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A MINIMIZATION PROBLEM 59

Figure 2.16 Graphical Solution for the M&D Chemicals Problem

B
600

500

Litres of Product B 400

300

200 2A + 3B= 1200

100
Optimal Solution 2A + 3B =800
(A = 250, B = 100)
A
0 100 200 300 400 500 600
Litres of Product A

3 Choose an arbitrary (but convenient) value for the objective function.
4 Draw a line on the graph showing the values of the decision variables that will
give this value for the objective function.
5 Using a ruler or straightedge, move the objective function line as close to the
origin as possible until any further movement would take the line out of the
feasible region altogether.
6 The feasible solution on this objective function line is the optimal solution.
7 Confirm the solution point mathematically using simultaneous equations.
You may have noticed that this is exactly the same procedure as for a maximization
problem except for the direction we push the objective function. For a maximization
problem we push the objective function as far away from the origin as possible. For a
minimization problem we push the objective function line as close to the origin as possible.

Surplus Variables
The optimal solution to the M&D Chemicals problem shows that the desired total
production of A + B ¼ 350 litres has been achieved by using all available processing
time of 2A +1B ¼ 2(250) + 1(100) ¼ 600 hours. In addition, note that the constraint
requiring that product A demand be met has been satisfied with A ¼ 250 litres. In fact,
the production of product A exceeds its minimum level by 250 125 ¼ 125 litres. This
excess production for product A is referred to as surplus. In linear programming termi-
nology, any excess quantity corresponding to a constraint is referred to as surplus.
Recall that with a constraint, a slack variable can be added to the left-hand side
of the inequality to convert the constraint to equality form. With a constraint, a
surplus variable can be subtracted from the left-hand side of the inequality to

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