Decision 1
Linear Programming
Exercise 5C: Question 20
(i) Let x represent the number of product X he carries [x = 0 or 1].
Let y represent the number of product Y he carries [x = 0 or 1].
Weight constraint: each product X weighs 16 kg, each product Y
weighs 10 kg;
maximum weight to be carried = 25 kg ⇒ 16x + 10y ≤ 25
Each product X is worth £210 and each product Y is worth £150, so
we need to
maximise R = 210x + 150y
subject to the above constraint
Draw in the line 16x + 10y = 25, joining (0, 2.5) and (1.5625, 0),
then shade out the unwanted region for the inequality, to the right
and above the line.
y
2
Maximum value for R:
1
R = 210 x 1 + 155 x 0 = 210
at (1, 0)
x
2
4
Three feasible points are (0, 0), (1, 0) and (0, 1). Optimal point (1,
0).
© MEI, 20/07/04
1/2
Decision 1
(ii) With a third product Z the weight
constraint is
16x + 10y + 13z ≤ 25
There are 23 = 8 combinations to
check,
of which 5 are feasible (see
table).
The optimal combination is (0, 1, 1),
which gives a value of £225.
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
weight value
0
0
13
75
10
150
23
225
16
210
29
26
39
(iii) With 57 different products, each
weighing between 1 kg and 20 kg, a
maximum of 25 products may be
carried, so number of combinations to
be checked = 57C25 ≈ 9.93 × 1015
The difficulty in solving the salesman’s problem is that an integer
programming problem is not really suitable for linear programming
techniques.
© MEI, 20/07/04
2/2