PS2_W11_Sol (1)

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 02/03/2011 for the course MSANDE 261 taught by Professor -1 during the Winter '11 term at Stanford.

Ask a homework question - tutors are online