Ch 10 Select HW Solutions

Ch 10 Select HW Solutions - Chapter 10 10.1 a. Assume 3:0...

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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 $49.28 219 10.8 a. Cost of H,- for stockout at warehouse i, ie. if shortage > 0 a cost H,- will be incurred. Let W,- = amount allocated to warehouse 1'. Expected cost for warehouse i = H I. L f3r (1:0)de N a we wish to minimize gG’Vl,W2,...,WN) = 2H]. j 33%)de ® i=3 w. subjectto 2w]. =W @andw;20 (3) z: 1, N. (2) can be handled by using a Lagrange multiplier M. We simply have to be careful about (3. N '° N Let h(Wl,W2,...,WN,M) =2Hijfx(xo)dxo +M[zwl. —W] i=1 w: 223 ——=0 ives (3 2M 3 2h 2—Wj= =‘anx(xo)+M 1:1, W M =lfx(x0)=_}-;— J=L JV (‘9 ® and G) are N+1 equations in N+l unknowns. We cannot easily eliminate M as is possible in some Lagrangian solutions. Another complexity is that for a given value of M, egn @ can have more than one solution. To solve for M and the wj’s, we can try different M values until ® is satisfied. If certain wj’s come out negative, set them to zero and re-solve with only the remaining wJ-‘s in the problem. b. Cost B; per unit short at warehouse 1'. Expected cost for warehouse i = Bi I (x0 — wi f l, (3:0)de wt As in (a) we minimize hl(w1,...,wN,M) = inflict, — wf)fi(xo};1xo +M[iwi —W] subjectto® =22_:=—Bj[1—jpxs(wj)]+M = 1—} szmj) 1% B —M J' i-ng(wj) = I j=1, 2, ...,N e Solve equations ® and @ for the NH unknowns, following the same procedure as in (a). 224 10.12 a. cu =0 .06 c0 = 0.02 Max # of papers = SOOIday Demand is normally distributed ' As in footnote 6 in section 10.3, we must find Q; to satisfy Equation 10.9 and then 3 compare ZQE with 500. I Using Table B.1, find page) = Kilt-$.51 = Edi pi “gr-PB: 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 =-.3Q-5+£ 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'g-Q—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 = 46-7)] = $11-67 (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 double-check 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 .

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Ch 10 Select HW Solutions - Chapter 10 10.1 a. Assume 3:0...

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