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Unformatted text preview: Chapter 10 10.1
a. Assume 3:0 and using equation (10.2): v =150, p = 210g = 150 — (.15)150 =127.50 SelectQ‘ to satisfy px<(Q)= p_v E .73
10—8 For uniform distribution between 75 & 125 X, — 75
so px((Xa)= _Q—75 =33: ‘=111.5 sa 112
50 Q y Strictly speaking one should evaluate proﬁt for each of Q = 111 and 112,
but the difference is likely to be negligible. _ 150—12150 _ 27 b. E t' 10.4: k ——'“—"
qua 10n( ) pu2() 210_127_50 From table 13.] k “=— 0.61 9‘ = 2+ka =110+0.61(15)=119 coats The expected proﬁt for this Q value, using equation (10.6) is: 119 — 1 10
E[P(119)]= (32.5)(110)—(22.5)(119)— (32.5)(15)G,, T = $6188.73
We know that the distribution is normal. Hence to evaluate E[P(112)] we use
equation. (10.?) to get E[P(112)]= (32.5)(1 10)— (22.5)(1 12)— (82.5)(15)Gn[%1—12]= $6139.45 '. The inside information was worth $49.28 219 10.8 a. Cost of H, for stockout at warehouse i, ie. if shortage > 0 a cost H, will be
incurred. Let W, = amount allocated to warehouse 1'. Expected cost for warehouse i = H I. L f3r (1:0)de N a we wish to minimize gG’Vl,W2,...,WN) = 2H]. j 33%)de ®
i=3 w. subjectto 2w]. =W @andw;20 (3) z: 1, N. (2) can be handled by using a Lagrange multiplier M. We simply have to be
careful about (3. N '° N
Let h(Wl,W2,...,WN,M) =2Hijfx(xo)dxo +M[zwl. —W]
i=1 w: 223 ——=0 ives (3
2M 3
2h
2—Wj= =‘anx(xo)+M 1:1, W
M
=lfx(x0)=_};— J=L JV (‘9 ® and G) are N+1 equations in N+l unknowns. We cannot easily eliminate M as
is possible in some Lagrangian solutions. Another complexity is that for a given
value of M, egn @ can have more than one solution. To solve for M and the wj’s, we can try different M values until ® is satisﬁed. If
certain wj’s come out negative, set them to zero and resolve with only the
remaining wJ‘s in the problem. b. Cost B; per unit short at warehouse 1'. Expected cost for warehouse i = Bi I (x0 — wi f l, (3:0)de wt As in (a) we minimize hl(w1,...,wN,M) = inﬂict, — wf)ﬁ(xo};1xo +M[iwi —W] subjectto®
=22_:=—Bj[1—jpxs(wj)]+M
= 1—} szmj) 1%
B —M J' ing(wj) =
I j=1, 2, ...,N e Solve equations ® and @ for the NH unknowns, following the same procedure as
in (a). 224 10.12 a. cu =0 .06 c0 = 0.02
Max # of papers = SOOIday
Demand is normally distributed ' As in footnote 6 in section 10.3, we must ﬁnd Q; to satisfy Equation 10.9 and then 3
compare ZQE with 500.
I Using Table B.1, ﬁnd page) = Kilt$.51 = Edi
pi “grPB: cu+ca ThenQi=£i+kcri i= 1,2,3 Vary M until EQ; = 500. The Qg’S found will be the number of papers stocked
each day. .02+ M
b. Re 0‘)  .08
IfM=0 k=.65 Q1=467 Q2=260 Q3=367
2Q; > 500
M = .05 =>k = 1.15 Q1: 285 Q2 :96 Q3 =185
2Q: = 555
Eventually, we ﬁnd that for M = .0533 k: 133
Q: 400 — 1.33 x 100 = 262 Quantities
Q2 = 200  1.38 x 90 = 76 l stocked on
Q3: 300 — 1.38 x 100 = 162 JTuesday 2Q, = 500 papers 0. The assumption that demand is normally distributed may not be valid.
The ﬁxed costs co and cu may be incorrect and may actually vary according to the
paper considered. As well, the main problem is that the Kiosk is too small to
satisfy demand. (1. It would be necessary to consider the tradeoff between the additional cost
of the larger Kiosk and the additional proﬁt to be made, taking into account larger
interest charges on the loaned money. IfM is the value of adding one unit to the
size of the Kiosk, then the approximate increase is 500 units it .0533 on Tuesdays,
for example. (=$31.6S) 229 10.14 a. Revenueﬂoaf sold = p = 20 x .04 = $0.80. 8 = $0.05 ﬂora) = U40 20 5 x0 5 60 ForQSSO
dxa 6° dxo
E[P] (Q)] = —.3Q — 5 + I: [8x0 + .05(Q — x0)]4—0+ £39 40
Q so
=.3Q5+£ mo+§Qdea+'3°dexo
40m0 40 20 40 Q __ _ E 2_ ﬂ _ ﬂ _
_ 3Q 5+80(Q 400)+ 40 (Q 20)+ 40 (60 Q) = —s.75+o.s7sg —..75Q2/30 For Q> 50 232 10.15 Erato] = (3)(50) + (.2)(Q — 50) — 5 + :xodxo + .OSQ Q .8Q 60 Wlmdxﬁﬁlaﬂ" __ _ _ E 2_ 92 _ y _
— 15 .2Q+10 5+40(Q 400)+ 40 20)+40 (60 2
= —13.75+.975Q — ‘759 b. —I—dE[P (Q)] = 0 Solve for Q0131; 46.7 loaves, which is in the valid range, is. Qupt S 50. dig'gQ—ﬂ = 0 leads to Q09, = 52 which is once again in the valid range. To decide which quantity to order, compute the associated expected proﬁts and
compare the results, choosing the better Qopt. E[P1(Q = 467)] = $1167 (for Q = 47 the same result is obtained)
E[P2(Q = 52)] = $11.60 Thus 47 loaves should be purchased each day. c. The expected demand is 40 loavesfday (mean of distribution). Calculate
expected proﬁt using E[P1(Q = 40)] and compare to E [P 1(Q = 47)]. The
difference is the daily loss in proﬁt. E[P1(Q = 40)] = $11.25 so the daily loss is
$0.42. (I. i) Demand distribution is not the same everyday
ii) Effects on sale of other items such as sandwiches
iii) Different types of bread
iv) Bread could be sold unchanged on the next day If the store purchases a lawn mower and it doesn’t sell during the regular season
then it must be discounted in September at a loss of $50. This is the overage cost
co because it is the cost of “over stocking” by one unit. On the other hand, if the
store does not purchase an additional unit and this unit would sell during the
regular season then the store forfeits a potential proﬁt of $125. This is the
underage cost cu because it is the cost of ‘ﬁinderstocking” by one unit. The 233 underage cost can also be thought of as the opportunity coast of forfeitin g an
additional sale. In other terms, p = $425, 1/ = $300, 3 = $250, B = 0. so, p—g+B 175 Using the probability distribution, we ﬁnd that the best order quantity is 3 units. It may be useful to doublecheck the answer by calculating the expected marginal
beneﬁt of purchasing the last unit. For example, there is a 45% chance the third
unit will sell, and thus there is a 55% chance it does not sell and must be
discounted Therefore, the expected marginal revenue of the third unit is (0.45)
($125) and the expected marginal cost of the third unit is (0.55) (3550).
Combining the marginal revenue and the marginal cost yields an expected beneﬁt
of $28.75. For the fourth unit there is a 25% chance it is sold and a 75% chance it
is discounted, thus an expected proﬁt of (0.25) (%125) + (0.75) ($50) = $6.25.
Therefore, the fourth unit should not be purchased. 10.18 a. The problem gives the following data: p $ 20.00
v $ 12.00
g $ 10.00
B 5 5.00
Equation 106 ives (k)=————12_10 =01333 Sok=111 andQ=600+
“g p”: 20—10+5 ' ’ ' ’ l.11(200) = 822. b. Reducing the stande deviation only changes the ﬁnal calculation of Q, which isnow711. ...
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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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