final aggregate inventory(1)
TRANSCRIPT
-
8/7/2019 Final Aggregate Inventory(1)
1/59
APRESENTATION
ON
Aggregate Inventory Management&
Distribution inventory management
:Presented ByPriyanka Gupta( )2010PMM124Anupama Kumari( )2009PMM119
-
8/7/2019 Final Aggregate Inventory(1)
2/59
Inventory & Flow of Materials
vRaw materialsvWork-in-process (WIP)vRaw and in-process (RIP)
vFinished goodsvDistribution inventoriesvMaintenance, repair, & operational supplies(MROs)
v
-
8/7/2019 Final Aggregate Inventory(1)
3/59
Inventory Objectives
Inventories must be coordinated to meetthree conflicting objectives:
Maximize customer service
Minimize plant operation costs
Minimize inventory investment
-
8/7/2019 Final Aggregate Inventory(1)
4/59
Inventory Costs
Inventory management costs
Item costs
Carrying costsOrdering costs
Stock out costs
Capacity-related costs
-
8/7/2019 Final Aggregate Inventory(1)
5/59
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
-
8/7/2019 Final Aggregate Inventory(1)
6/59
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
-
8/7/2019 Final Aggregate Inventory(1)
7/59
Aggregate Inventory
ManagementAIM involves
Flow & kind of inventory needed
Supply & demand patterns
Functions inventory performs
Objectives of inventory management
Costs associated with inventory
-
8/7/2019 Final Aggregate Inventory(1)
8/59
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)
-
8/7/2019 Final Aggregate Inventory(1)
9/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
10/59
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
-
8/7/2019 Final Aggregate Inventory(1)
11/59
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
-
8/7/2019 Final Aggregate Inventory(1)
12/59
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
-
8/7/2019 Final Aggregate Inventory(1)
13/59
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
-
8/7/2019 Final Aggregate Inventory(1)
14/59
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 $187.5 598 25Total 49
-
8/7/2019 Final Aggregate Inventory(1)
15/59
This leads us directly to the LIMIT order quantities:
Inventory LIMIT carrying percentageIL= IT(HL/HT)2
=0.35(35/49)2=0.1786
Item LMIT Quantity QL
Approximately Yearly SetupHours HLA 1.4*845=1183 17B 1.4*598=837 18Total 35
-
8/7/2019 Final Aggregate Inventory(1)
16/59
Based on this implied carrying cost
percentage of 17.86%, we calculate the TotalRelevant Cost in the table with the help of theseformulas:-
Annual holding cost= (Q/2)*ILcAnnual ordering cost= (D/Q)S
TRC=annual holding cost+annual ordering cost
TRC=(Q/2)h+(D/Q)S
-
8/7/2019 Final Aggregate Inventory(1)
17/59
Lot size Annual HoldingCost ($)
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
-
8/7/2019 Final Aggregate Inventory(1)
18/59
Optimal,given I=35%
Agg
regate
AnnualSetupC
ost($
)
Aggregate Annual Holding Cost ($)
Exchange curve
-
8/7/2019 Final Aggregate Inventory(1)
19/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
20/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
21/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
22/59
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
-
8/7/2019 Final Aggregate Inventory(1)
23/59
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)
-
8/7/2019 Final Aggregate Inventory(1)
24/59
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
-
8/7/2019 Final Aggregate Inventory(1)
25/59
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
-
8/7/2019 Final Aggregate Inventory(1)
26/59
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
-
8/7/2019 Final Aggregate Inventory(1)
27/59
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
-
8/7/2019 Final Aggregate Inventory(1)
28/59
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
-
8/7/2019 Final Aggregate Inventory(1)
29/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
30/59
Multi location inventories
-
8/7/2019 Final Aggregate Inventory(1)
31/59
Multi location systemabsorbent system
-
8/7/2019 Final Aggregate Inventory(1)
32/59
Multi location system
Coalescent systems: have material coming together into one enditem.Series system : have locations feeding each other in a direct path.
-
8/7/2019 Final Aggregate Inventory(1)
33/59
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 .
-
8/7/2019 Final Aggregate Inventory(1)
34/59
Inventory holding cost. Setup costs and ordering costs. Stockout cost. System stability costs : cost associated with
overreaction to changes demand rates.
-
8/7/2019 Final Aggregate Inventory(1)
35/59
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:
-
8/7/2019 Final Aggregate Inventory(1)
36/59
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
-
8/7/2019 Final Aggregate Inventory(1)
37/59
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.
-
8/7/2019 Final Aggregate Inventory(1)
38/59
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
-
8/7/2019 Final Aggregate Inventory(1)
39/59
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
-
8/7/2019 Final Aggregate Inventory(1)
40/59
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
-
8/7/2019 Final Aggregate Inventory(1)
41/59
Distribution inventory
system
-
8/7/2019 Final Aggregate Inventory(1)
42/59
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
-
8/7/2019 Final Aggregate Inventory(1)
43/59
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
-
8/7/2019 Final Aggregate Inventory(1)
44/59
One origin with severaldestination
Destination
Origin
S l i i i h
-
8/7/2019 Final Aggregate Inventory(1)
45/59
Several origins with onedestination
Origins
Destination
S l i i h ith
-
8/7/2019 Final Aggregate Inventory(1)
46/59
Several origins each withseveral destination
Origins
destination
-
8/7/2019 Final Aggregate Inventory(1)
47/59
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
-
8/7/2019 Final Aggregate Inventory(1)
48/59
Several origins with aconsolidation terminal to
several destinationOrigins
destination
consolidated terminal
Algorithm for sol ing
-
8/7/2019 Final Aggregate Inventory(1)
49/59
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
-
8/7/2019 Final Aggregate Inventory(1)
50/59
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
-
8/7/2019 Final Aggregate Inventory(1)
51/59
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
-
8/7/2019 Final Aggregate Inventory(1)
52/59
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
-
8/7/2019 Final Aggregate Inventory(1)
53/59
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 /
-
8/7/2019 Final Aggregate Inventory(1)
54/59
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
-
8/7/2019 Final Aggregate Inventory(1)
55/59
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 /
-
8/7/2019 Final Aggregate Inventory(1)
56/59
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 /()
-
8/7/2019 Final Aggregate Inventory(1)
57/59
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 /()
-
8/7/2019 Final Aggregate Inventory(1)
58/59
References
PRODUCTION,PLANNING ANDINVENTORY CONTROL
--S.L. NARASIMHAN --D.W. McLEAVEY --P.J. BILLINGTON
(PRENTICE HALL OF INDIAPRIVATE LIMITED)
-
8/7/2019 Final Aggregate Inventory(1)
59/59
Thank you