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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 50day 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 4month 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 =1110 ' 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=£51=—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 " 2503075
EO d=.15 = —==951
Q( ) 2.85.2 90019 .2 + 503075
.2 900 TRC(900) = = 341.75 10001.7.2 + 503075
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 .
 Spring '08
 Wilson

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