Final Aggregate Inventory(1)

8/7/2019 Final Aggregate Inventory(1)

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APRESENTATION

ON

Aggregate Inventory Management&

Distribution inventory management

:Presented ByPriyanka Gupta( )2010PMM124Anupama Kumari( )2009PMM119


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Inventory & Flow of Materials

vRaw materialsvWork-in-process (WIP)vRaw and in-process (RIP)

vFinished goodsvDistribution inventoriesvMaintenance, repair, & operational supplies(MROs)

v


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Inventory Objectives

Inventories must be coordinated to meetthree conflicting objectives:

Maximize customer service

Minimize plant operation costs

Minimize inventory investment


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Inventory Costs

Inventory management costs

Item costs

Carrying costsOrdering costs

Stock out costs

Capacity-related costs


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Functions of Inventories

Meet anticipated demandSmooth production requirements

Decouple components of productiondistribution systemProtect against stock outs (provide customerservice)

Take advantage of order cyclesHedge against price increasesExploit quantity discountsPermit operations


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Aggregate Inventory

ManagementAggregate inventory management (AIM) is

concerned with managing inventories according totheir classifications (raw material, work-in-process, finished goods, etc.) & the function theyperform.

AIM is financially oriented & concerned with costs

& benefits of carrying the classifications ofinventories


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Aggregate Inventory

ManagementAIM involves

Flow & kind of inventory needed

Supply & demand patterns

Functions inventory performs

Objectives of inventory management

Costs associated with inventory


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It is a technique developed for handling

EOQs in a aggregate and dealing with theproblem of constraint of EOQ equation. It providesa means to calculate directly the proper lot size fora family of items to meet some constraints.

Then LIMIT order quantities are calculatedwith the help of formulas A & B

LOT SIZE INVENTORYMANAGEMENT INTERPOLATION

TECHNIQUES (LIMIT)


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FORMULA

LIMIT formula A = IL= IT(HL/HT)2

LIMIT formula B =

M=HT

/HL

=(IT

/IL

)WhereHL=Di*hi/QLi=Total set up hours for present LIMITorderquantities.HT=Di*hi/QTi=Total set up hours for trial order quantities.IT = Inventory carrying cost for trial order quantities.IL = Inventory carrying cost used for LIMIT order quantities.


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Example 1 Suppose that D=10000, S=$125, I=0.25,C=$10, what is the trial lot size? If we were limitedto 5 setup in the year, find M with the help of LIMITformula.

Solution:QT= (2DS/Ic) = (2*10000*125/0.25*10)

=1000Setup in the year= D/Q=10


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limited setup in the year=5D/QL=5QL=2000QL= (2DS/ILc) = (2*10000*125/IL*10)=2000

IL=6.25%Relationship between the trial and the LIMIT lotsizeQL=M*QTWhere M=(IT/IL) 2000=M*1000M=2


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Example-2The two item in our inventory have been managed in a seat ofthe pants fashion for several years. The following table showsthe current situation:

Current setup costs=$62.50 per hourCarrying cost percentage= IT=35%How can this situation be handled using LIMIT?

Item AnnualUsage D

SetupHours per

order h

Unit Costof item c

PresentOrder

QuantityQ

Yearly SetupHours HA 10000 2 10 769 26

B 5000 3 15 1667 9Total 35


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Yearly limited Setup Hours HHP=Di *hi/QPi

For A, HP=(10000*2/769)=26For B, HP=(5000*3/1667)=9Total HP=35hours,

Cost per Setup SFor A=$62.50*2=$125.00For B=$62.50*3=$187.50

Trial lot size QT=(2DS/Ic)For A={(2*10000*125)/(0.35*10)}=845For B={(2*5000*187)/(0.35*15)}=598


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Approximate Yearly Setup Hours HTHT=Di *hi/QTi

For A, HT=(10000*2/845)=24For B, HT=(5000*3/598)=25

M=HT/HL=1.4QL=M*QT=1.4*QT

Item Cost per Setup

S

Trial Q Approximate Yearly Setup

Hours HTA $125.0 845 24B $187.5 598 25Total 49


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This leads us directly to the LIMIT order quantities:

Inventory LIMIT carrying percentageIL= IT(HL/HT)2

=0.35(35/49)2=0.1786

Item LMIT Quantity QL

Approximately Yearly SetupHours HLA 1.4*845=1183 17B 1.4*598=837 18Total 35


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Based on this implied carrying cost

percentage of 17.86%, we calculate the TotalRelevant Cost in the table with the help of theseformulas:-

Annual holding cost= (Q/2)*ILcAnnual ordering cost= (D/Q)S

TRC=annual holding cost+annual ordering cost

TRC=(Q/2)h+(D/Q)S


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Lot size Annual HoldingCost ($)

Annual Setup Cost($)

TotalRelevant

Costs ($)

QP-A 769 687 1625 2312

QP-B 1667 2233 563 2796Total 2920 2188 5108QT-A 845 755 1479 2234QT-B 598 801 1568 2369

Total 1556 3047 4603QL-A 1183 1057 1057 2114QL-B 837 1120 1120 2240Total 2177 2177 4354


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Optimal,given I=35%

Agg

regate

AnnualSetupC

ost($

)

Aggregate Annual Holding Cost ($)

Exchange curve


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19

Lagrange Multipliers

The method of Lagrange multipliers provides astrategy for finding the maxima and minima ofa function subject to constraints.

It gives a set of necessary conditions to identify

optimal points of equality constrained optimizationproblems.

This is done by converting a constrained problemto an equivalent unconstrained problem with thehelp of certain unspecified parameters known asLagrange multipliers.


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consider the optimization problemmaximize f(x,y)subject to g(x,y)=c.the Lagrange function defined byL(x,y,)=f(x,y)+*{g(x,y)-c}

Figure 1: Find x and y to maximize f(x,y) subjectto a constraint (shown in red) g(x,y) = c.

Figure 2: Contour map of Figure 1. The red lineshows the constraint g(x,y) = c. The blue linesare contours of f(x,y). The point where the red

line tangentially touches a blue contour is oursolution.


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Lagrange multiplier technique:-L(x,y,)=f(x,y)+g(x,y)(the Lagrange function)

Lx=fx+gx=0 (1) =-(fx/gx)Ly=fy+gy=0 (2)

=-(fy/gy)L=g(x,y)=0 (3)

eq. (1) & (2) gives the slope condition: fx/gx=fy/gy fx/fy=gx/gy =dx/dyeq. (3) provides satisfaction of the constraints.


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Example-3

Solve the following problemusing the Lagrange multiplier method:Suppose that we have a profit function:-Max f(x,y)=x+3ySubject to g(x,y)=x2+y2-10=0


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SolutionExamine the profit line:

=x+3yIn terms of y y=( /3)-(1/3)x slope=-1/3dy/dx=-1/3

also g(x,y) solve in terms of x

y 2=10-x 2 y=10-x 2

dy/dx=-(x/y)


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equating the slope of the profit line and theconstraint line gives -(1/3)=-(x/y)

y=3x; but y2=10-x2 x=1 and y=3Conclusion:-profit are maximized at (x,y)=(1,3)

with profit of x+3y=10.

Apply the Lagrange multiplier technique:-L(x,y,)=x+3y-(x2+y2-10)Lx=1-2x=0Ly=3-2y=0L=x2+y2-10=0


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solve for : =(1/2x)=(3/2y) y=3xalso y=10-x 2

3x=10-x 2

x=1, y=3

gives profit f(1,3)=1+3*3=10


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Example-4

Applying the Lagrange multiplier techniquein example-3 and specify the minimization from thetotal setup cost & inventory holding costs.

Solution:-for minimize the total setup costs (D1S1/Q1)+(D2S2/Q2)

=(10000*125/Q1)+(5000*187.5/Q2)subject to an inventory investment constraintQ1*(I c1)/2+ Q2*(I c2)/2 =2920


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Q1*0.1786*10/2+ Q2*0.1786*15/2=2920

apply Lagrange functionaL=(D1S1/Q1)+(D2S2/Q2)+[Q1*(I c1)/2+ Q2*(I c2)/2]diff with respect to Q1,Q2 and we obtainL/ Q1= D1S1/Q1^2+( I c1)/2=0

L/ Q2= D2S2/Q2^ 2+( I c2)/2=0L/ =Q1*(I c1)/2+ Q2*(I c2)/2=2920

solve eq we obtain=0.555Q1=1587Q2=1122


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minimum value of inventory holding costs asQ1*h1/2+ Q2*h2/2=(1587*10*0.1786)/2+(1122*15*0.1786)/2=2920

setup cost are

(D1S1/Q1)+(D2S2/Q2)=10000*125/1587+5000*187.50/1122=1624


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Distribution inventory

management Reality customers are not conveniently located

next to the factory. Often inventory must stored in several locations. The main issues are :1. Where to have warehouses and what to stock.2. How to replace stocks, given the answer to the

first issue.


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Multi location inventories


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Multi location systemabsorbent system


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Multi location system

Coalescent systems: have material coming together into one enditem.Series system : have locations feeding each other in a direct path.


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Measures of multi location inventorysystem

Fill rate : fill rate or percent unit service, gives the averagefraction of unit demand satisfied from stock on hand.

Fills: number of unit demanded and satisfied per unit time. Fills =fill rate * demand

Expected number of backorder: is the time weighted averagenumber of backorders outstanding at a stocking location.Including times of zero backorders, this measures dependson fill rate.

Expected number of backorder=

expected delay*demand rate Expected delay : average time necessary to satisfy a unit of

demand .


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Inventory holding cost. Setup costs and ordering costs. Stockout cost. System stability costs : cost associated with

overreaction to changes demand rates.


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Example:-D(annual demand)=1000unitQ(order quantity)=100unit

B(maximum back order)=10unitwhat is the expected number of back orders.

Solution:Average back order position=

B/2=10/2=5unitPercentage time when back orders are

possible =B/Q=10/100= 0.1expected number of back orders=

(B/Q)*(B/2)=0.1*5=0.5unit

Example:


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Example:-D(annual demand)=1000unitQ(order quantity)=100unitB(maximum back order)=50unit , what is the

expected delay or the average time to satisfy a unitdemand. Working days=250

Demand rate(d)=1000/250=4units /

dayExpected delay=[(B/Q)*(B/2)]/d =[(50/100)*(50/2)]/4 =3.125days


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Centralization of inventories

Order decision rules and safety stock rules together todemonstrate their combined pressure to centralizeinventories.

TRC=(DS/Q)+(Qh/2)+h(SS)

D=annual demandS=setup costQ=ordered quantityh=holding cost/unit/year.= standard deviation of lead time.


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k= number of standard deviation of lead timedemand used to determine safety stock.

SS= safety stock.TRC=total relevant cost.SS=kQ=(2DS/h)TRC=(2ShD)+hk


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N=stocking point.Di=annual demand.i=standard deviationThe decentralized relevant cost would be

TRC={(2Sh )} + {hk i}

Assuming that h and k , we could centralize theseinventories at one location. Ignoring transportation costs,the total relevant costs would be

TRC=(2Sh)D+hk where =( ) Di

=

N

i 1

Di =

N

i 1

D =

N

i

iD1

=

N

i 1


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Example: two inventory locations have annual demands and costs shown.S h D k D100 10 1000 1.64 50 31.62100 10 2000 1.64 50 44.72

The decentralized system is given byTRC={(2Sh Di)} + {hk i}TRC ={(2*100*10(1000+2000))} + {10*1.64(50+50)} =5054The centralized system is given byTRC=(2Sh)Di+hk = ( 1 *1 + 2*2 ) =70.7TRC ={(2*100*10(3000))} + {10*1.64(70.7)} =3609

=

N

i 1

=

N

i 1


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Distribution inventory

system


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Level decompositon systems

Set aggregate service level objectives for all items at anechelon.

e.g. The objectives at the main distribution center might

be 95% service, interpreted as a 95% fill rate. Withn items in the inventory, the problem can be statedas follows:

minimize (unit value on the item i)(safety stock on

the item i)Subjected to (item demand rate/aggregate demand

rate)(item fill rate)0.95

=

n

i1

=

n

i 1


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Multiechelon systems

Sometimes called Differentiated distribution oritem decomposition systems, focus oneffective safety stock.

Applied to low demand rate items becausemathematics of system is complex.

O i i ith l


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One origin with severaldestination

Destination

Origin

S l i i i h


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Several origins with onedestination

Origins

Destination

S l i i h ith


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Several origins each withseveral destination

Origins

destination


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Multiple origins and multiple

destinations. When multiple sources shipment to several

destinations, however solving these problem

becomes hopelessly difficult to solve This type of problem becomesmanageable if we assume that allorigins ship their products to asingle consolidated terminal andthat all items are distributed todestination as demanded from the

consolidated terminal.

Several origins with a


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Several origins with aconsolidation terminal to

several destinationOrigins

destination

consolidated terminal

Algorithm for sol ing


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Algorithm for solvingproblems

Formulas:

Qic=

Qcj =wheredijk= quantity of demand from origin i for destination j for product kPk= price/unit of kDi,k= demand at source ifor item k from all destinations j=Sic= cost of load from source i to consolidation terminal c

)(

50)(

j k

ijk

j k

ijkk

ic

d

dp

S

I

dj k

ijk

)/(

50)(

i k i k

ijkijkk

ck

ddp

S

I

di k

ijk

j

ijkd


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Sck = cost of load from consolidation terminal c to destinations kWic=capacity of vehicle from source i to consolidation terminal c

Wck=capacity of vehicle from consolidation terminal c to destinations kTic=lead time/travel time from source i to consolidation terminal cTck=lead time/travel time from consolidation terminal c to destinations kFic= total quantity of items flowing per period from source i to consolidation

terminal =

Fck= total quantity of items flowing per period from consolidation terminal c todestinations k =I = inventory carrying percentage.

The shipping quantity from source i to consolidation terminal c is given by sic

min[Qic,Wic]The shipping quantity from consolidation terminal c to destinations is given by scj min[Qcj,Wcj]Qic =economical shipment quantity from source i to consolidated terminal c.Qcj = economical shipment quantity from consolidated terminal c to destination j.

j k

ijkd

i k

ijkd


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Example: The demand for the products at destination 1 and 2 andsource of these are presented in the table. The capacity ofvehicles, relevant setup costs, lead time between locations, andother pertinent data in tables.Assume that the inventory carrying charges amount to 20% andthat the firm operates fifty periods (weeks) per year.Find the economical quantity to ship from each source to the

terminal and from the terminal each destination.Destination demand and origin capacitydemand / period at

Product Cost per unit Destination1 Destination2 Sourcelocation20 8 4 1

2 25 6 10 13 25 5 8 24 30 6 8 2


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Setup cost, vehicle cost, and lead timedata

Fromto

Source 1terminal

Source 2terminal

Terminaldestination

1

Terminaldestinatio

n 2Setupcost 45 25 30 35

Vehiclecapacity

150 200 150 100

Lead time(days)

4 2 3 4

Solution: economical quantity to ship from each


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Solution: economical quantity to ship from eachsource to the terminal.Step1:calculate the total annual demand for items 1 and 2

formula=[ ] *50[8+6+4+10]*50=1400unit Eq1Step2:calculate the average cost per part at source1.Formula=

[(12*20)+(16*25)]/28=22.86 $ Eq2Step3: Calculate the economical shipment quantity flowing from

source 1 to the consolidated terminal.Qic=[{ *50 }/{I( )}] Eq3

Q1c=[(45*1400)/(0.20*22.86)]=118unitFind the minimum of Q1c and W1c.sic=min[118,150]=118unit

j k

ijkd

j k j k

ijkijkk ddp /

)(j k

ijkic

dS

j k j k

ijkijkk ddp /


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Step5: calculate the quantities of individual items 1 and 2flowing from origin 1 to the consolidated terminal.

Q1c1=( ) =(118*12)/28=51unitQ1c2=(118*16)/28=67unitStep6: repeat the calculation for source 2. following the same

procedure, we obtain the quantities of individual items 3 and 4flowing from 2 to the consolidated terminal.

Q2c=78unit W2c=200 s2c=min[78,200]=78unitQ2c3= (78*13)/27=38unitQ2c4=(78*14)/27=40unit

j k

ijk

j

jc dds/)(

111


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The same procedure can be followed to obtain the shipmentquantities from the consolidation terminal to destination 1 and2, respectively.

Step1: calculate the total annual demand for all items atdestination1.[ ]*50[8+6+5+6]*50=1250unit

Step2:calculate the average cost per part at destination 1.[ ][(20*8)+(25*6)+(25*5)+(30*6)]/25=24.6$Step3:calculate the economical shipment quantity for the total

flow from the consolidation terminal to destination.Qcj= [{ }/{I( )}]

i k

ijkd

i k i k

ijkijkk ddp /

50)( i k

ijkck dS i k

ijkijk

i k

k ddp /


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Qc1= [(30*1250)/(0.2*24.6)]=87.3Step4: find minimum of Qc1 and Wc1sc1=min[87.3,150]=87.3unitStep5:calculate the quantities of individual items 1 through 4

flowing the consolidated terminal to terminal 1.Qc11=(sc1 ) =(88*8)/25=28unitQc12=(88*6)/25=21unitQc13=(88*5)/25=18unitQc14=(88*6)/25=21unitStep6:repeat the calculation for destination 2. following the same

procedure, we obtain the quantities of items 1 through 4 fromthe consolidation terminal to terminal2.

i i k

kii dd 111 /()


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Qc2= 102unitSc2=min[102,100]=100unitQc21= (sc2 ) ) =(100*4)/30=13unitQc22 =(100*10)/30=33unitQc23 =(100*8)/30=27unitQc24 =(100*8)/30=27unit

i k

ki

i

i dd 221 /()


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References

PRODUCTION,PLANNING ANDINVENTORY CONTROL

--S.L. NARASIMHAN --D.W. McLEAVEY --P.J. BILLINGTON

(PRENTICE HALL OF INDIAPRIVATE LIMITED)


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Thank you

Documents

Final Aggregate Inventory(1)