Midterm_in_97_word

# Midterm_in_97_word - cHanéﬂ 1.——I-—-—-l Cr 4 cl...

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Unformatted text preview: cHanéﬂ 1- _.——I-—-—-_-l- Cr 4. cl = \$25,000, p = \$.40, c0 = 3.15, v = P _ c, 25.15110 .40 _. .15 = 100,000 lb per month 10. Break-even volume as percentage of capacity - % - m- 833 33 3% 120,000 I ' 22. a.) Cf: \$350,000 cu = \$12,000 p = \$13,000 Cf P _ Ctr _ 350, 000 ‘ 13,000 — 12, 000 v: = 53.33 or 59 students hJZ = (75)(13.000) - 350,000 -- (75011000) = \$100,000 c.)Z = [35){25,000] — 350.000 -— (35)(12,000) = 105,000 This is approximately the same as the proﬁt for 75 students and a lower tuition in part (b). caapTea 11— 6. Binomial distribution: Hr = 3) = .0574 _‘—-—-—-— Pass 14. a) PM“) = 3” pram = .225 Fall PtFN} = .025 Pass PtPS} = 34 Fall FIFE} = 00 b) . Test __ Marginal District Pass Fail Probabilities North .225 .025 .250 South .340 .060 .400 Central .332 .018 .350 Marginal Probabilities .89? .103 1.00 e) P(tail) - P(FN)_+_P(E] + retro} = .103 26. .2967 = .5934 CHAPTER. 12 f—ﬂ 2. a) Drive-up window; maximum payoff = \$20,000 It) Breakfast; maximum of minimum payoffs = \$4,000 8. a) Purchase motel; maximal: payoff = \$20,000 11) Purchase theater; maximin payoff = \$5,000 c} _____——————————— Shortage Stable Surplus #__—————— Motel 14,030 0 0 Restaurant 4.000 "3,000 14,000 Theater 0 9,00) 15,000 ______.__—————— Select either motel or restaurant (both have minimum regret values of \$14,000). (1) Motel: 20,000L4] ._ 8.000(0) = \$3.200; restaurant: 8.000(4) + 2030(0) = \$4,400; theater: 6,003.4) + 5.000(0) = \$5,400; select theater. e) Motel: - E.000(.33) + 15,000L33) + 20.000(.33) = \$3,910; restaurant: 2,000(.33) + 3,000(.33) + 6.000(33) = \$5,230; theater: 6,000.33) + 60001.33] + 5,000(.33) = \$5,610; select motel. 12. a) Minimin: South Korea 152 China 17.6 Taiwan 14.9 Philippines 13.8 Mexico 12.5 (— minimum Select Mexico b) Minimax: South Korea 2] .1' China 19.0 (— minimum Taiwan 19.2 Philippines 225 Mexico 25.0 Select China c) Humicz {a = .40): South Korea: 152010) + 21.7(60) = 19.10 China: 17.6(.40) + 19.0(.60) = 18.44 Taiwan: 14.9(.40) + l9.2(.60) = 11.48 (— minimum Philippines: 13.8(.40) + 22.5(.60) = 19.02 Mexico: 12.5(.40) + 25 .0(.60) = 20.0 Select Taiwan d} Equal likelihood.- South Korea: 21.7(.33) + 19.1{.33) + 15.2(.33} = 18.48 China: l9.0(.33) + 185(33} + 17.6633) = 18.18 Taiwan: 19.2(33) + 11111.33) + 14.9(.33) = 16.90 (— minimum Philippines: 22.5(33) + 16.8(.33) + t3.3(.33) = 17.52 Mexico: 25.0(.33) + 21.2(.33] + 125(33): 19.37 Select Taiwan HRH-97,512— ! Li W ‘ 6. a) The first three columns are from Problem 3. Select as many rz‘s as there are breakdowns from a different random number stream. _—___—_—,_..—_————- MW 1 20 1 58 2 2 31 2 47,23 2+l=3 3 93 5 69.35.21,4l,14 2+2+l+2+1= 4 24 2 59,28 2+1=3 5 01 0 — 0 6 56 3 13,09,20 1+1+1-3 7 43 3 73,77,29 2+2+1=5 8 00 5 7289,81,20.85 2+3+3+1+3=12 9 58 3 59,72,88 2+2+3=7 10 27 2 11,89 1+3=4 11 74 4 \$7,551,616.53 3+2+2+2=9 12 16 4 45,56,22,49 2+2+1+2_1 13 79 4 03,82,55,27 1+3+2+1=7 ‘14 77 4 4924,8105 2+1+3+1F7 15 48 3 81,0178 3+1+2=6 16 81 4 92,36,53,04 3+2+2+l=3 17 92 4 95,7961.“ 3+2+2+2-9 18 48 3 3145,13 2+2+1=5 19 64 3 65,3130 2+2+1=5 20 06 0 — 0 Total repair time - 110 hr; ,u. =110/20- 5.50br/week. b) It could bias the results. 11‘ a high random number is selected—for example, 98—this results in a high number of breakdowns (i.e., 5). 1f the same random number is used, it will result in a high repair time [i.e., 3 hr). Thus, a relationship will result wherein high number of breakdowns equals high repair times. and vice versa. The effect in this model will not be too bad since several repair time random numbers are selected for each breakdown. it) Average weekly breakdown cost: 110 hours >< \$50 = 35.500; #- = \$5,500/20 = \$275.00 per week. 61} Breakdowns Cumulative Random per week Probability Numbers 0 .20 01—20 1 .50 21—50 2 .70 51—70 3 .85 71—85 4 .95 86—95 5 1.00 96-99, 00 Week r1 Breakdowns r, Repair Time 1 20 0 — 0 2 31 1 58 2 3 98 5 47,23,69.3S,21 2+1+2+2+l=8 4 24 1 ' 41 2 S 01 0 — 0 6 56 2 14, S9 1 + 2 = 3 7 48 1 28 1 S 00 5 [3,09,20,73,]?7 l+1+1+2+2=7 9 58 2 29, 72 1 + 2 = 3 10 23' 1 89 3 11 '14 3 81.2085 3+1+3=7 12 76 3 59,72,88 2+2+3=7 Week r. Breakdowns r1 Repair Time 13 79 3 11,89,87 1+3+3=7 14 77 3 59,66,53 2+2+2-6 15 48 1 45 2 16 81 3 56,22,49 2+1+2-5 17 92 4 03,82,55,27 l+3+2+1r=7 13 43 1 49 2 19 64 2 24,83 1+3=4 2D 06 0 — 0 'lbtal repair time - 76 hr; average weekly breakdown coat 7'6 X 550 - \$5,800; tr. - \$3,800,720 - \$190 reduction in average weekly repair cost - 1275 190 = \$85. Since the maintenance program costs \$150 and only 585 would be saved. it should not be put into effect. However. the results should be applied with some hesitancy, since they were derived for only one actual simulation. This whole simulation process should be repeated at number of times. Note: In part :1 the same random number streams were used as in part a. This was done to replicate as much as possible the conditions of the ﬁrst simulation since the results were to be compared. However, if many simulations were conducted. this would not have been necessary. cH-ﬁ-PT—Ef; l5 ______________.-— 2. a) andb) 3-Month Weighted Month Demand Moving Average Moving Average 1 8.00 — __ 2 12.00 —— — 3 7.00 -— — 4 9.00 9.00 8.77 5 15.00 9.33 8.70 6 11.00 10.33 12.06 7 10.00 11.67 12.08 8 12.00 12.00 10.93 9 —- 11.00 11.22 c) Three month MAD = 1.6; weighted 3-month MAD = 2.15. The 3-month moving average forecast appears to be more accurate. 6. a) and b) Exponentially Adjusted Exponentially Smoothed Forecast Smoothed Forecast Month Demand {a = .30) (a = .30, ﬂ = .20) Oct 800 800.00 — Nov 725 800.00 800.00 Dec 630 777.50 773.00 Jan 500 733.25 720.80 Feb 645 663.27 639.32 Mar 690 657.79 637.53 Apr 730 667.45 653.18 May 810 686.21 678.55 Jun 1200 723.35 724.64 Jul 980 866.34 895.98 E -— 900.44 930.96 c) Exponentially Smoothed MAPD = 12828617710 - .l66 ,— 16.63% Adjusted forecast MAPD = 12645977710 = .1640 = 16.4% The adjusted exponentially smoothed forecast appears to be more accurate but only slightly. ...
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