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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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