Aggregate Inventory

8/2/2019 Aggregate Inventory

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Inventory FundamentalsWhat is inventory?

Materials & supplies that a business carries

either for sale or to provide inputs to theproduction process

Those stocks or items used to support

production (raw materials and WIP), supportingactivities (maintenance, repair, & operatingsupplies), & customer service (finished goods &spare parts)


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Inventory & Flow of Materials

Raw materialsWork-in-process (WIP)Raw and in-process (RIP)

Finished goodsDistribution inventoriesMaintenance, repair, & operational supplies(MROs)


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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 costs

Ordering costs

Stockout costs

Capacity-related costs


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Functions of Inventories

Inventory serves as a buffer between:

supply & demand

customer demand & finished goods finished goods & component availability

requirements for an operation & the output from

the preceding operation parts & materials to begin production and the

supplies of materials


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Aggregate Inventory

Management Aggregate inventory management (AIM) is

concerned with managing inventories

according to their classifications (rawmaterial, work-in-process, finished goods,etc.) & the function they perform

AIM is financially oriented & concerned w/

costs & benefits of carrying the classificationsof inventories


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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 w/ inventory


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It is a technique developed for handling EOQs in a aggregate anddealing with the problem of constraint of EOQ equation. It provides ameans to calculate directly the proper lot size for a family of items tomeet some constraints.Then limit order quantities are calculated with the help of formulas A& B

Limit formula A = Ccb = Cca = ( Hb/ha)2Limit formula B = M = Ha/Hb

Where Hb = Total set up hours for present quantities.Ha = Total set up hours for trial order quantities.Cca = Inventory carrying cost for trial order quantities.

Ccb = Inventory carrying cost used for limit order quantities.

LOT SIZE INVENTORY MANAGEMENTINTERPOLATION TECHNIQUES (LIMIT)


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Example

D=10000S=$125

I=0.25

C=$10

what is the trial lot size?

Solution:

QT= (2DS/Ic)

= (2*10000*125/0.25*10)

=1000

Setup in the year=

D/Q=10


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If Limit order quantity

QL=2000

QL= (2DS/ILc)

= (2*10000*125/IL*10)=2000

IL=6.25%

Relationship between the trial and the limit lot size

QL=M*QT

Where M=(IT/IL)

2000=M*1000

M=2

We have

HT=Di *hi/QTi; HL=Di *hi/QLi

Also

M=HT/HL= (IT/IL)


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ExampleThe 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=$62.50HL=35hours, IT=35%How can this situation be handled using LIMIT?

Item AnnualUsage

D

SetupHours

perorder h

UnitCost of

item c

PresentOrder

Quantity Q

YearlySetup

Hours H

A 10000 2 10 769 13*2=26

B 5000 3 15 1667 3*3=9

Total 35


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M=HT/HL=1.4

QL=M*QT=1.4*QT

This leads us directly to the LIMIT order quantities:

Item Cost perSetup S

Trial Q Approximate YearlySetup Hours HT

A $125.0 845 24

B $187..5 598 25

Total 49

Item LMIT Quantity QL Approximately YearlySetup Hours HL

A 1.4*845=1183 17

B 1.4*598=837 18

Total 35


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Inventry limit carrying percentage

IL= IT(HL/HT)^2=0.35(35/49)^2=0.1786

Lotsize

Annual HoldingCost ($)

Annual SetupCost ($)

TotalRelevantCosts ($)

QP-A 769 687 1625 2312

QP-B 1667 2233 563 2796Total 2920 2188 5108

QT-A 845 755 1479 2234

QT-B 598 801 1568 2369

Total 1556 3047 4603

QL-A 1183 1057 1057 2114

QL-B 837 1120 1120 2240

Total 2177 2177 4354


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Optimal,given I=35%

QT

QL

QR

QP

1600

2100

2600

3100

1500 1700 1900 2100 2300 2500 2700 2900 3100 3300 3500

(1556,3047)

(2177,2177) (2920,2188)

(2920,1623)Agg

regate

AnnualSetupCost($

)

Aggregate Annual Holding Cost ($)

Exchange curve


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16

Lagrange Multipliers

The method of Lagrange multipliers gives a set ofnecessary conditions to identify optimal points of

equality constrained optimization problems.

This is done by converting a constrainedproblem to an equivalent unconstrainedproblem with the help of certain unspecified

parameters known as Lagrange multipliers.


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The classical problem formulation

minimize f(x1, x2, ..., xn)

Subject to h1

(x1

, x2

, ..., xn

) = 0

can be converted to

minimize L(x, l) = f(x) - l h1(x)

where

L(x, v) is the Lagrangian function

l is an unspecified positive or negative constant called

the Lagrangian Multiplier


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Finding an Optimum usingLagrange Multipliers

New problem is: minimize L(x, l) = f(x) - l h1(x)

Suppose that we fix l = l* and the unconstrained minimum of

L(x; l) occurs at x = x* and x* satisfies h1(x*) = 0, then x*minimizes f(x) subject to h1(x) = 0.

Trick is to find appropriate value for Lagrangian multiplier l.

This can be done by treating l as a variable, finding theunconstrained minimum of L(x, l) and adjusting l so thath1(x) = 0 is satisfied.


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Method

1. Original problem is rewritten as: minimize L(x, l) = f(x) - lh1(x)

2. Take derivatives of L(x, l) with respect to xi and set themequal to zero.

If there are n variables (i.e., x1, ..., xn) then you will get n equations with n+ 1 unknowns (i.e., n variables xi and one Lagrangian multiplier l)

3. Express all xi in terms of Langrangian multiplier l

4. Plug x in terms of l in constraint h1(x) = 0 and solve l.

5. Calculate x by using the just found value for l.

Note that the n derivatives and one constraint equationresult in n+1 equations for n+1 variables!


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Multiple constraints The Lagrangian multiplier method can be used for any number

of equality constraints.

Suppose we have a classical problem formulation with kequality constraints

minimize f(x1, x2, ..., xn)

Subject to h1(x1, x2, ..., xn) = 0

hk(x1, x2, ..., xn) = 0

This can be converted inminimize L(x, l) = f(x) - lT h(x)

where

lT is the transpose vector of Lagrangian multpliers and has

length k


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21

In closing

Lagrangian multipliers are very useful in sensitivity analyses(see Section 5.3)

Setting the derivatives of L to zero may result in finding asaddle point. Additional checks are always useful.

Lagrangian multipliers require equalities. So a conversion ofinequalities is necessary.

Kuhn and Tucker extended the Lagrangian theory to includethe general classical single-objective nonlinear programming

problem:minimize f(x)

Subject to gj(x) 0 for j = 1, 2, ..., J

hk(x) = 0 for k = 1, 2, ..., K

x = (x1, x2, ..., xN)


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Example

Suppose that we have a profit function andMax f(x,y)=x+3y

Subject to g(x,y)=x2+y2-10=0

Specify the minimization problem with the help oflagrngian function


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SolutionExamine the profit line: =x+3y

In terms of y y=(/3)-(1/3)x slope=-1/3 dy/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)

equating the slope of the profit line and the constraint linegives -(1/3)=-(x/y) y=3x; but y2=10-x2

x=1 and y=3


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Conclusion

profit are maximized at (x,y)=(1,3) with profit of x+3y=10.

Lagrange multiplier technique:-

L(x,y,)=f(x,y)+ g(x,y)(the Lagrange function)

Lx=fx+gx=0

=-(fx

/gx

) Ly=fy+gy=0

=-(fy/gy)

Thus fx/gx=fy/gy

fx/fy=gx/gy =dx/dy

L(x,y,)=x+3y-(x2+y2-10)


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Lx=1-2x=0

Ly=3-2y=0

L=x2+y2-10=0

solve for :

=(1/2x)=(3/2y)

y=3x

also y=10-x2

3x=10-x 2

x=1, y=3

gives profit

f(1,3)=1+3*3=10for minimize the total setup costs

(D1S1/Q1)+(D2S2/Q2)

=(10000*125/Q1)+(5000*187.5/Q2)


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subject to an inventory investment constraint

Q1*(I c1)/2+ Q2*(I c2)/2=2920

Q1

*0.1786*10/2+ Q2

*0.1786*15/2apply Lagrange function

L=(D1S1/Q1)+(D2S2/Q2)+[Q1*(I c1)/2+ Q2*(I c2)/2]

diff with respect to Q1,Q2 and we obtain

L/Q1= D1S1/Q1^2+( I c1)/2=0L/Q2= D2S2/Q2^ 2+( I c2)/2=0

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

solve eq we obtain

=0.555

Q1=1587

Q2=1122


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minimum value of inventory holding costs as

Q1*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 inventorymanagement

Reality customers are not conveniently locatednext 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

Factory warehouse.

Regional distribution centre. Local service centre.

Customer.


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Multi location system

absorbent system

7

64

3

52

1


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Multi location system

3

2

1

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 inventory

system

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 depends onfill rate.

Expected number of backorder=expected delay*demand rate

Expected delay : average time necessary to satisfy a unit ofdemand .


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Inventory holding cost.

Setup costs and ordering costs.

Stockout cost. System stability costs : cost associated withoverreaction to changes demand rates.


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Example:-

D(annual demand)=1000

Q(order quantity)=100

B(maximum back order)=50what is the expected delay or theaverage time to satisfy a unit demand. Working days=250

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

Expected 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 centralize inventories.

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

D=annual demandS=setup cost

Q=ordered quantity

h=holding cost/unit/year.

= standard deviation of lead time.


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k= number of standard deviation of lead time

demand used to determine safety stock.

SS= safety stock.

TRC=total relevant cost.

SS=k

Q=(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 beTRC=(2Sh)D+hk where=( ) Di

N

i 1

D i

N

i 1

D

N

i

iD1

N

i 1


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Distributioninventory

system


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Level decompositon systems

Set aggregate service level objectives for all items at an

echelon.

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 on

effective safety stock.

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


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One origin with several destination

Destination

Origin


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Several origins with one destination

Origins

Destination


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Several origins each with several

destination

Origins destination

l h l d


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Several origins with a consolidation

terminal to several destinationOrigins destination

consolidated

terminal


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Multiple origins and multiple destinations.

When multiple sources shipment to severaldestinations, however solving these problem

becomes hopelessly difficult to solve

This type of problem becomes manageable ifwe assume that all origins ship their products

to a single consolidated terminal and that all

items are distributed to destination asdemanded from the consolidated terminal.


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Algorithm for solving problems

Formulas:

Qic=

Qcj =

wheredijk= quantity of demand from origin i for destination j for product k

Pk= price/unit of k

Di,k= demand at source i for 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

ij k

)/(

50)(

i k i k

ijkijkk

ck

ddp

S

I

di k

ij k

j

ijkd


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Sck = cost of load from consolidation terminal c to destinations k

Wic=capacity of vehicle from source i to consolidation terminal c

Wck=capacity of vehicle from consolidation terminal c to destinations k

Tic=lead time/travel time from source i to consolidation terminal cTck=lead time/travel time from consolidation terminal c to destinations k

Fic= total quantity of items flowing per period from source i to consolidationterminal

=

Fcj= total quantity of items flowing per period from consolidation terminal c to

destinations k=

I = inventory carrying percentage.

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

min[Qic,Wic]

The shippping quantity from consolidation terminal c to destinations is given by

min[Qcj,Wcj]

j k

ijkd

i k

ijkd


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Example: The demand for the products at destination 1 and 2 and

source of these are presented in the table. The capacity of

vehicles, relevant setup costs, lead time between locations, and

other pertinent data in tables.Assume that the inventory carrying charges amount to 20% and

that 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 capacity

demand / period at

Product Cost per unit Destination1 Destination2 Source location

1 20 8 4 1

2 25 6 10 1

3 25 5 8 2

4 30 6 8 2


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Setup cost, vehicle cost, and lead time data

From

to

Source 1

terminal

Source 2

terminal

Terminal

destination

1

Terminal

destination

2

Setup

cost

45 25 30 35

Vehicle

capacity

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 each

source to the terminal.

Step1:calculate the total annual demand for items 1 and 2

formula=[ ] *50

[8+6+4+10]*50=1400 Eq1

Step2:calculate the average cost per part at source1.

Formula=

[(12*20)+(16*25)]/28=22.86 Eq2

Step3: Calculate the economical shipment quantity flowing from

source 1 to the consolidated terminal.

Qic=[{ *50 }/{I( )}] Eq3

Q1c=[(45*1400)/(0.20*22.86)]=118

Find the minimum of Q1c and W1c.

S1c=min[118,150]

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 2 flowing

from origin 1 to the consolidated terminal.

Q1c1=( )=(118*12)/28=51

Q1c2=(118*16)/28=67

Step6: 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=78

S2c=min[78,200]=78

Q2c3= (78*3)/27=38

Q2c4=(78*14)/27=40

j k

ijk

j

jc ddS/)(

111


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The same procedure can be followed to obtain the shipment

quantities from the consolidation terminal to destination 1 and

2, respectively.

Step1: calculate the total annual demand for al items atdestination1.

[ ]*50

[8+6+5+6]*50=1250

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 totalflow from the consolidation terminal to detination.

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)]=87.3

Step4: find minimum of Qc1 and Wc1

Sc1

=min[87.3,150]=87.3Step5:calculate the quantities of individual items 1 through 4

flowing the consolidated terminal to terminal 1.

Qc11=(Sc1 )

=(88*8)/25=28Qc12=(88*6)/25=21

Qc13=(88*5)/25=18

Qc14=(88*6)/25=21

Step6:repeat the calculation for destination 2. following the sameprocedure, we obtain the quantities of items 1 through 4 from

the consolidation terminal to terminal2.

i i k

kii dd 111 /()


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Qc2= 102

Sc2=min[102,100]=100

Qc21= (Sc2 ) )

=(100*4)/30=13

Qc22 =(100*10)/30=33

Qc23 =(100*8)/30=27Qc24 =(100*8)/30=27

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 INDIA PRIVATELIMITED)


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Thank you

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Aggregate Inventory