Ch 9 Assigned HW Solutions

# Ch 9 Assigned HW Solutions - 9 15 a From Equation 9.10 2...

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Unformatted text preview: 9. 15 a. From Equation 9.10 2 β 2 ' y =L=m=1250 units 2b 2(400) β 2 , 2 ' b. current m = Γ©bβa = = 0.05 < 0.075 Β« vra 0.8 Β° 0.2 Β° 1000 Use EOQ expressed as a time supply, based on current a. ~sz I 2-β50 - - T = β= βββββββββββ=o.791 I ββ9 Dvr (1000)(0.8)(0.2)β - years . We need an order quantity that will last 0.791 years; 213 1000 / /// slope = 400 \ / 0.791 2.5 time >aβ=1000 β 400 0 .791 = 683.6 Q1: hatched area _ _ = 0,791[W) 2 =666 units at time 0. , > 2 4 New m =βiΒ§9Β₯ο¬O)βββ3 = 0.156 > 0.075 (0.8)(0.2)(683.6) Q2 remaining requirements area of dotted triangle = 1/2(683.6)(2.5 - 0.791) = 584 units at time 0.791 years 9.16 . . β a. Item X. Touse Figure 9.1 we need the values of A/E(i) and E(t)vr. A=\$4.00 r=.30/yr c,.=.50 v=\$6 D=10 E(t) = 5 = average transaction size. .1202. E(i) E(i) = 1/2 yr. E(r)vr=5Β°6Β°.3=9 A/E(i) = 4/1/2 = 8 = .50 line, do not stock item X. } this point falls above the C; Item Y. E(i) = .2 years E(t)vrβ= 2 0 6 - .3 = 3.6 A/E(i) = 4/2 = 20. 214 This point falls-belowthe C; = .50 line of Figure 9.1, therefore stock item Y. . ~ b. The infrequent large transaction of X makes it relatively attractive to not stock. On the average, 5 units are satisο¬ed by each replenishment and no stock charges are involved. Holding costs would be higher on the average" for item X than the replenishment cost A. β74???β3'9β5 β β β v x 9.19 a. pβ2 (k) = βQβ : Jiβ = 0.033 D(TBS) 3600) According to PoissOn distribution, k=1.83 s=5EL+kG,, 2L=DL, 0L=JDL s=DL+kJDL =36+1.'83Β«/3_6Β§46 _ Q = 12 = b, pu2(k)ββ-D(TBS) β36(2) .16 According to Poisson distribution, k = 0.97 's = DL+ kJDL = 36 + 97% .242 According to Poisson distribution, k=2. 13 s = DL+kw/DLA= 36+2.13Β«/%a49 217 9.20 D(v - g) EOQ = 2AD V" vr w = 2000- (mo; _ 600[4 β mm] = 306 . 4(2) _ , 4(.2) m . ' β a V6 + β1- 7 _ Find the largest integer less than log LL] +log[ D a D + 1 (Va +7) 1 I o 2 + _βΒ°'Β°7 log 0'14 + log|:_lo__:| E76 05 + 0.07 3o+o.14 0.14 { W =IβEOQβ 9.21 218 ...
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Ch 9 Assigned HW Solutions - 9 15 a From Equation 9.10 2...

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