Decision making is an important aspect of the Paper F5 syllabus, and questions on this topic will be common. The range of possible questions is considerable, but this article will focus on only one: linear programming. The ideas presented in this article are based on a simple example. Suppose a profit-seeking firm has two constraints: labour, limited to 16,000 hours, and materials, limited to 15,000kg. The firm manufactures and sells two products, X and Y. To make X, the firm uses 3kg of material and four hours of labour, whereas to make Y, the firm uses 5kg of material and four hours of labour. The contributions made by each product are $30 for X and $40 for Y. The cost of materials is normally $8 per kg, and the labour rate is $10 per hour. The first step in any linear programming problem is to produce the equations for constraints and the contribution function, which should not be difficult at this level. In our example, the materials constraint will be 3X + 5Y ≤ 15,000, and the labour constraint will be 4X + 4Y ≤ 16,000. You should not forget the non-negativity constraint, if needed, of X,Y ≥ 0. The contribution function is 30X + 40Y = C Plotting the resulting graph (Figure 1, the optimal production plan) will show that by pushing out the contribution function, 66 student accountant March 2008 the optimal solution will be at point B – the intersection of materials and labour constraints. The optimal point is X = 2,500 and Y = 1,500, which generates $135,000 in contribution. Check this for yourself (see Working 1). The ability to solve simultaneous equations is assumed in this article. The point of this calculation is to provide management with a target production plan in order to maximise contribution and therefore profit. However, things can change and, in particular, constraints can relax or tighten. Management needs to know the financial implications of such changes. For example, if new materials are offered, how much should be paid for...
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