ECO00030M

UNIVERSITY OF YORK

DEPARTMENT OF ECONOMICS AND RELATED STUDIES

ECO00030M Management Decision Analysis

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 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
media that they are considering using.

Category Audience reached
Cost per
Maximum number
TV 250,000 £6,000 10
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

Priority level 3 goal

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

s1 s2 s3 s4 Maximum Regret
d1 6 0 9 11 11
d2 9 9 0 10 10
d3 10 10 13 0 13
d4 0 4 14 4 14

Minimax regret approach: select d2.

(2) Let

Min P1( d1
 ) + P2( d2
 ) + P3( d3
 ) + P4( d4
 ) + P5(d5+)
s.t.
x1  10 TV
x3  5 News
X4  15 Inter
25x1 + 5x2 + 20x3 + 30x4 - d 1
+
+ d1
 = 200 Goal 1
0.7x1 - 0.3x2 - 0.3x3 - 0.3x4 - d2
 + d2
 = 0 Goal 2
-0.2x1 + 0.8x2 - 0.2x3 - 0.2x4 - d3
 + d3
 = 0 Goal 3
30x4 - d4
 + d4
 = 100 Goal 4
60x1 + 4x2 + 50x3 + 20x4 - d5+ + d5- = 350 Goal 5

x1, x2, x3, x4, d1
 , d1
 , d2
 , d2
 , d3
 , d3
 , d4
 , d4
 ,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 Price per pair
0-49 £36
50-99 £32
100-149 £30
150 or more £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

2
* o
h
DC
Q
C

Order Quantity

Ch

Q*
Q to obtain
Discount

TC
0-49 (0.20)(36) = 7.20 100 * —
50-99 (0.20)(32) = 6.40 106 * ----
100-149 (0.20)(30) = 6.00 109 109 36,657
150 or more (0.20)(28) = 5.60 113 150 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.

Cities
Member
London

Birmingham

Manchester Edinburgh
A 64 70 95 54
B 76 80 40 70
C 82 84 50 76
D 100 90 60 84
E 65 55 80 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 54
B to 3 40
C to 1 82
D to 5 0
E to 2 55
Minimal costs 231

It’s a unique solution.

5. [15 marks]
(a)

The matrix of transition probabilities below deals with consumers’ brand
loyalty to supermarket Arrison and supermarket Besco.
Current
Purchase
Next Purchase
Arrison Besco
Arrison 0.90 0.10
Besco 0.05 0.95

(i) What are the projected market shares for the two supermarkets? [3
marks]

95.05.
10.90.
],[],[ 2121 
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:
Current
Purchase
Next Purchase
Arrison Besco Censberry
Arrison 0.70 0.10 0.20
Besco 0.10 0.75 0.15
Censberry 0.30 0.20 0.50

(i)

What are the new long-run market shares? [6 marks]

1 = 0.701 + 0.102 + 0.303 
2 = 0.101 + 0.752 + 0.203 
3 = 0.201 + 0.152 + 0.503 
also
1 + 2 + 3 = 1 

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%. 