wait

and see while the other agent experiment (substitutability) but they need to provide some

effort to keep the other agent experimenting (complementarities)

Therefore uniqueness in my model could exist because complementarities are not

too strong Then more general payoff functional forms could yield to multiplicity of

equilibria For example imagine that the externalities come as synergies such that the

payoff is π times γ if both are successful and zero otherwise This makes complementarities

stronger because now agents receive a positive payoff iff both are successful Nevertheless

wait and see incentives does not disappear They might be even stronger because

experimentation is more risky since receiving a positive payoff is less likely to happen

so the costs of experimentation are relatively higher Is not clear if there exist a unique

symmetric equilibrium with such payoff functions but the insights provided by our model

make me think that uniqueness depends on a balance between complementarities and

substitutability forces Furthermore insights from coordination games literature suggest

that the uniqueness in settings with stronger complementarities will be possible to obtain

only for low enough γi (Bergemann and Morris 2012) The general payoff functions that

make possible to obtain a unique symmetric equilibrium is an interesting research agenda

More agents My model is restricted to two agents since the non linear structure

of sub-gamesrsquo payoffs makes difficult to obtain analytical solutions for the non-cooperative

equilibrium Therefore is very hard to characterize the sub-games for games with more

than two agents21 Despite that is possible to make simulations for equilibrium with more

agents is very hard to obtain uniqueness or sufficiency results which are the interesting

results

Is not clear what force becomes stronger with the inclusion of more agents On

one hand the expected externality is higher so incentives to experiment today should

be stronger On the other hand incentives to wait and see could become stronger too

because each agent is not the only that incentivizes the other to experiment Therefore

as N rarr infin we could find more efficient or inefficient equilibria To study under what

conditions non-cooperative equilibria becomes more (or less) efficient as there are more

21Note that in a model with three agents the non-cooperative equilibrium of my model is the sub-gamethat two unsuccessful agents play after a first breakthrough

20

agents is a relevant question not only from a theoretical perspective but also from a policy

perspective

Commitment We have focussed on two extreme solutions A cooperative solution

where agents have perfect commitment and the non-cooperative equilibrium where agents

canrsquot commit to anything Is interesting to study efficiency gains from intermediate levels

of commitment For example to commit to deadlines or wages

Following Bonatti and Horner (2011) is possible that deadlines could make

experimentation more efficient by compromising to a deadline agents could exert the

interior levels of effort in a more efficient way (ie provide maximal effort for less time)

Also commitment to wages is an interesting mechanism science is not direct from Bonatti

and Horner (2011) Note that a successful agent has willingness to pay to the other agent

to exert effort for more time than T S Hypothetically at T S the unsuccessful agent could

offer to the successful agent to exert more effort for a time interval [T S T prime] in exchange

for the total value of the expected discounted externality that is generated by that effort

profile with initial prior pjT prime Since the successful agent is risk neutral he would be

indifferent between accepting the offer or rejecting it This intuition suggest that if agents

could commit to wages they could reach more efficient levels of experimentation

9 Concluding remarks

I considered a simple model of strategic experimentation with externalities on risky

armrsquos payoffs and unobservable actions I have shown that that the combination of a

dynamic structure and learning lead to a unique symmetric MPBE such that agents

coordinate their experimentation

In equilibrium agents coordinate in a way that they exert too little effort in a

dynamically inefficient way relative to the first best The magnitude of efficiency losses

is mainly defined by the common prior about projectsrsquo quality the lower the prior the

higher the efficiency losses Even more for low enough priors agents exert zero effort

even if it is socially efficient to provide maximal effort Also efficiency losses are (almost)

decreasing in the level of the externality and are minimized for interior levels of patience

The main message of this work is that the dynamic structure and the learning

component in experimentation coordination games could lead to equilibrium uniqueness

making unnecessary to choose an equilibrium refinement criteria My model has many

limitations it is only a first step in order to understand the problem of experimentation

coordination Nevertheless it provide several opportunities for future research An

interesting research agenda is to study if the main results hold for more general payoff

functions or for an arbitrary number of agents

21

References

Bergemann D and Morris S (2012) Robust mechanism design The role of private

information and higher order beliefs volume 2 World Scientific

Bessen J and Maskin E (2009) Sequential innovation patents and imitation The

RAND Journal of Economics 40(4)611ndash635

Bonatti A and Horner J (2011) Collaborating American Economic Review

101(2)632ndash663

Chen Y (2012) Innovation frequency of durable complementary goods Journal of

Mathematical Economics 48(6)407ndash421

Georgiadis G (2015) Projects and team dynamics Review of Economic Studies

82(1)187

Griliches Z (1992) The Search for RampD Spillovers The Scandinavian Journal of

Economics pages S29ndashS47

Heidhues P Rady S and Strack P (2015) Strategic experimentation with private

payoffs Journal of Economic Theory 159531ndash551

Kamihigashi T (2001) Necessity of transversality conditions for infinite horizon

problems Econometrica 69(4)995ndash1012

Keller G Rady S and Cripps M (2005) Strategic experimentation with exponential

bandits Econometrica 73(1)39ndash68

Klein N and Rady S (2011) Negatively correlated bandits Review of Economic Studies

page rdq025

Mailath G J and Samuelson L (2006) Repeated games and reputations long-run

relationships Oxford university press

Malerba F Mancusi M L and Montobbio F (2013) Innovation international

RampD spillovers and the sectoral heterogeneity of knowledge flows Review of World

Economics 149(4)697ndash722

Moroni S (2015) Experimentation in organizations Manuscript Department of

Economics University of Pittsburgh

Park J (2004) International and intersectoral RampD spillovers in the OECD and East

Asian economies Economic Inquiry 42(4)739ndash757

Salish M (2015) Innovation intersectoral RampD spillovers and intellectual property

rights Working paper

22

Seierstad A and Sydsaeter K (1986) Optimal control theory with economic applications

Elsevier North-Holland Inc

Shan Y (2017) Optimal contracts for research agents The RAND Journal of Economics

48(1)94ndash124

Steurs G (1995) Inter-industry RampD spillovers what difference do they make

International Journal of Industrial Organization 13(2)249ndash276

23

Appendix

A Calculating continuation payoffs

A1 Expression for ΠS(pjt)

I am going to derive an expression for

ΠS = maxuiτ

983133 infin

t


983125 τt (pisλuis+r)dsdτ


First note that using pτ with a little abuse of notation

eminus983125 τ primeτ psλusds = eminus

983125 pτ primepτ

dp(1minusps) =

1minus pτ1minus pτ prime

using Corollary 1 uτ = 1 until T S and 0 otherwise Wlg I change the integration

interval from [tinfin] to [0 T S] with pi0 = pit I can rewrite the objective function as

c

983133 TS

0

(piτλπ + γ

cminus 1)

1minus pit1minus piτ

eminusrτdτ

Note that


=1minus pit

1minus pitpit+(1minuspit)eλτ

= piteminusλτ + (1minus pit)

Then

24

c983125 TS

0(pite

minusλτλπ+γc

minus piteminusλτ minus (1minus pit))e

minusrτdτ

hArr c983125 TS

0(pite

minusλτ (λ(π+γ)c

minus 1)minus (1minus pit))eminusrτdτ

hArr c

983069pit

983059λ(π+γ)minusc

c

983060 983147minus eminus(λ+r)τ

λ+r

983148TS

0minus (1minus pit)

983147minus eminusrτ

r

983148TS

0

983070

hArr c983153

pitλ+r


c

983060 9831471minus eminus(λ+r)TS

983148minus (1minuspit)

r

9831471minus eminusrTS

983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084

Define β = λ(π+γ)minuscc

Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084

A2 Expression for ΠDE(pjt)

Now I am going to find an expression for

ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus

983125 τt (pjsλujs+r)dsdτ

using the same techniques and notation as before

25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r

(1minus eminus(λ+r)TS)

hArr pjtγ

1+ rλ

9830631minus βminus(1+ r

λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064

Define γ = γ(1 + rλ) Finally

ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078

B Proof of Proposition 1

The cooperative problem is to maximize

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to

pmt = minuspmt(1minus pmt)λumtdt

pm0 = p

for m = i j Where

ΠSP (pmt) = maxumτ

983133 infin

t



is the expected value of experimentation in one of the projects after a breakthrough

in the other project at some arbitrary t Define ω = λ(π+2γ)minuscc

from Lemma 1 is direct

that (ignoring the individual subscript) the cooperative solution of ΠSP is to experiment

according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064

is direct that T SP gt T S for any t that is after a breakthrough the cooperative

solution involves strictly more effort than the non-cooperative With the same technics

used in Appendix A we can rewrite

ΠSP (pt) = c

983083pt

λ+ r

983077ω minus ωminus r

λ

983061pt

1minus pt

983062minus(1+ rλ)983078minus (1minus pt)

r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084

Now I compute the cooperative solution in absence of a breakthrough Since both

agents are symmetric the solution is to set uit = ujt = 1 while

ptλ983043π + ΠSP (pt)

983044minus c ge 0

Since p lt 0 and partΠSPpartpt gt 0 existTC such that the equation is satisfied with

equality and is optimal to stop experimenting in both projects Is not possible to obtain

an analytical solution for TC but the computational solution is trivial

Finally since ΠSP gt 0 it follows that TC gt TA

C Proof of Proposition 2

This proof relies on the Pontryaginrsquos principle and was inspired by the proof of

Theorem 1 of Bonatti and Horner (2011)

C1 Preliminaries

Since agentrsquos objective function (8) is complex I am going to simplify the problem

before solving it First note that I can rewrite expminus983125 t

0(pisλuis + pjsλujs + r)ds =

(1minusp)2

(1minuspit)(1minuspjt)eminusrt (See Appendix A)

27

On the other hand since pt = minuspt(1 minus pt)λut I have ptλut = minuspt(1 minus pt) and

ut = minusptpt(1minus pt)λ I can rewrite (8) as

983133 infin

0

983069minuspit

(1minus pit)(π + ΠDE(pjt)) +

pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2

(1minus pit)(1minus pjt)eminusrtdt

subject to


pm0 = p

for m isin i j Now I am going to work with some terms of the objective function in

order to make the optimal control problem easier we are going to ignore the constants

Consider the first term

(1minus p)2983133 infin

0

minuspit(1minus pit)2

π

(1minus pjt)eminusrtdt

integrating by parts


0


π

(1minus pjt)eminusrtdt = C minus (1minus p)2

983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt

Now I am going to work with the second term


0


ΠDE(pjt)

(1minus pjt)eminusrtdt =


0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt

983062minus(1+rλ)983078eminusrtdt



0


ΠDE(pjt)


C minus (1minus p)2983133 infin

0

γ

(1minus pit)

983077pjt

1minus pjt(λujt + r) + βminus(1+ r

λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt

Now I focus on the term


0

pitpit(1minus pit)2

c

λ(1minus pjt)eminusrtdt

28



0

pitpit(1minus pit)2

c

λ(1minus pjt)eminusrtdt =

C + (1minus p)2983133 infin

0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt

Now I can rewrite the objective function (ignoring constants)

983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r

983078983084eminusrtdt

I make the further change of variables xmt = ln((1minuspmt)pmt) and rearrange someterms Agent irsquos problem

983133 infin

0

983083minus (1 + eminusxit)

983147983043r(1 + eminusxjt) + eminusxjtλujt

983044 983059π minus c

λ

983060+ γ

983059eminusxjt(λujt + r) + e

rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit


983044+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to

xmt = λumt xmo = ln((1minus p)p)

given an arbitrary function ujt The hamiltonian of this problem is

H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084eminusrt + ηitλuit

wlg I am going to work with the current value hamiltonian (See Seierstad and

29

Sydsaeter 1986 Ch24 Exercise 6)22 Define ηit = ertηit Then

Hc =η0


983147 983043r(1 + eminusxjt) + eminusxjtλujt

983044 983059π minus c

λ

983060

+ γ983059eminusxjt(λujt + r) + e

rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)

Note that the change of variables has two advantages First it makes easier to work with

the hamiltonian because the control variable uit is no longer in the objective function

and the state variable is no longer in the evolution of the state variable xit = λuit On

the other hand note that the evolution of the state variable can be interpreted as the

marginal effect of instant effort in the function of beliefs This gives a straightforward

interpretation for many of the optimality conditions that will arise later

C2 Necessary conditions

Is direct that that this is not an abnormal problem (See Seierstad and Sydsaeter

1986 Ch24 Note 5) so η0 = 1 By Pontryaginrsquos principle there must exist a continuous

function ηmt for all m isin i j such that

(i) Maximum principle For each t ge 0 uit maximizes Hc hArr ηitλ = 0

(ii) Evolution of co-state variable The function ηit satisfies

ηit = rηit minuspartHc

partxit

(iii) Transversality condition If xlowast is the optimal trajectory limtrarrinfin ηit(xlowastt minusxt) le 0 for

all feasible trajectories xt (Kamihigashi 2001)23

C3 Candidate (interior) equilibrium

From (i) since λ gt 0 the candidate of equilibrium satisfies ηit = 0 for all t ge 0

Then condition (ii) becomes

minuspartHc

partxit

= 0 (12)

22All references to Seierstad and Sydsaeter (1986) Ch 2 apply to finite horizon problems Since allextensions to infinite horizon are straightforward and standard in the literature I am going to omit them

23This is the same transversality condition used by Bonatti and Horner (2011)

30

minus partHc

partxit= minuseminusxit


983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ


983044minus eminusxjtλujtc

983063e

rλxit

βminus rλ

λ+ rminus eminusxit

β

λ+ r

983064

Since I am interested in the symmetric equilibrium ujt = uit and xit = xjt

0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit

983061π + γ minus c

λminus cβ

λ+ r

983062983064minus eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064

minuseminus(1minus rλ)xit

983063γβminus(1+ r

λ) r

λ(λuit minus λ) + λuitc

βminus rλ

λ+ r

983064+

cr

λ

If the solution is interior the following must hold

partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)

the intuition is that the agent derives disutility from changing the state variable with full

effort (xit = λ) Therefore the upper bound of effort is not binding That is both agents

can reach a level of indifference between exerting effort in an interval [0 dt) and [dt 2dt)

with an interior level of effort By continuity of the exponential function is easy to see

that condition (13) can not be satisfied for a high enough p (low enough x0) we address

this issue later

Now I characterize the interior equilibrium Since λuit = xit the optimal belief

function xlowastt is characterized by the following non-linear ODE


λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)

this differential equation has not analytical solution Therefore I solve it by

numerical methods (See Appendix D)

C4 Corner solutions

As discussed before condition (13) is not satisfied if p is high enough In order tomake this more clear consider

partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064

+eminus(1minus rλ )xit


λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0

983063r(π + γ minus c

λ) +

983061r(λπ minus c)

λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064

+eminus(1minusrλ)xi0

983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31

note that x isin (minusinfininfin) with limprarr1 x = minusinfin and partxpartp lt 0 Letrsquos consider a

p gt 12 which implies xit lt 0 Also if Assumption 1 is satisfied the first and third RHS

terms in brackets of (16) are positive If Assumption 2 is satisfied the arguments of the

three exponential functions are positive This implies that RHS is strictly positive and

an interior equilibrium is not possible24

If p lt 12 then x gt 0 and the value of the RHS is not clear Note that as x

increases the value of the exponential functions decreases Then since there is a cost

minuscrλ by continuity of the exponential function existp such that for every pt lt p (13) is

not binding Is not possible to obtain a closed form for p but is easy to find numerically

using the implicit function (16)

Therefore given π γ c l r if p is high enough condition (13) dos not hold and we

have a corner solution uit = 1 (See Appendix D for codes that allows a numerical analysis)

This implies that in the symmetric equilibrium for p ge p agents exert maximal effort

until a time t at which belief p is reached and for every pt lt p they exert interior levels

of effort that satisfies the non-linear ODE (14)

Finally as in the interior equilibrium agents exert a strictly positive amount of

effort is possible that a beliefmacrp is reached at which

partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)

and the solution is to exert zero effort for all t gtmacrt (in the numerical analysis is not possible

to find a set of parameters for whichmacrp does not exists see Appendix D)

C5 Uniqueness

Despite I can not find a closed form for our optimal path candidate xlowastit I can

proof some characteristics of it that makes easier to proof uniqueness of our symmetric

equilibrium candidate

Effort is (weakly) decreasing in time Assume for the sake of contradiction

that for an arbitrary t uit lt uit+dt If at t condition (13) does not hold the solution is

uit = 1 and by the upper bound of u ut+dt gt ut is a contradiction On the other hand if

condition (13) holds then partHCpartxit = 0 for an interior uit But partHcpartxit is decreasing

in x and xit gt 0 implies xit lt xit+dt Therefore if uit+dt ge uit then partHcpartxit+dt lt 0

which is a contradiction Note that for all t gt t effort must be strictly decreasing

Since effort is weakly decreasing xlowastit is also weakly decreasing (strictly decreasing for

24Assumption 1 does not guarantee that the RHS second term in brackets is positive but even if itsnegative the effect is dominated by the others

32

all t gt t) Then there are two possibilities xlowastit converges and agents exert effort forever

or existmacrp and xit reaches a value less than infinity (pt gt 0) where agents stop experimentation

We now use the trajectory (ηlowastjt ηlowastit x

lowastjt x

lowastit) and the necessary conditions to

eliminate other admissible trajectories Note that the optimal trajectories are

ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t

= 0 if t isin (macrt t)

lt 0 if t gemacrt

Clearly ifmacrt does not exists ηt = 0 for all t ge t

WhilepartHc(ujtxitxjt)

partxit

983055983055983055u=1

ge 0 any strategy uit ∕= 1 violates necessary condition (i)

Therefore I need to prove that my symmetric equilibrium candidate is unique for pt le p

bull Consider paths that start with ηit lt 0 for t ge t From necessary condition (i)

uit = 0 Condition (ii) implies that ηit lt 0 Clearly xit = xit for all t ge t Since

in our reference path xlowast is greater than xt in the limit and ηit rarr minusinfin It follows

that this path violates the transversality condition

bull Now consider paths that start with ηit gt 0 By an analogous argument uit = 1

and ηit gt 0 Then xit gt xlowastit for all t gt t and ηit rarr +infin Therefore this path

violates the transversality condition

Since I am restricting the analysis to symmetric equilibrium the above arguments

eliminate any other possible trajectory In particular any trajectory that differ from

the equilibrium candidate must start with a co-state variable gt 0 or lt 0 then the

same arguments starting at tprime eliminate the alternative candidate path Therefore the

equilibrium candidate is the only symmetric equilibrium of the game

C6 Sufficiency

From (15) the maximized hamiltonian is concave in xt because if part2Hcpartx2it gt 0

then partHcpartxit lt 0 for all u isin (0 1] but values of xt that makes that possible are not

reached in the symmetric equilibrium

Therefore from the Arrow sufficiency theorem necessary conditions are also

sufficient (Seierstad and Sydsaeter 1986 Ch 3 Theorem 17)

33

D MATLAB codes

All numerical exercises presented on this work can be replicated for any combination

of parameters with the MATLAB codes available here The readmepdf document explains

the function and how to use each mfile

34

1 Introduction

There are many industries where the profitability of a firmrsquos innovation depends on

other firmsrsquo innovations and vice versa For example a new and more powerful computer

hardware would be more valued by customers if new and more demanding PC software

is created but that potential software could be used only with more powerful hardware

Also decisions on how much to invest to create a new product should consider potential

innovations of intermediate goods producers that might lower their production costs as

well as the investment in RampD to reduce production costs should consider that new

products might boost intermediate goods demand in the future For example while

many tech-companies were investing in RampD to reduce the size of lithium batteries for

smartphones and other devices the mining industry was investing in RampD to reduce

lithiumrsquos production costs1

The consideration of possible future innovations by other firms in the RampD decisions

of a firm can be understood as an externality on the experimentation payoffs This

means that the success of a firmrsquos RampD increments net payoff of other firmrsquos RampD If

firms invest more when there are externalities than when there are none then they are

coordinating their experimentation The necessary conditions that lead to the coordination

of experimentation could be crucial to understanding the success of some innovation-based

industries as well as the failure (or non-existence) of others

This work provides a simple model of experimentation to understand how

externalities on experimentation payoffs affect agentsrsquo behavior I model experimentation

as a variation of the continuous-time two-armed exponential bandit model of Keller et al

(2005) In particular I follow the main variations used by Bonatti and Horner (2011)

I consider the strategic behavior of experimenting agents when externalities on the

payoffs are exogenous and actions are unobservable The goal is to determine how payoff

externalities change the dynamics of investment provision An example of this is when

firms have to decide whether to dedicate their resources to selling their usual products or

investing in RampD to try to develop a new product that will have a greater market value

(or lower production cost) if other firmsrsquo RampD is also successful I model this situation

as one where two agents have to decide whether to allocate resources to experiment on

a risky project (risky arm) of unknown quality or into a safe one (safe arm) that pays

a fixed amount each period I consider the safe arm as the experimentation cost of

opportunity The risky arm can be bad and yield zero profits or good and yield a profit

Each agentrsquos risky arm quality is independent from the others and it can be learned only

by experimentation The strategic component arises because if the risky arm is good it

1Although the motivation of this work is theoretical it is worth mentioning that identification andmeasurement of RampD spillovers have been an active line of empirical research in the last decades Seefor example Griliches (1992) Malerba et al (2013) Steurs (1995) and Park (2004)

2






















criterion












provide

3

















risky arms



















4






3 Model

















r gt 0



r


983133 t

0

eminusrsuisds

983062



r


983133 t

0

eminusrsuisds

983062


5














pit =peminusλ

983125 t0 uiτdτ


(1)


equation

pit =dpitdt




both projects





externality




τuitds

6

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34






















criterion












provide

3

















risky arms



















4






3 Model

















r gt 0



r


983133 t

0

eminusrsuisds

983062



r


983133 t

0

eminusrsuisds

983062


5














pit =peminusλ

983125 t0 uiτdτ


(1)


equation

pit =dpitdt




both projects





externality




τuitds

6

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

















risky arms



















4






3 Model

















r gt 0



r


983133 t

0

eminusrsuisds

983062



r


983133 t

0

eminusrsuisds

983062


5














pit =peminusλ

983125 t0 uiτdτ


(1)


equation

pit =dpitdt




both projects





externality




τuitds

6

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34






3 Model

















r gt 0



r


983133 t

0

eminusrsuisds

983062



r


983133 t

0

eminusrsuisds

983062


5














pit =peminusλ

983125 t0 uiτdτ


(1)


equation

pit =dpitdt




both projects





externality




τuitds

6

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34














pit =peminusλ

983125 t0 uiτdτ


(1)


equation

pit =dpitdt




both projects





externality




τuitds

6

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

41 Autarky




ΠA(p) = maxut

983133 infin

0



subject to










uAt =

983099983103

9831011 if t le TA

0 if t gt TA

where

TA = λminus1

983063ln

983061λπ minus c

c

983062minus ln

9830611minus p

p

983062983064(3)









7


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


pa =c

λπ (4)









arbitrary time t

ΠS = maxuiτ

983133 infin

t







the effort profile

uSiτ =

983099983103

9831011 if τ le T S + t

0 if τ gt T S + t

where

T S = λminus1

983063ln


c

983062minus ln

9830611minus pitpit

983062983064 (6)





8






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34






ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084






ΠS(pjt)


983133 infin

t

pjτλujτeminus





ΠDE(pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



ΠEi = max

uit

983133 infin

0

983069983063pitλ

983043π + ΠDE(pjt)

983044minus c


S(pit)

983070eminus



9

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

subject to














W (p) =983131

I

ΠE =

983133 infin

0

983069983063pitλ


983044minus c

983064uit

+

983063pjtλ


983044minus c

983064ujt

983070eminus


subject to



function as

W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSC(pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSC(pit)

983044minus c

983064ujt

983070eminus


(9)

where

ΠSC(pmt) = maxumτ

983133 infin

t




10









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34









uCit =

983099983103

9831011 if t le TC

0 if t gt TC






uSCτ =

983099983103

9831011 if τ le T SC + t

0 if τ gt T SC + t

where

T SC = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064



set of parameters








11

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

















lowastit)



according to

ulowastit =

983099983103

9831011 if pt gt p

xitλ if pt le p





λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(10)





12


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


macrt











macrt at which pt =

macrp For




of parameters

0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit








13









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34









0 05 1 15 2 25 3 35 4Time t

0

01

02

03

04

05

06

07

08

09

1

Effo

rt u it

auta

rky

(a) Effort

0 05 1 15 2 25 3 35 4Time t

0

005

01

015

02

025

03

035

04

045

05

Belie

f pit















983125infin0


14
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34
































EL =

983125infin0(uC


0uCt dt


macrp exists


again

15


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


(NC) solutions19












0

05

1

15

2

25

3

35

4

45

5

Agg

regat

eEor

t

(a)


05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)










16

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) γ = 005

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) γ = 025

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) γ = 075

0 05 1 15 2 25 3Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) γ = 1










17


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


0 02 04 06 08 1Patience r

01

02

03

04

05

06

07

08

09

1

11

Aggre

gate

Eort

(a)

0 02 04 06 08 1Patience r

05

055

06

065

07

075

08

085

09

095

1

E

cien

cyLoss

(b)






















18





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34





setting

0 01 02 03 04 05 06Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(a) r = 001

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(b) r = 025

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(c) r = 075

0 01 02 03 04 05Time t

0

01

02

03

04

05

06

07

08

09

1

Eort

uit

(d) r = 1


8 Discussion




19































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34































results








20


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


perspective


































21

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

References






101(2)632ndash663




82(1)187










page rdq025












22




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34




48(1)94ndash124



23

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

Appendix




ΠS = maxuiτ

983133 infin

t






983125 pτ primepτ

dp(1minusps) =




c

983133 TS

0

(piτλπ + γ

cminus 1)


eminusrτdτ

Note that


=1minus pit



Then

24

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

c983125 TS

0(pite

minusλτλπ+γc


minusrτdτ

hArr c983125 TS

0(pite



hArr c

983069pit


c


λ+r

983148TS

0minus (1minus pit)


r

983148TS

0

983070

hArr c983153

pitλ+r


c



r


983148983154

Note that

eminusaTS

=


c

pit1minus pit

983062minusaλ

then

c

983083pit

λ+ r

λ(π + γ)minus c

c

9830771minus


c

pit1minus pit

983062minus(1+ rλ)983078

minus(1minus pit)

r

9830771minus


c

pit1minus pit

983062minus rλ

983078983084


Finally

ΠS(pit) = c

983083pit

λ+ r


λ

983061pit

1minus pit


r

9830771minus

983061β

pit1minus pit

983062minus rλ

983078983084



ΠDEi (pjt) = γ

983133 infin

t

pjτλujτeminus



25

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

λγ983125 TS

0pjte

minus(λ+r)τdτ

hArr λγpjtλ+r


hArr pjtγ

1+ rλ


λ)983059

pjt1minuspjt

983060minus(1+ rλ)983064


ΠDEi (pjt) = pjtγ

9830771minus

983061β

pjt1minus pjt

983062minus(1+ rλ)983078



W (p) =

983133 infin

0

983069983063pitλ

983043π + ΠSP (pjt)

983044minus c

983064uit

+

983063pjtλ

983043π + ΠSP (pit)

983044minus c

983064ujt

983070eminus


Subject to


pm0 = p

for m = i j Where


983133 infin

t







according

26

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

uSPτ =

983099983103

9831011 if τ le T SP + t

0 if τ gt T SP + t

Where

T SP = λminus1

983063ln


c

983062minus ln

9830611minus ptpt

983062983064




ΠSP (pt) = c

983083pt

λ+ r


λ

983061pt

1minus pt


r

9830771minus

983061ω

pt1minus pt

983062minus rλ

983078983084




983044minus c ge 0








C1 Preliminaries




(1minusp)2


27



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34



983133 infin

0

983069minuspit


pitpit(1minus pit)

c

λ+ pjtλujtΠ

S(pit)

983070(1minus p)2


subject to


pm0 = p





0


π




0


π


983133 infin

0

π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064eminusrt



0


ΠDE(pjt)



0


pjt(1minus pjt)

γ

9830771minus

983061β

pjt1minus pjt




0


ΠDE(pjt)



0

γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078eminusrtdt



0

pitpit(1minus pit)2

c


28



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34



0

pitpit(1minus pit)2

c



0

c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064eminusrt


983133 infin

0

983083minus π

(1minus pit)

983063r

(1minus pjt)+ λujt

pjt(1minus pjt)

983064

minus γ

(1minus pit)

983077pjt


λ)

983061pjt

1minus pjt

983062minus rλ r

λ(λujt minus λ)

983078

+c

λ

9830631

(1minus pit)+ ln(

pit1minus pit

)

983064 983063r

1minus pjt+

pjt1minus pjt

λujt

983064

+pjt

1minus pjtλujtc

983077pit

1minus pit

β

λ+ r+

983061β

pit1minus pit

983062minusrλ 9830611

rminus 1

λ+ r

983062minus 1

r



983133 infin

0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r


Subject to



H = η0



983044 983059π minus c

λ

983060+ γ


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

minus c

λxit



983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r



29


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


Hc =η0



983044 983059π minus c

λ

983060


rλ βminus(1+ r

λ ) r

λ(λujt minus λ)

983060 983148minus c

λxit


983044

+ eminusxjtλujtc

983063eminusxit

β

λ+ r+ e

rλxitβminus r

λ

983061λ

r(λ+ r)

983062minus 1

r

983064983084+ ηitλuit

(11)














partxit






minuspartHc

partxit

= 0 (12)



30

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

minus partHc



983044 983059π minus c

λ

983060+ γ


rλxjtβminus(1+ r

λ ) r

λ(λujt minus λ)

983060983148

+c

λ



983063e

rλxit

βminus rλ


β

λ+ r

983064


0 = minuseminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r


983063r(π minus 2c

λ)minus c

λλuit

983064



λ) r


βminus rλ

λ+ r

983064+

cr

λ


partHc(uit xit)

partxit

983055983055983055u=1t=0

lt 0 (13)






this issue later




λ) + eminusxit

983043π minus 2c

λ


λ)xitβminus(1+ r

λ)γ minus c

λ

983046


λminus cβ

λ+r


λ)xit


λ) + cβminus r

λ

λ+r

983060+ eminusxit c

λ

(14)



C4 Corner solutions


partHc

xit= eminus2xit

983063r983059π + γ minus c

λ

983060+ λuit


λminus cβ

λ+ r

983062983064+ eminusxit

983063r(π minus 2c

λ)minus c

λλuit

983064



λ ) r


βminus rλ

λ+ r

983064minus cr

λ

(15)

rArr partHc

partxit

983055983055983055u=1t=0

= eminus2xi0


λ) +


λ+ r

983062983064+ eminusxi0

983063r

983061π minus 2c

λ

983062minus c

983064


983077λc

λ+ r

983061c

λ(π + γ)minus c

983062 rλ

983078minus cr

λ

(16)

31


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34


















partHc(ujt xit xjt)

partxit

983055983055983055uisin(01]

lt 0 (17)



C5 Uniqueness












32




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34




lowastjt x



ηlowastt =

983099983105983105983105983103

983105983105983105983101

gt 0 if t le t


lt 0 if t gemacrt



partxit

983055983055983055u=1















C6 Sufficiency






33

D MATLAB codes




34

D MATLAB codes




34

Documents

Coordinating experimentation - Fernando Ochoa