# OPTIMISATION CONCEPT

Optimisation is the act of choosing the best alternative out of the available ones. It describes how decisions or choices among alternatives are taken or should be made. All such optimisation problems have 3 elements:

a) Decision Variables: These are variables whose optimal values have to be determined. For example, a production manager wants to know at what level to set output in order to achieve maximum profit or maximum sales revenue. Here output is the decision or choice variable. Similarly labour, machine, time and raw materials are choice variables if a works manager wants to know what amount of these are to be used so as to produce a given output level at minimum cost. The quantity of any choice variable must be measurable (20kg, 5 labourers, 10 hours, etc.).

b) The Objective Function: It is a mathematical relationship between the choice variables and some variables whose values are to be maximised or minimised. For example, the objective function could relate profit to level of output or cost to amount of labour, machine, time, raw materials, etc. in the above example.

c) The Feasible Set: The available set of alternatives is called a feasible set.

A solution to an optimisation problem is that set of values of the choice variables which is in the feasible set and which yields maximum or minimum of the objective function over the feasible set.

###### Unconstrained Optimisation Technique

For unconstrained optimisation problem involving single independent variable, we need to satisfy some “conditions”. In economics, the necessary (first order) condition is called the equilibrium condition and sufficient (second order) condition is called the stability condition (continuation of state of equilibrium). There may be equilibrium but it may not be stable. Thus first order condition does not guarantee second order condition. This is summarised in Table-1.

First order Necessary Maximisation Conditions Some economic uses of these conditions are discussed below. Given a firm’s demand function, P = 45 – 0.5 Q and the average cost function, AC = Q2 – 8Q + 5 + 2/Q, we have to find the level of output Q which

a) Maximises total revenue

b) Maximises profits.