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Unformatted text preview: 1 ∗ ∗ · TL ) ≥ f(2i · TL ) ≤ f(2i +1 · TL ) More formally: consider all the integers i such that the cost corresponding to
2i · TL is no greater than the cost corresponding to a cycle of length 2i+1 · TL ;
that is, f(2i · TL ) ≤ f(2i+1 · TL ). Then i∗ is the smallest value of i among all these
values.
J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 36 / 55 Computing i∗
The property that f(2i · TL ) ≤ f(2i+1 · TL ) implies the following:
f(2i · TL )
K
+ g · 2i · TL
2 · TL
K
K
− i+1
2i · TL
2 · TL
)
(
K
1
1
·
− i+1
TL
2i
2
i K
g
√
K
2g
J. G. Carlsson, U of MN ISyE ≤ f(2i+1 · TL )
K
≤
+ g · 2i+1 · TL
i+1
2 · TL
≤ g · 2i+1 · TL − g · 2i · TL
(
)
≤ g · TL · 2i+1 − 2i
(
)
2i+1 − 2i
( )2
2i
) = T2 ·
≤ T2 · (
= T2 · 2 · 2i
L
L
L
i+1
1/2
1/2i − 1/2i+1
√
1
K
i
≤ TL · 2 =⇒
≤ 2i
TL 2g
Lecture 10: Inventory Models November 7, 2013 37 / 55 Computing i∗
Recall that for the regular EOQ with no Po2 policy, the optimal value of T
(which we’ll call T∗ ) was given by
EOQ
T∗ =
EOQ √ K/g We just established that for the Po2 policy, the optimal i∗ is the smallest integer
that satisfies
√
1
K
≤ 2i
TL 2g
√
Substituting for K/g in the above, we find that the optimal i∗ for the Po2
policy is the smallest integer i that satisfies
T∗
EOQ
2i ≥ √
2TL J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 38 / 55 Example Say demand is 10 units per week, fixed costs are $10, and holding cost rate is $2
per unit per year
The original EOQ formula would have an order size of
√
√
2aK
2 · 520 · 10
∗
Q=
=
≈ 72.11
h
2
which costs us √
2 · 520 · 10 · 2 = $144.22/year Rewriting this in terms of cycle length T, with g = ah/2 = 520, we would have
√
T∗ = K/g ≈ 0.139 years ≈ 7.21 weeks
EOQ J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 39 / 55 Example
How much worse are we if we restrict ourselves to a number of weeks that is a
power of 2?
We have TL = 1/52 since there are 52 weeks in a year
We want the smallest integer i such that
T∗
0.139
EOQ
2i ≥ √
=√
≈ 5.10 =⇒ i∗ = 3
2TL
2 · 1/52
so we should place orders every 8 weeks instead of every 7.21
The cost is now
∗
K
10
+ g · 2i · TL =
+ 520 · 8 · 1/52 = $145/year
8 · 1/52
2 · TL i∗ which is an increase of only $0.78 over our earlier annual costs, or 0.54%
Did we just get lucky that the increase was so small?
J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 40 / 55 Analysis of Po2 model
We established earlier that the optimal i∗ for the Po2 policy is the smallest
√
∗
integer i that satisfies 2i ≥ TEOQ/ 2TL
∗ It can also be shown that i∗ is the largest integer i that satisfies 2i ≤
(how?), and therefore,
√∗
T∗
2TEOQ
EOQ
i∗
√
≤2 ≤
TL
2TL
⇓
√∗
∗
T∗
EOQ
√
≤ 2i · TL ≤
2TEOQ
2 √∗
2TEOQ/TL In other words, the optimal Po2 reorder interval is at least
√
√
∗
1/ 2 · T∗
2 · T∗ = 1.414 · T∗
EOQ
EOQ ≈ 0.707 · TEOQ and at most
EOQ J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 41 / 55 Analysis of Po2 model
It turns out that the cost of the E...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14

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