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Final

# ADMS 3330 Fall 2009 Final Exam Solutions.pdf

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ADMS 3330 Fall 2009 EXAM Solutions Multiple Choice 1 2 3 4 5 6 7 8 9 10 A B D D D B B B B D 11 12 13 14 15 16 17 18 19 20 A C C B A C A B A C 21 22 23 24 25 26 27 28 29 30 B A C B C B D C B A 31 32 33 34 35 B E D 0.64 A Page 1 of 6 ADMS 3330 Fall 2009 EXAM Solutions Problem 1 A. Constraints 2 and 3. B. X1 + X2 + X3 + X4 + X5 + X6 + S1 = 22 X1 + X2 + X3 - S2 = 8 X4 + X5 + X6 - S3 = 10 X4 + X5 + X6 + S4 = 17 S1, S4 are slack variables, S2, S3 are surplus variables C. Invest \$6 million in Project 1 Invest \$0 million in Project 2 (or simply, do not invest in project 2) Invest \$2 million in Project 3 Invest \$0 million in Project 4 Invest \$10 million in Project 5 Invest \$4 million in Project 6 The optimal total return on investment is \$1.78 million D. Constraints 1 and 2 (Slack = 0) E. Constraint 3 Surplus Requirement Exceeded Constraint 4 Slack Partial Utilization F. No. Because 9% (0.09) is outside optimality range for the rate of return on project 3. G. It means that the rate of return on project 2 needs to improve (increase) by 1% (0.01) before the amount to invest in project 2 can assume a positive value in the optimal solution. H. Decrease Dual Price = 0.08 Decrease in investment = 22 – 20.5 = 1.5million Reduction in current optimal objective function value = 0.08*1.5 = \$0.12 million I. Remain the same (Constraint 4) 14million is the lower limit. The dual price of 0 applies.Therefore there is no change in the current optimal objective function value. J. Increase Dual price is 0.010 (Constraint 5) Amount of Increase = 7.5 – 6 = 1.5 Increase in optimal objective function value = 0.010*1.5 = \$0.015 million or \$15,000 K. Cannot tell because the lower limit is 4 (Constraint 10) and a decrease to \$3million is outside of feasibility range. The model needs be re-solved. Problem 2 Page 2 of 6 ADMS 3330 Fall 2009 EXAM Solutions a) Service Time Relative (s) Freq Cumulative RN 20 10% 10% 00 to 09 40 20% 30% 10 to 29 60 40% 70% 30 to 69 80 25% 95% 70 to 94 100 5% 100% 95 to 99 Service System Customer Arrival Start Waiting RN Service Completion Time in Idle Time time Time Time Time system Time 1 0 s 0 0 76 80 80 80 0 2 60 s 80 20 69 60 140 80 0 3 120 s 140 20 82 80 220 100 0 4 180 s 220 40 2 20 240 60 0 5 240 s 240 0 15 40 280 40 0 6 300 s 300 0 94 80 380 80 20 Note: The table could be set up in a totally different order. b) Average waiting time: 𝟎 + 𝟐𝟎 + 𝟐𝟎 + 𝟒𝟎 + 𝟎 + 𝟎 𝟖𝟎 = = 𝟏𝟑.𝟑𝟑 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟔 𝟔 c) 20s (last column of second table) d) proportion of customers that wait = 3/6 = 50%. Page 3 of 6 ADMS 3330 Fall 2009 EXAM Solutions Problem 3 (a) Decision Variables Xi= number of thousands of tons of producti i = 1, 2, 3 W = 1 if Management chooses 2 units of extended capacity for Machinery W, 0 otherwise 1 W 2 1 if Management chooses 7 units of extended capacity for Machinery W, 0 otherwise Y1 = 1 if Management chooses 5 units of extended capacity for Machinery Y, 0 otherwise Y2= 1 if Management chooses 15 units of extended capacity for Machinery Y, 0 otherwise Z1= 1 if Management chooses 12 units of extended capacity for Machinery Z, 0 otherwise Z2 = 1 if Management chooses 32 units of extended capacity for Machinery Z, 0 otherwise p = the profit contribution per 1000 tons of productj j = 1, 2, 3 j (b) Objective Function Max p X1 1p X +2 2X – 3 3 – 40W -150Y – 82Y – 301 – 90Z 2 1 2 (c) 6X 1 5X +22X ≤ 23 + 2W + 7W 1 2 (Machinery W’s capacity) 6X 1 5X + 2X - 2W3 - 7W ≤ 26 2 (Rewritten) 7X 1 3X +21X ≤ 23 + 5Y + 15Y1 2 (Machinery Y’s capaci
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