ECO00030M
UNIVERSITY OF YORK
Graduate Qualifying Examinations 2016
DEPARTMENT OF ECONOMICS AND RELATED STUDIES
ECO00030M Management Decision Analysis
(Sketch Answers)
Time allowed: TWO hours
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.
Turn Over
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 10x1  5 x2  x3
Subject to
2 x1 + x 2 + x3  8
x1 + 2 x 2 - x3  12
5 x1 + 2 x3  10
a)
b)
c)
d)
e)
x1 , x2 , x3  0
Use the simplex method to solve the maximization problem. [25 marks]
Is the optimal solution you obtained above unique? If unique, why? If not
unique, then what is (are) the alternative solution(s)? [5 marks]
Which constraint(s) is (are) non-binding? [4 marks]
Determine the range of optimality for C 2 , i.e., the coefficient of x 2 . [5 marks]
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
10x1  5x2  x3  0s1  0s 2  0s3  Ma1
2 x1
x 2 + x3 - s1 + a1 = 8
= 12
x1 + 2 x 2 - x3 + s 2
5 x1 + 2 x3 + s3 = 10
x1 , x2 , x3 , s1 , s 2 , s3 , a1  0
Subject to
+
First tableau:
Basis
A1
S2
S3
CB
-M
0
0
Zj
Cj-Zj
X1
10
2
1
5
-2M
10+2M
X2
5
1
2
0
-M
5+M
X3
1
1
-1
2
-M
1+M
S1
0
-1
0
0
M
-M
S2
0
0
1
0
0
0
S3
0
0
0
1
0
0
A1
-M
1
0
0
-M
0
8
12
10
-8M
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:
Zj
X1
10
0
0
1
10
X2
5
1
2
0
-M
X3
1
1/5
-7/5
2/5
-M+4
S1
0
-1
0
0
M
S2
0
0
1
0
0
S3
0
-2/5
-1/5
1/5
2M/5+2
A1
-M
1
0
0
-M
Cj-Zj
0
5+M
M-3
-M
0
-2M/5-2
-M-5
Basis
A1
S2
X1
CB
-M
0
10
Here, X2 is the entering variable, A1 is the leaving variable. 1 is the pivot element.
4
10
2
204M
Third tableau:
Basis
X2
S2
X1
CB
5
0
10
Zj
Cj-Zj
X1
10
0
0
1
10
0
X2
5
1
0
0
5
0
X3
1
1/5
-9/5
2/5
5
-4
S1
0
-1
2
0
-5
5
S2
0
0
1
0
0
0
S3
0
-2/5
3/5
1/5
0
0
A1
-M
1
-2
0
5
-M-5
4
2
2
40
Here, S1 is the entering variable, S2 is the leaving variable. 2 is the pivot element.
Fourth tableau:
Basis
X2
S1
X1
CB
5
0
10
Zj
Cj-Zj
X1
10
0
0
1
10
0
X2
5
1
0
0
5
0
X3
1
-7/10
-9/10
2/5
1/2
1/2
S1
0
0
1
0
0
0
S2
0
1/2
1/2
0
5/2
-5/2
S3
0
-1/10
3/10
1/5
3/2
-3/2
A1
-M
0
-1
0
0
-M
5
1
2
45
Here, X3 is the entering variable, X1 is the leaving variable. 2/5 is the pivot element.
Final tableau:
Basis
X2
S1
X3
CB
5
0
1
Zj
Cj-Zj
X1
10
7/4
9/4
5/2
45/4
-5/4
X2
5
1
0
0
5
0
X3
1
0
0
1
1
0
S1
0
0
1
0
0
0
S2
0
1/2
1/2
0
5/2
-5/2
S3
0
1/4
3/4
1/2
7/4
-7/4
A1
-M
0
-1
0
0
-M
8.5
5.5
5
47.5
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:
Basis
X2
S1
X3
CB
C2
0
1
Zj
Cj-Zj
X1
10
7/4
9/4
5/2
(7C2+10)/4
(30-7C2)/4
X2
C2
1
0
0
C2
0
X3
1
0
0
1
1
0
S1
0
0
1
0
0
0
S2
0
1/2
1/2
0
C2/2
-C2/2
S3
0
1/4
3/4
1/2
(2+C2)/4
(2+C2)/4
A1
-M
0
-1
0
0
-M
8.5
5.5
5
8.5C2+5
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.
Decision alternative
D1
D2
D3
D4
S1
14
11
10
20
State of nature
S2
S3
19
16
10
25
9
12
15
11
S4
5
6
16
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
TV
Radio
Newspaper
Internet
Audience reached
per advertisement
250,000
50,000
200,000
300,000
Cost per
advertisement
£6,000
£400
£5000
£2,000
Maximum number
of advertisement
10
8
5
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]
(1)
Regret or Opportunity Loss Table
d1
d2
d3
d4
s1
6
9
10
0
s2
0
9
10
4
s3
9
0
13
14
s4 Maximum Regret
11
11
10
10
0
13
4
14
Minimax regret approach: select d2.
(2) Let
x1 = number of TV advertisements
x2 = number of radio advertisements
x3 = number of newspaper advertisements
x4 = number of internet advertisements
Min
P1( d1 )
+
P2( d 2 )
P3( d 3 )
+
+
P4( d 4 )
+
P5(d5+)
s.t.
25x1
+
5x2
+
20x3
+
30x4
-
d +1
+
d1




=
0.7x1
-
0.3x2
-
0.3x3
0.3x4
-
d 2
+
d 2
=
0 Goal 2
+
0.8x2
-
0.2x3
-
-

3
+

3
=
0 Goal 3

4
=
100 Goal 4
=
350 Goal 5
x1
x2
x3
X4
-0.2x1
60x1
+
4x2
+
50x3
+
0.2x4
30x4
-
20x4
-
d

4
d
d5+
+
+
d
d
d5-
10
8
5
15
200
TV
Radio
News
Inter
Goal 1
x1, x2, x3, x4, d1 , d1 , d 2 , d 2 , d 3 , d 3 , d 4 , d 4 ,d-5,d5+  0
3. [20 marks] AKKO Shoe Stores (AKKO) carries a basic black dress shoe for men that
sells at an approximate constant rate of 300 pairs of shoes every three months. AKKO’s
current buying policy is to order 300 pairs each time an order is placed. It costs RYF £30
to place an order. The annual holding cost rate is 20%. With the order quantity of 300,
AKKO obtains the shoes at the lowest possible unit cost of £28 per pair. Other quantity
discounts offered by the manufacturer are as follows. What is the minimum cost order
quantity for the shoes? What are the annual savings of your inventory policy over the
policy currently being used by AKKO?
Order quantity
0-49
50-99
100-149
150 or more
Price per pair
£36
£32
£30
£28
D = 4(300) = 1,200 per year
Co = $30
I = 0.20
C = $28
Annual cost of current policy: (Q = 300 and C = $28)
TC = 1/2(Q)(Ch) + (D/Q)Co + DC
= 1/2(300)(0.2)(28) + (1200/300)(30) + 1200(28)
= 840 + 120 + 33,600 = 34,560
Evaluation of Quantity Discounts
Q* 
2 DCo
Ch
Order Quantity
0-49
50-99
100-149
150 or more
Ch
(0.20)(36) = 7.20
(0.20)(32) = 6.40
(0.20)(30) = 6.00
(0.20)(28) = 5.60
Q*
Q to obtain
Discount
TC
100
106
109
113
*
*
109
150
—
---36,657
34,260
*Cannot be optimal since Q* > the threholds.
Reduce Q to 150 pairs/order. Annual savings is £300; note that shoes will still be
purchased at the lowest possible cost (£28/pair).
Section B (15 marks)
Candidates should attempt to answer only ONE question in this section.
4.
[15 marks] The board of a club consists of five members, A, B, C, D, and
E, who have to represent their club at four important meetings in the cities
London, Birmingham, Manchester, and Edinburgh. At each of these
meetings at least one member of the board should be present, and each
member can make at most one trip. Their travel costs are shown in the
following table. Consider how the cities can be assigned to the members
of the board in the cheapest way.
London
Member
A
B
C
D
E
64
76
82
100
65
Cities
Birmingham Manchester
70
80
84
90
55
95
40
50
60
80
Edinburgh
54
70
76
84
76
(a)
Formulate the above problem as a Linear Programming problem. [5 marks]
(b)
Use the Hungarian method to find an assignment that minimises the total
time to be used. (Note that you should carefully show the steps of the
calculations.) [10 marks]
a)
Min 64XA1 + 70XA2 + 95XA3 + 54XA4 + 76XB1 + 80XB2 + 40XB3 + 70XB4 + 82XC1 + 84XC2 + 50XC3
+ 76XC4 + 100XD1 + 90XD2 + 60XD3 + 84XD4 +65XE1 + 55XE2 + 80XE3 + 76XE4
s.t. 1) XA1 + XA2 + XA3 + XA4 =1
2) XB1 + XB2 + XB3 + XB4 =1
3) XC1 + XC2 + XC3 + XC4 =1
4) XD1 + XD2 + XD3 + XD4 =1
5) XE1 + XE2 + XE3 + XE4 =1
6) XA1 + XB1 + XC1 + XD1 + XE1=1
7) XA2 + XB2 + XC2 + XD2 + XE2=1
8) XA3 + XB3 + XC3 + XD3 + XE3 =1
9) XA4 + XB4 + XC4 + XD4 + XE4=1
10) Xij = 0 or 1 for all i,j
b) A dummy city 5 with cost 0 is added, subtract 0 from each row to obtain
1
2
3
4
5
A
64
70
95
54
0
B
76
80
40
70
0
C
82
84
50
76
0
D
100
90
60
84
0
E
65
55
80
76
0
Subtract 64 from column 1, 55 from column 2, 36 from column 3, 54 from column 4 and 0 from
column 5 to obtain
1
2
3
4
5
A
0
15
55
0
0
B
12
25
0
16
0
C
18
29
10
22
0
D
36
35
20
30
0
E
1
0
40
22
0
Four lines to cover all zeros. Subtract 10 from all remaining uncovered elements and add 10 to the
numbers covered twice to obtain
1
2
3
4
5
A
0
15
55
0
10
B
12
25
0
16
10
C
8
19
0
12
0
D
26
25
10
20
0
E
1
0
50
22
10
Still minimum four lines to cover all zeros. Subtract 8 from all remaining uncovered elements and
add 8 to the numbers covered twice to obtain
1
2
3
4
5
A
0
15
63
0
18
B
4
17
0
8
10
C
0
11
0
4
0
D
18
17
10
12
0
E
1
0
58
22
18
Minimum five lines to cover all zeros, hence optimal solution is obtained:
A to 4
B to 3
C to 1
D to 5
E to 2
Minimal costs
It’s a unique solution.
54
40
82
0
55
231
5. [15 marks]
(a)
The matrix of transition probabilities below deals with consumers’ brand
loyalty to supermarket Arrison and supermarket Besco.
Next Purchase
Arrison
Besco
Current
Purchase
(i)
0.90
0.10
Arrison
Besco
0.05
0.95
What are the projected market shares for the two supermarkets? [3
marks]
.90 .10
[ 1 ,  2 ]  [ 1 ,  2 ]

.05 .95
Hence,1 = 1/3, 2 = 2/3.
(ii)
Suppose in the market there are 6,000 consumers. How many
customers will switch supermarkets on the next purchase after a
large number of periods? [3 marks]
P(switching) = (1/3)(0.1) + (2/3)(0.05) = 1/15
So, 6000*(1/15)=400 consumers will switch brands.
(b)
Suppose now a new supermarket, Censberry, is founded such that the
following transition probabilities exist:
(i)
Current
Purchase
Next Purchase
Arrison
Besco
Censberry
Arrison
Besco
Censberry
0.70
0.10
0.30
0.10
0.75
0.20
0.20
0.15
0.50
What are the new long-run market shares? [6 marks]
1
2
=
3
also
1 + 2 + 3 = 1
+
=
0.701
0.101
+
+
0.102
0.752
=
0.201
[1]
+
0.303
0.203
+
0.152
+
0.503
[3]
[2]
[4]
Using equations 1,2 and 4, we have 1 = 0.38, 2 = 0.36, and 3 = 0.26.
(ii)
Which supermarket will suffer more from the introduction of the
new supermarket? [3 marks]
The Markov analysis shows that A now has the largest market share. In fact,
its market share has increased by nearly 5%. Supermarket B will be hurt by
the introduction of the new supermarket, C, by about 30%.