Linear programming - formulation
You will recall from the Two Mines example that the conditions for a mathematical model to be a linear program (LP) were:
LP's are important - this is because:
We will return later to the simplex algorithm for solving LP's but for the moment we will concentrate upon formulating LP's.
Some of the major application areas to which LP can be applied are:
We consider below some specific examples of the types of problem that can be formulated as LP's. Note here that the key to formulating LP's is practice. However a useful hint is that common objectives for LP's are minimise cost/maximise profit.
Financial planning
A bank makes four kinds of loans to its personal customers and these loans yield the following annual interest rates to the bank:
The bank has a maximum foreseeable lending capability of £250 million and is further constrained by the policies:
Formulate the bank's loan problem as an LP so as to maximise interest income whilst satisfying the policy limitations.
Note here that these policy conditions, whilst potentially limiting the profit that the bank can make, also limit its exposure to risk in a particular area. It is a fundamental principle of risk reduction that risk is reduced by spreading money (appropriately) across different areas.
Financial planning solution
We follow the same approach as for the Two Mines example - namely
Note here that as in all formulation exercises we are translating a verbal description of the problem into an equivalent mathematical description.
A useful tip when formulating LP's is to express the variables, constraints and objective in words before attempting to express them in mathematics.
Variables
Essentially we are interested in the amount (in £) the bank has loaned to customers in each of the four different areas (not in the actual number of such loans). Hence let
xi = amount loaned in area i in £m (where i=1 corresponds to first mortgages, i=2 to second mortgages etc)
and note that xi >= 0 (i=1,2,3,4).
Note here that it is conventional in LP's to have all variables >= 0. Any variable (X, say) which can be positive or negative can be written as X1-X2 (the difference of two new variables) where X1 >= 0 and X2 >= 0.
Constraints
(a) limit on amount lent
x1 + x2 + x3 + x4 <= 250
Note here the use of <= rather than = (following the general rule we put forward in the Two Mines problem, namely given a choice between an equality and an inequality choose the inequality (as this allows for more flexibility in optimising the objective function)).
(b) policy condition 1
x1 >= 0.55(x1 + x2)
i.e. first mortgages >= 0.55(total mortgage lending) and also
x1 >= 0.25(x1 + x2 + x3 + x4)
i.e. first mortgages >= 0.25(total loans)
(c) policy condition 2
x2 <= 0.25(x1 + x2 + x3 + x4)
(d) policy condition 3 - we know that the total annual interest is 0.14x1 + 0.20x2 + 0.20x3 + 0.10x4 on total loans of (x1 + x2 + x3 + x4). Hence the constraint relating to policy condition (3) is
0.14x1 + 0.20x2 + 0.20x3 + 0.10x4 <= 0.15(x1 + x2 + x3 + x4)
Note: whilst many of the constraints given above could be simplified by collecting together terms this is not strictly necessary until we come to solve the problem numerically and does tend to obscure the meaning of the constraints.
Objective
To maximise interest income (which is given above) i.e.
maximise 0.14x1 + 0.20x2 + 0.20x3 + 0.10x4
In case you are interested the optimal solution to this LP (solved using the package as dealt with later) is x1= 208.33, x2=41.67 and x3=x4=0. Note here this this optimal solution is not unique - other variable values, e.g. x1= 62.50, x2=0, x3=100 and x4=87.50 also satisfy all the constraints and have exactly the same (maximum) solution value of 37.5