PS2_W11_Sol (1)

PS2_W11_Sol (1) - MS&E 261 Winter 2010-11 Prof. Warren...

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Unformatted text preview: MS&E 261 Winter 2010-11 Prof. Warren H. Hausman SOLUTIONS TO PROBLEM SET #2 1. a) A period is three months. Holding cost per year is $500, which means that the cost for a 3-month period is: 500/4 = $125 = c0 cu = 250 (emergency shipment cost) Critical ratio = * = .667 ⇒ z = .44. Hence Q = σz + µ = (6)(.44) + 60 = 62.64 ≈ 63 cars b) cu = 150 c0 = 125 F(Q ) = .5454 ⇒ z = .11 Q = σz + µ = (6)(.11) + 60 = 60.66 ~ 61 cars. c) This corresponds to an infinite horizon problem with lost sales. From part 3 of Appendix B at the end of the chapter. cu = lost profit = $3,500 c0 = holding cost = 125 Critical ratio = = .9655 * * Critical ratio = = .5454 * ⇒ z = 1.82 which gives Q = (6)(1.82) + 60 = 70.92 ~ 71 cars. 1 Note: The stated solutions below iterate to find the optimal Q values. As we said in class, use of the EOQ value for Q is a close approximation and is suggested for all calculations both for homework and exams. Thus, there was no need to iterate in any of the questions below. 2. a)h = (1.50)(.28) = .42 K = 100 p = 12.80 λ = (280)(12) = 3360 µ = (280)(5) = 1400 σ = 77 = 172.18 EOQ = Q0 = 2Kλ (2)(100)( 3360) = 1265 = h .42 1 - €(R1) = F z1 = 2.24, Q1 = 1 - F(R2) = z2 = 2.23 L(z2) = .004486 L(z1) = .0044, = .0124 n(R1) = .75 = 1324 = .0129 n(R2) = .77 Q2 = Close enough to stop. = 1326 z = 2.23, R = σz + µ = (172.18)(2.23) + 1400 =1784 Optimal (Q,R) = (1326,1784) 2 b) G(Q,R) = = $253.39 h = $439.74 = $24.97 c) σ = 0 ⇒ EOQ solution ⇒ cost = = $531.26 G(Q,R) = $718.10 ⇒ cost of uncertainty = $718.10 - 531.26 = $186.84 yearly. 3. Weekly demand has mean 38 and standard deviation LTD has µ = 38 x 3 = 114 and σ = a) 1 - F(R) = z = 2.59, R = σz + µ = 165 b) Must determine optimal Q by iteration EOQ = 198 = Q0 1 - F(R0) = = .0018838 = 19.75 = .004757 z = 2.90, L(z) = .000542, n(R0) = σL(z) = .0107 Q= = 204 3 This value of Q turns out to be optimal. (must iterate once more to obtain R = 171). c) Use formula G(Q,R) = h[Q/2 + R - µ] + Kλ/Q + pλn(R)/Q Substitute (Q, R) = (500,165) from part (a) and (Q, R) = (204,171) from part (b) Obtain G(500,165) = $2606.75 G(204,171) = $1965.16 ⇒ Δ cost= $641.59 yearly 4 ...
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This note was uploaded on 02/03/2011 for the course MSANDE 261 taught by Professor -1 during the Winter '11 term at Stanford.

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