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11-1 Inventory Management
William J. Stevenson
Operations Management
8th edition
11-2 Inventory Management
CHAPTER11
Inventory Management
McGraw-Hill/IrwinOperations Management, Eighth Edition, by William J. StevensonCopyright © 2005 by The McGraw-Hill Companies, Inc. All rights
reserved.
11-3 Inventory Management
Economic order quantity model
Economic production model
Quantity discount model
Economic Order Quantity ModelsEconomic Order Quantity Models
11-4 Inventory Management
Only one product is involved
Annual demand requirements known
Demand is even throughout the year
Lead time does not vary
Each order is received in a single delivery
There are no quantity discounts
Assumptions of EOQ ModelAssumptions of EOQ Model
11-5 Inventory Management
The Inventory CycleThe Inventory CycleFigure 11.2
Profile of Inventory Level Over Time
Quantityon hand
Q
Receive order
Placeorder
Receive order
Placeorder
Receive order
Lead time
Reorderpoint
Usage rate
Time
11-6 Inventory Management
Total CostTotal Cost
Annualcarryingcost
Annualorderingcost
Total cost = +
Q2H D
QSTC = +
TC= Total annual costQ= Order quantity in unitsH= Holding cost per unitD= Annual DemandS= Ordering cost
11-7 Inventory Management
Cost Minimization GoalCost Minimization Goal
Order Quantity (Q)
The Total-Cost Curve is U-Shaped
Ordering Costs
QO
An
nu
al C
os
t
(optimal order quantity)
TCQ
HD
QS
2
Figure 11.4C
11-8 Inventory Management
Deriving the EOQDeriving the EOQ
Using calculus, we take the derivative of the total cost function and set the derivative (slope) equal to zero and solve for Q. The total cost curve reaches its minimum where the carrying and ordering costs are equal.
QDyearperordersofNo
DQcycleorderofLength
SQ
DtorderingAnnual
/ .
/
cos
Cost Holding Annual
Cost) Setupor der Demand)(Or 2(Annual =
H
2DS = QOPT
QOPT= Optimum order quantityQ= Order quantity in unitsH= Holding cost per unitD= Annual DemandS= Ordering cost
11-9 Inventory Management
EOQ MODEL EXAMPLEEOQ MODEL EXAMPLE
A local distributor for a national tire company expects to sell approximately 9600 steel-belted radial tires of a certain size and tread design next year. Annual carrying cost is $16 per tire, and ordering cost is $75. The distributor operates 288 days a year.
D= $ 9600 H= $ 16 S= $ 75 a) What is the EOQ?
b) No. Of orders per year=D/Q=9600/300=32
tires 30016
2(9600)75
H
2DS = QOPT
11-10 Inventory Management
EOQ MODEL EXAMPLEEOQ MODEL EXAMPLE
D= $ 9600 H= $ 16 S= $ 75 c) Length of order cycle= Q/D= 300/9600
=1/32 of a year*288 =9 work days. d) Total Cost=Carrying cost+Ordering cost
=(Q/2)H+(D/Q)S
=(300/2)16+(9600/300)75
=2400+2400
=$ 4800
11-11 Inventory Management
Only one item is involved Annual demand is known Usage rate is constant Usage occurs continually Production rate is constant Lead time does not vary No quantity discounts
Economic Production Quantity AssumptionsEconomic Production Quantity Assumptions
11-12 Inventory Management
Economic RunEconomic Run (Batch) (Batch) Size Size
p
QtimeRun
u
QtimeCycle
upMaximumI
SQDHI
CostSetupCostCarryingTC
up
p
H
DSQ
p
p
p
)(p
Qinventory
/2
2
pmax
maxmin
Qp= Optimum production quantityH= Holding cost per unitD= Annual DemandS= Setup costP= Production or delivery rateU= Usage rate
11-13 Inventory Management
Economic RunEconomic Run (Batch) (Batch) Size Size Example Example
A toy manufacturer uses 48000 rubber wheels per year for its popular dump truck series. The firm makes its own wheels, which it can produce at a rate of 800 per day. The toy trucks are assembled uniformly over the entire year. Carrying cost is $ $1 per wheel a year. Setup cost for a production run of wheels is $45. The firm operates 240 days per year. D= 48000 S=$45 H=$1 per year p=800 wheels per day u= 48000 wheels per 240 days or 200 wheels per day.a)Optimal run size
b) Minimum total annual cost
wheelsup
p
H
DSQp 2400
200800
800
1
45)48000(22
1800$452400
48001
2
1800
2
1800)200800(800
2400)(
p
Q
maxmin
pmax
SQ
DH
ITC
wheelsupI
11-14 Inventory Management
Economic RunEconomic Run (Batch) (Batch) Size Size Example Example
D= 48000 S=$45 H=$1 per year p=800 wheels per day u= 48000 wheels per 240 days or 200 wheels per day.
c)
Thus, a run of wheels will be made every 12 days.
d)
Thus, each run will require three days to complete.
dayperwheelswheelsu
QtimeCycle p 200/ 2400
daysdayperwheelswheelsp
QtimeR p 3 800/ 2400un