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Unformatted text preview: Exercise 2  Solution for Problem 1
Max. Z = 30W + 20F subject to 1/2W + 1/4F <= 160 hours Mach. 1 1/2W Mach. 1/3W + 1/3F <= 160 hours Mach. 2 1/3W Mach. W + F => 200 hours Min. Production => Min. W => 250 units F => 500 units => Desired but not Desired necessary necessary Exercise 2  Solution for Problem 1
With demand constr. With Mi n. p rod . Ma ch. 1 No Feasible Solution h h.. ac ac M M 2 2 Exercise 2  Solution for
Problem 1
Without demand constr. Mi n. p rod . Ma ch. 1 Feasible Solution h h.. ac ac M M 2 2 Exercise 2  Solution to
Problem 2
Ace Manufacturing Co. a) Max. Z = 50R + 75S subject to 1.2R + 1.6S <= 1600 Assembly .8R + .5S <= 700 Paint .8R 1.5R + .7S <= 1200 Inspection 1.5R R => 150 Demand for regular => S => 90 Demand for super Exercise 2  Solution to Problem
2 B C D E A Exercise 2  Solution to Problem 2
Evaluate Corner Points: Point A (150, 90) Z = 50(150) + 75(90)=14,250 Point B (150, 887.5) Z = 50(150) + 75(887.5) *** = 74,062.50 Point C (470.6, 647) Z = 50(470.6) + 75(647) = 72,059 Point D (579, 474) Z = 50(579) + 75(474) = 64,473.60 Point E (758, 90) Z = 50(758) + 75(90)=44,650 Exercise 2  Solution to Problem
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b) Slack Variables: 1 .2 (1 5 0 ) + 1 .6 (8 8 7 .5 ) = 1 6 0 0 vs Assembly Slack = 0 Assembly 1600 1600 .8 (1 5 0 ) + .5 (8 8 7 .5 ) = 5 6 3 .7 5 vs Paint Slack = 136.25 Paint 700 700 1 .5 (1 5 0 ) + .7 (8 8 7 .5 ) = 8 4 6 .2 5 vs Inspection Slack = 353.75 Inspection 1200 1200 150 = 150 vs 150 Regular surplus = 0 87.5 = 887.5 vs 90 Super surplus = 797.5 8 887.5 Super Exercise 2  Solution to Problem
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c) Demand for Super product can increase by the Demand amount of the slack (surplus) = 797.5 amount Demand for Regular product cannot increase at Demand all before optimal point changes; it can increase optimal to the point (470.6,647) before the optimal intersection changes. intersection See graph Exercise 2  Solution to Problem
2
Regular Inspection Paint Assembly Super Exercise 2  Solution to Problem
R Regular e g Inspection u Paint l a Assembly r 2 Super Exercise 2  Solution to Problem
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New point (470.6,647) Plug into R>150 ==> R=470.6 Plug ==> Allowable increase = 470.6150 = 320.6 Exercise 2  Solution to Problem
d) Determining Shadow Prices For Paint time and Inspection time, slack > 0; For slack Therefore, SP = 0 for Paint and Inspection Therefore, SP For Assembly time, slack = 0; thus SP > 0. For slack thus SP 1.2R + 1.6S = 1601 Original Z = $74,062 Original R = 150 New Z = $74,109 SP = $47 R = 150; S =888.125 150; 2 ExerciseUpperSolution to Problem e) Finding 2  and Lower Bounds for e)
2 Assembly Assembly R Regular e g Inspection u Paint l a Assembly r Super Exercise 2  Solution to Problem
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e) Finding Upper and Lower Bounds for e) Assembly Assembly 1.2R + 1.6S = 1600 Original constraint Upper Bound: Use pt. (150, 1160) (150, 1.2 (150) + 1.6 (1160) = 2036 2036 Lower Bound: Use pt. (150, 90) Lower (150, 1.2 (150) + 1.6 (90) = 324 1.2 324 Exercise 2  Solution to Problem 2 Exercise 2  Solution to Problem
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X1 Heavy duty X2 Regular X3 Extra duty (a) (d) (b) (c) Exercise 2  Solution to Problem
3 (d) (e) (c) Exercise 2  Solution to Problem
3
Special Duty (X4) requires: 1.5 hrs. Manuf. @ $21.33/hr. = $32 1.5 $21.33 0.75 hrs Testing @ $10/hr. = $7.50 0.75 $10 1 ball/can @ $0/can = $ 0 ball/can $0 Total cost = $39.50 (MC) Revenue = $33.00 (MR) Since MC > MR, do not produce Since Special Duty balls Special Use SPs Exercise 1  Solution to model 3mo. M.A. (a)
1a Exercise 1  Solution to (b) 5mo. M.A. model
1b Exercise 1  Solution to (c)
1c α = .2 E.S. model Exercise 1 1d α = .6 E.S.(d) model Solution to Exercise 1  Solution to (e) Exercise 1  Solution to (e) Exercise 1  Solution to (e) Exercise 1  Solution to (e) Exercise 1  Solution to (e)
Comparison of MADs and Biases 3mo. MA 5mo. MA ES α=.2 ES α=.6 5mo. MAD Bias 7764.0 5368.1 8446.1 8062.7 7310.5 3274.1 6945.2 3601.1 Regression  sales Exercise 1  Solution to (f) Exercise 1  Regression
Graph
Sales vs. Time RegressionMkt. Exercise 1  Solution to (f) Pr. Exercise 1  Regression
Graph
MarketPrice vs. Time RegressionCPI Exercise 1  Solution to (f) Exercise 1  Regression Graphs
CPI vs. Time RegressionGNP Exercise 1  Solution to (f) Exercise 1  Regression graphs
GNP vs. Time Exercise 1 – Double Smoothing
Period 1 2 3 4 5 6 7 8 9 10 Sales 35725 47180 54965 63220 66315 57730 62700 60025 74590 83900 F' F" F'  F" bt Final F Error 65000 65789 60953.6 62001.44 60815.58 69080.23 77972.09 65000 0 65000 789 0 65473.4 4519.8 473.4 62761.52 760.08 2711.88 62305.47 1489.9 456.048 61411.53 7668.696 893.938 66012.75 11959.34 4601.218 66578 8848 56907.2 5792.8 58529.48 1495.52 58869.63 15720.37 75854.99 8045.011 94532.65 MAD= Bias= 7980.34 4441.14 ...
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This note was uploaded on 10/20/2010 for the course BUAD BUAD306 taught by Professor Kydd during the Spring '07 term at University of Delaware.
 Spring '07
 KYDD

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