ECO00030M DEPARTMENT OF ECONOMICS AND RELATED STUDIES
DEPARTMENT OF ECONOMICS AND RELATED STUDIES
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ECO00030M
Candidates should attempt to answer ALL questions of Section A and ONE
question from Section B.
The use of hand-held, battery-operated, electronic calculators will be permitted subject to
the regulations governing their use which are specified by the University.
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ECO00030M Management Decision Analysis
Section A (85 marks)
Candidates should attempt to answer BOTH questions in this section.
1. [45 Marks] Consider the following LP problem:
Maximise 321 510 xxx
Subject to 2 1x + 2x + 83 x
1x + 2 2x - 3x 12
5 1x + 2 103 x
321 ,, xxx 0
a) Use the simplex method to solve the maximization problem. [25 marks]
b) Is the optimal solution you obtained above unique? If unique, why? If not
unique, then what is (are) the alternative solution(s)? [5 marks]
c) Which constraint(s) is (are) non-binding? [4 marks]
d) Determine the range of optimality for 2C , i.e., the coefficient of 2x . [5 marks]
e) Find the dual price for the first, second and third constraints from the final
simplex tableau you get, respectively. Interpret these dual prices? [6 marks]
a) In tableau form: (note that M is very large positive number)
Maximise 1321321 000510 Masssxxx
Subject to 2 1x + 2x + 3x - 1s + 1a = 8
1x + 2 2x - 3x + 2s = 12
5 1x + 2 3x + 3s = 10
0,,,,,, 1321321 asssxxx
First tableau:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 5 1 0 0 0 -M
A1 -M 2 1 1 -1 0 0 1 8
S2 0 1 2 -1 0 1 0 0 12
S3 0 5 0 2 0 0 1 0 10
Zj -2M -M -M M 0 0 -M -8M
Cj-Zj 10+2M 5+M 1+M -M 0 0 0
10+2M is the most positive number in Cj-Zj row, X1 is the entering variable. Compare the ratios in
this column, we find S3 is the leaving variable. 5 is the pivot element.
Second tableau:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 5 1 0 0 0 -M
A1 -M 0 1 1/5 -1 0 -2/5 1 4
S2 0 0 2 -7/5 0 1 -1/5 0 10
X1 10 1 0 2/5 0 0 1/5 0 2
Zj 10 -M -M+4 M 0 2M/5+2 -M 20-
4M
Cj-Zj 0 5+M M-3 -M 0 -2M/5-2 -M-5
Here, X2 is the entering variable, A1 is the leaving variable. 1 is the pivot element.
Third tableau:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 5 1 0 0 0 -M
X2 5 0 1 1/5 -1 0 -2/5 1 4
S2 0 0 0 -9/5 2 1 3/5 -2 2
X1 10 1 0 2/5 0 0 1/5 0 2
Zj 10 5 5 -5 0 0 5 40
Cj-Zj 0 0 -4 5 0 0 -M-5
Here, S1 is the entering variable, S2 is the leaving variable. 2 is the pivot element.
Fourth tableau:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 5 1 0 0 0 -M
X2 5 0 1 -7/10 0 1/2 -1/10 0 5
S1 0 0 0 -9/10 1 1/2 3/10 -1 1
X1 10 1 0 2/5 0 0 1/5 0 2
Zj 10 5 1/2 0 5/2 3/2 0 45
Cj-Zj 0 0 1/2 0 -5/2 -3/2 -M
Here, X3 is the entering variable, X1 is the leaving variable. 2/5 is the pivot element.
Final tableau:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 5 1 0 0 0 -M
X2 5 7/4 1 0 0 1/2 1/4 0 8.5
S1 0 9/4 0 0 1 1/2 3/4 -1 5.5
X3 1 5/2 0 1 0 0 1/2 0 5
Zj 45/4 5 1 0 5/2 7/4 0 47.5
Cj-Zj -5/4 0 0 0 -5/2 -7/4 -M
Optimal solution: X1=0, X2=8.5, X3=5, S1=5.5, S2=S3=A1=0, and the value of objective function is
47.5.
b) Here, a non-basic variable X1 has non-zero value in the Cj-Zj row, so there is no alternative solution.
c) The first constraint is nonbinding (S2=S3=0) whereas the second and third constraints are binding.
d) Range of optimality of C2:
X1 X2 X3 S1 S2 S3 A1
Basis CB 10 C2 1 0 0 0 -M
X2 C2 7/4 1 0 0 1/2 1/4 0 8.5
S1 0 9/4 0 0 1 1/2 3/4 -1 5.5
X3 1 5/2 0 1 0 0 1/2 0 5
Zj (7C2+10)/4 C2 1 0 C2/2 (2+C2)/4 0 8.5C2+5
Cj-Zj (30-7C2)/4 0 0 0 -C2/2 -
(2+C2)/4
-M
The optimal solution remains optimal for C2>=30/7.
e) For “less than or equal to” constraint, the dual price is found as the Zj value of the corresponding slack
variable for the constraint in the final tableau. For “greater than or equal to” constraint, the dual price is
the negative of the Zj value of the corresponding surplus variable for the constraint in the final tableau.
Hence, it is 0 for the first constraint, but for the second and third constraints, the dual prices are 5/2 and
7/4, respectively.
A dual price for a constraint is the increase in the objective function value resulting from a one unit
increase in its right-hand side value. Here, if the RHS value of the 2nd and 3rd constraints are increased by
one unit, respectively, the value of the objective function is increased by 5/2 and 7/4, respectively. But for
the 1st constraint, since it is non-binding, a unit increase in the RHS would not help at all, and hence the
dual price is 0.
2. [20 marks] Solve the following two decision making problems.
(1) A decision maker is faced with four decision alternatives and four states of nature,
and has the following profit payoff table.
State of nature
Decision alternative S1 S2 S3 S4
D1 14 19 16 5
D2 11 10 25 6
D3 10 9 12 16
D4 20 15 11 12
Suppose the decision maker knows nothing about the probabilities of the four states of
nature, what is the recommended decision using the minimax regret approach? [5 marks]
(2) A committee in charge of promoting a College Basketball League tournament is
trying to determine how best to advertise the event during the three months prior to the
tournament. The committee obtained the following information about the four advertising
media that they are considering using.
Category Audience reached
per advertisement
Cost per
advertisement
Maximum number
of advertisement
TV 250,000 £6,000 10
Radio 50,000 £400 8
Newspaper 200,000 £5000 5
Internet 300,000 £2,000 15
The committee established the following goals for the campaign.
Priority level 1 goal
Goal 1: Reach at least 2 million people.
Priority level 2 goal
Goal 2: The number of TV advertisements should be at least 20% of the total number of
advertisements.
Priority level 3 goal
Goal 3: The number of radio advertisements should not exceed 20% of the total number
of advertisements.
Priority level 4 goal
Goal 4: The total audience reached by Internet advertisements should be at least 1 million.
Priority level 5 goal
Goal 5: Limit the total expenditure for advertising to £35,000.
Formulate a goal programming model for this problem (but do not solve it). [15 marks]