A requirement for the use of LP is that all cost functions are linear. This is often the case for operation, where the use of a resource is typically proportional to the time of use or throughput. In designing new plant the `7/10 power law' tends to apply to capital costs and so o.f.s are nonlinear and Lp cannot be used directly.

A company has two plants in which it can make a product. Plant A is an older plant on which production costs are £ 170/te and for the newer plant B they are £ 150/te.

Transport costs to the three main customers P, Q and R are significant and depend on their distances from the plants. These costs in £/te are as shown below.

Plant: A B Customer: P Q R P Q R Cost: 25 60 75 20 50 80

Plant A has a maximum production rate of 1.7 te/day and plant B of 0.75 te/day. Customers will take as much product as is offered (all pay the same price) but their minimum requirements, listed below, must always be satisfied.

Customer: P Q R Minimum: 0.9 0.7 0.4

Find the company's optimum production and distribution schedule.

Let the production for P, Q and R at A be x1 to x3 respectively, and the same at plant B be x4 to x6. The function to be minimised is then:

min: P = 170 (x1+x2+x3) + 150 (x4+x5+x6) + 25x1 + 60x2 + 75x3 + 20x4 + 50x5 + 80x6 ;Plant capacity imposes the following inequality constraints:

x1+x2+x3 <= 1.7 ; x4+x5+x6<= 0.75 ;The customers' minimum requirements give:

x1 + x3 >= 0.9 ; x2 + x4 >= 0.7 ; x3 + x6 >= 0.4 ;

Three products A, B and C might be manufactured using this capacity. the requirements in hours/batch of each are listed below.

Product: A B C Reactor 0.8 0.2 0.3 Crystalliser 0.4 0.3 - Centrifuge 0.2 - 0.1There is a sales limit equivalent to 20 batches/week for product C, but none for A or B.

The profit per batch is £20, £6 and £8 for A, B and C respectively. find the optimum weekly production schedule.

Letting the number of batches of A, B and C be x1, x2 and x2, the o.f. is:

max: 20 x1 + 6 x2 + 8 x3;Equipment constraints are then:

0.8 x1 + 0.2 x2 + 0.3 x3 <=20; 0.4 x1 + 0.3 x2 <= 10 ; 0.2 x1 + 0.1 x3 <= 5 ;And C production limit gives:

x3<=20 ;Strictly speaking the number of batches must be an

* Solving as an LP gives optimum as 13.75 batches of A, 15 of B and 20
of C for a profit of £525/week. Rounding A production up to 14 batches
violates the first constraint, requiring 20.2 hours of reactor
time per week. Rounding down to 13 batches gives a near-optimal
o.f. of 510. A
true MILP solution (section 5.4.3) gives a better o.f. of 524
with 14 batches of both A and B.
*

Note that this `solution' simply says that the equipment required will be available for the total number of hours required each week. it does not say what the operating schedule should be, nor indeed, since it will normally be necessary to have the equipment available at particular times and in a given sequence, whether a schedule is in fact feasible.

Next - Section 5.4.3: MILP Problems

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