Ch 5 Assigned HW Solutions

Ch 5 Assigned HW - 2A A 2 Dv 5.6 a T = — or — = T — EOQ Dvr r EOQ 2 Here i(7 days)2(200 units/day x 0.03 Slunit r 2 = 14}r $ days b At EOQ

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Unformatted text preview: 2A A 2 Dv 5.6 a. T = — or — = T — EOQ Dvr r ( EOQ) 2 Here, i _ (7 days)2 (200 units/day) x 0.03 Slunit r 2 = 14'}r $ - days. b. At EOQ with no discount, the cost per day is TRC(EOQ) =JZAD_Vr+Dv mm ' At bfeakpoint | TRCCQap)=W+é£+Dv(1—d) ...(2) by I! To solve these explicitly we need separate values of A and r. However, we can find a1imiting(A, r) pair for which Ernie is indifferent between (25,, and the ' EOQ. We set 95 ' ' A TRC£EOQ) = mag” ), and use r '= 133. Thus, 1/2A(6.00)A! 147 + 6.00 = W + “200) + 4,00 2(147) 10,000 which solves for A = $4.82 4'82 = .0328 $f$fday = 11.9'3’r $f$fy12 andr: If (A, r) are lower than this pair, then TRC (pr) < TRC (EOQ). This is likely to be the case since the r value is very high. However, shelf life may be a problem in that Q? represents a 50-day supply. 5.12 Item I D, v,- D,v, D112, 1 7,200 2.00 14,400 120 2 4,000 0.90 3,600 60 3 500 5.00 2,500 60 4 _ 100 0.81 3 1 9 Total 20,581 ' 239 a. The EOQ aggregate exchange curve is defined by R 2 I (TAGS)><(N) =§[Z Jag] =28,560 (Equation 5.29) i=1 Moreover, associated with each point (N, TACS) on the curve is a value of Afr given by T—AAS—S = Al r. (Equation 5.30) 100 I Thus points on the curve include the following: N TACS = Air N 10 2856 285.6 12 2380 - 198.3 14 2040 16 _ 1785 111.6 18 1587’r 20 1428 8.33 3430 411.9 Required in part c. 11.10 2573 ' 231.8 - Required in parts d and e. 4,000 TACS ' ' ‘ ' - ' ' ' ' _ ' ' ' ' F ‘ _ ' ' ' ‘- ". Current [ POSition 3,000 i I I ' ' ‘ _ _ ‘ ' ' " ‘ _ ' ' ' ' * "' r"'“""l.Proposod . . Position I I 2,000 : I 1,000 10 12 14 16 N (# of replenishments) 101 E=K§§=200=>TACS=ZOOXN r TACS = fl = ZOON N N=J§§§9=1L95 200 TACS = 200 x1195 = 2390 c. Current rule: order 4-month supply .'.Ni=%=3 N=3xr1=3><4=12 TACS = Z 9"“ Qf = % A! r = (3430)2 {28,560 = 3430! 8.33 = 3430f8.33 = 411.9 where 8.33 and 411.9 are frorn part a. d. Proposed rule: 3 months supply Ni=%=4 N=4><4=16 TACS =2 ini 22%; Div!- =¥=2573 i=1 i=1 i _ 28560 = 2573 = 2313 r _ (16)2 11.10 c. The proposed option has TACS = 2573 and N = 16. They could achieve the same TACS by reducing the number of replenishments to: N_ 28560 _ 23560 =11-10 ' TACS ‘ 2573 Therefore, they should decrease the number of orders per year to 11.1.14 = 2.775 orders per year per item. 102 Or, he could retain his proposed number of replenishments (N,- = 4, N=16), and reduce TACS TAGS = 28560 = 28560 N 16 TACS = 1785 WithAI'r: 111.6 asinpartd. 5.13 a. The answer depends on the specific rules used. This question is based on the MIDAS D Case, which indicates, for that particular case, the comptroller’s order quantity me is given by Qcom = (2 + 0.2L) D152. The estimated cost savings (ECS) in using EOQ considerations are (2 + 0.2L)Dvr + 52A ———— 2ADvr ...(l) 104 (2+0.2L) ECS = TRC(Q,,,,,,) - TRC(EOQ) = using Equations 5.3 and 5.5. b. With the given data QM = [2 + 0.2(10)]900!52 = 69 units While EOQ = 1’ 2A1) = 49 units ‘W' Using (1) we find ECS = $124.68 - $117.53 = $7.10 per year. 5.14 a. D = 300 notebooks! term v = $050 A = $4.00 1’ term r = $0.08 I 3/ term 2 x $4.00 1‘ term x 300 notebooks/ term E = f __—_.__—. = 245 b ks 09 $050 x $0.08 ; s 1 term 00 103 b TRC(EOQ) = 1/2 x $4.00! term x 300 notebook: I term x $0.50 x $0.08 1’ $ i term = $9.80 p=£5-1=—0.183. 300 2 PCP=50[ P ]= 2.06% -1+p EOQwiscomt) = 2 X$4.00 I term :4 300 notebooks I term 3 316 books . $050 X (1 — 0.40) X $0.08 2' 5 I term Since EOQ(discount) > 300, the best order quantity is 316 books. Q0 ‘ = 2 x $4.00 1’ termx 300 notebooks! term = 283 books " $0.50><($0.08!$Henn—0.02) h = 0.10 cubic feet w = $2 ! zerm/ cubic foot Q _ 2 x $4.00 I term x 300 notebooks! term _ 74 books “P ' 2x010 cubic feetx$2ltennlcubic foot+$050x$0.08l$ 3 term E O Q: = '2 x $4.00! termx 300notebookslterm = 224 books $0.60x $0.08}r $ 1 term f. 5.20 108 5.21 b. -TRC/yr= Evr+2fi D1 +gvr+fiA D‘ . ' 4 Q D2+DI dI‘RC _ 3er2 + M), _ _ 2DID2A _ 0 = Q. _ SDIDZA dQ 4(1)2 +131) 92(1)2 +01) vr(3D2 +01) Altemately, one could have set carrying costfyear = ordering costfyear g = W =21.91'=: 22 V 2(45+30) Q. = {3(30x30x20) = szD = 24.49 = 25 2(90+30) vr d. EOQ = [ESE—'30? = 5J30750 a 371 . l EOQ(d = .05) = [M = 900 = 2 0 .95 - .2 " 2-50-3075 EO d=.15 = —==951 Q( ) 2-.85-.2 900-19 -.2 + 50-3075 .2 900 TRC(900) = = 341.75 1000-1.7-.2 + 50-3075 2 1000 TRC(1000)= = 323.75 Order 1000 trays at a time 109 1000 t = —— = .325 = . m, 3075 years 3 9 months ...
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This note was uploaded on 11/03/2008 for the course IE 251 taught by Professor Wilson during the Spring '08 term at Lehigh University .

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Ch 5 Assigned HW - 2A A 2 Dv 5.6 a T = — or — = T — EOQ Dvr r EOQ 2 Here i(7 days)2(200 units/day x 0.03 Slunit r 2 = 14}r $ days b At EOQ

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