MS 401
Production and Service Systems Operations
Spring 2009-2010
Inventory Control – II
Deterministic Demand: EOQ Extensions
Slide Set #4
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Lead time
Inventory
Q
Reorder
point
R= τλ
Time
Order
given
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τ
Order
Arrives
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In-Class Exercise: Nahmias 4.15
•
•
•
•
•
•
David Gold orders salamis from New York and sells them
The demand for salamis is pretty steady at 175 per month
The salamis cost David 1.85 each
The fixed cost of ordering salamis from New York is $200
It takes three weeks to receive an order
The annual cost of capital is 22%, the cost of shelf space is 3% of
value and the cost of taxes and insurance is 2% of value
a. Find the optimal order quantity of salamis and how often David should
order
b. How many salamis should David have on hand when he orders?
c. Suppose that salamis sell $3 each. What is David’s annual profit from
this business?
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In-Class Exercise: Nahmias 4.15
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Sensitivity
• What is the cost of using a suboptimal Q?
• At the optimal solution,
G(Q*) = Kλ/ Q* + h Q*/2
• For any other Q
G(Q) 1  Q * Q 

 

G(Q*) 2  Q Q * 
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Sensitivity
• Let Q*=250. What if we order Q=300?
G(Q) 1  Q * Q 

 

G(Q*) 2  Q Q * 
• G(Q) /G(Q*) = 0.5 (0.83333 +1.2) = 1.01666
• We can conclude that G(Q) is relatively insensitive to
errors in Q
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Finite Production Rate
Inv. Level
Q  PT1
H / T1  P  
H
H  Q (1   / P )
K hH K  hQ
G (Q )  


(1   / P )
T
2
Q
2
h '  h(1   / P )
-λ
P-λ
T1
T2
Time
2 K
Q* 
h'
T
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Example: Nahmias 4.3
• Read-only memory (EPROM) producer. Demand flat at
2,500 units per year. Production rate is 10,000 units per
year. Costs $50 to initiate a production run. Each unit costs
$2 to manufacture. Annual interest rate is 30%.
•
•
•
•
Optimal size of a production run?
Length of a production run?
Average annual cost of holding and setup?
Maximum level of on-hand inventory?
– maximum dollar investment in inventory?
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Quantity Discount Models
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Relaxing EOQ: Quantity Discounts
• When the unit ordering cost depends on the order size
• Two most popular types:
– All-units: the discount is applied to all units in an order
– Incremental: the discount is applied only to the items
beyond the breakpoint
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All-Units Discounts: Example
.30Q

C (Q)  .29Q
.28Q

for
0  Q  500
for 500  Q  1000
for
1000 Q
• Somewhat irrational
– 499 bags cost $149.70
– 500 bags cost $145.00
• What is the optimal order quantity, if
λ=600/year, K=$8, I=20% ?
• Can we use EOQ directly?
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All-Units Discounts
Copyright © 2001 by The McGraw-Hill Companies, Inc
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The Average Annual Cost Function
G(Q)
G0(Q)
G1(Q)
G2(Q)
Q (0) 
2K 
 400
Ic0
Q (1) 
2K 
 406
Ic1
Q
(2)
2K 

 414
Ic2
Q
500
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1000
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All-Units Discounts: Solution Technique
1. Determine the largest realizable EOQ value
– compute the EOQ for the lowest price first, and continue with the
next higher price
– stop when the EOQ value is realizable
2. Compare the value of the average annual cost at the
largest realizable EOQ and at all the price breakpoints
that are greater than the largest realizable EOQ
3.
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In-Class Exercise: Nahmias 4.24
• In the calculation of an all-units discount schedule, you
first compute the EOQ values for each of the three order
costs, and you obtain:
Q0=800, Q1=875, Q2=925
The all-units discount schedule has breakpoints at 750 and
925. Based on this information only, can you the optimal
order quantity?
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Incremental Quantity Discounts
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Incremental Quantity Discounts
• The first 500 units cost 30 cents each,
the second 500 units cost 29 each,
the remaining cost 28 cents each
0.30Q


C (Q)   150  0.29(Q  500)  0.29Q  5
295  0.28(Q  1000)  0.28Q  15

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for
0  Q  500
for 500  Q  1000
for
1000  Q
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Incremental Quantity Discounts
Copyright © 2001 by The McGraw-Hill Companies, Inc
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The Unit Cost
0.30


c(Q)  C (Q) / Q   0.29  5 / Q
0.28  15 / Q

for
0  Q  500
for 500  Q  1000
for
1000 Q
• The unit cost is a function of Q (for Q>=500). Hence, we
cannot use the EOQ formula
• Instead, we use C(Q)/Q. The average annual cost function is:
K
Q
G (Q)  c 
 Ic
Q
2
C (Q) K
C (Q) Q


I
Q
Q
Q 2
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Finding the Minima for Each Interval
• Next, for each of the three intervals:
– we substitute the relevant C(Q)/Q expression into G(Q) expression
– find the Q value that minimizes G(Q) and check if it is realizable
• For the interval Q < 500 , we have
G0 (Q)  (600)(0.30)  (8)(600) / Q  (0.20)(0.30)Q / 2
• which is minimized at
2*8*600
Q 
 400
0.2*0.3
• This value is realizable as it satisfies Q < 500
(0)
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Finding the Minima for Each Interval
• For the interval 500<= Q <1000, we have
G1 (Q)  (600)(0.29  5 / Q)  (8)(600) / Q  (0.20)(0.29  5 / Q)(Q / 2)
 (0.29)(600)  (13)(600) / Q  (0.20)(0.29)Q / 2  (0.20)(5) / 2
• which is minimized at
2*13*600
Q 
 519
0.2*0.29
(1)
• This value is realizable as it satisfies 500<= Q <1000
• Note that
–
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Finding the Minima for Each Interval
• For the interval 1000<= Q, we have
G2 (Q)  (600)(0.28  15 / Q)  (8)(600) / Q  (0.20)(0.28  15 / Q)(Q / 2)
 (0.28)(600)  (23)(600) / Q  (0.20)(0.28)Q / 2  (0.20)(15) / 2
• which is minimized at
Q
(2)
2*23*600

 702
0.2*0.28
• This value is NOT realizable
• Then,
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Average Annual Cost Function
Copyright © 2001 by The McGraw-Hill Companies, Inc
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Incremental Discounts: Solution Technique
1. Determine an algebraic expression for C(Q) corresponding
to each price interval. Use that to determine an algebraic
expression for C(Q)/Q
2. Substitute the expression derived for C(Q)/Q into the
defining equation for G(Q). Compute the minimum value
of Q corresponding to each price interval separately.
3. Determine which minima computed in (2) are realizable
(that is, fall into the correct interval). Compare the values
of the average annual costs at the realizable EOQ values
and pick the lowest.
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Resource-Constrained
Multiple Product Systems
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Resource-Constrained Multiple Product Systems
•
•
•
There is more than one product and the budget to invest in
inventory is limited
Suppose that we are selling three different kinds of goods
with the following data
Item 1
Item 2
Item 3
Demand
1,850
1,150
800
Variable Cost
50
350
85
Set-up Cost
100
150
50
Suppose we don’t want more than $30,000 invested in
inventory at any time. Interest rate is 25%. What should
be the ordering quantities?
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Budget Constraint
•
If the budget constraint is satisfied when the EOQ values
are used, then the EOQs are optimal
•
The EOQ values are calculated to be
– EOQ1=172, EOQ2= 63, EOQ3=61
•
The budget is exceeded when using the EOQs:
172*50+63*350+61*85 = $35,835 > 30,000
•
Hence, we need to reduce the lot sizes…
–
but, how much?
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Budget Constraint
•
•
•
In general, the budget constraint is expressed as
c1Q1+ c2Q2+…+ cnQn<=C
If the constraint is not active, than EOQ is optimal
If the constraint is active:
– if we include the following assumption,
c1/h1= c2/h2=…= cn/hn
(which requires the same I to be used for all products)
then the optimal solution is to scale the EOQ values so that the
budget constraint holds
C
Qi*  m EOQi
where m  n
 ci EOQi 
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i 1
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Space Constraint
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Space Constraint
•
Suppose that each item occupies an area of wi
•
The space constraint is
w1Q1+ w2Q2+…+ wnQn <= W
•
•
Mathematically similar to the budget-constraint
If the constraint is not active, the EOQ values are optimal
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Space Constraint
• If the constraint is active
– the simplifying condition requires
wn
w1 w2

 ...
h1 h2
hn
which is not realistic
• When the condition is not met, solve the problem using the
Lagrangian function. The optimal lot sizes are found as
• where θ is the Lagrange multiplier
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An EOQ Model for Production Planning
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An EOQ Model for Production Planning
• Determine the optimal procedure to produce n products on
a common machine
– to minimize the total cost of holding and setups
– to guarantee that no stock outs occur
• Assumptions:
– Rotation cycle policy
– Setup costs are not sequence-dependent
– The following relation holds for feasibility:
n

j 1
j
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Pj  1
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An EOQ Model for Production Planning
• Why not produce the EOQ for each item?
–
• Let T be the cycle time.
• The lot size for product j satisfies
• The objective is to minimize
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Q j   jT
hence T  Qj /  j
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An EOQ Model for Production Planning
n
Optimization yields
T* 
2 K j
j 1
n
 h 
j 1
j
j
 s
n
If setup times (sj) need to be considered,
j 1
j
 O j / Pj   T
n
which leads to the constraint
T
s
j 1
j
1     j / Pj 
n
 Tmin
j 1
In this case, the cycle time would be the larger of T* and Tmin
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