Lecture10 (1)

# Then i is the smallest value of i among all these

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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 Re-writing 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 re-order 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.

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