Linear programming example 1991 UG examA company manufactures four products (1,2,3,4) on two machines (X and Y). The time (in minutes) to process one unit of each product on each machine is shown below: Machine X YProduct 1 10 27 2 12 19 3 13 33 4 8 23The profit per unit for each product (1,2,3,4) is £10, £12, £17 and £8 respectively. Product 1 must be produced on both machines X and Y but products 2, 3 and 4 can be produced on either machine.The factory is very small and this means that floor space is very limited. Only one week's production is stored in 50 square metres of floor space where the floor space taken up by each product is 0.1, 0.15, 0.5 and 0.05 (square metres) for products 1, 2, 3 and 4 respectively.Customer requirements mean that the amount of product 3 produced should be related to the amount of product 2 produced. Over a week approximately twice as many units of product 2 should be produced as product 3.Machine X is out of action (for maintenance/because of breakdown) 5% of the time and machine Y 7% of the time.Assuming a working week 35 hours long formulate the problem of how to manufacture these products as a linear program.SolutionVariablesEssentially we are interested in the amount produced on each machine. Hence let:xi = amount of product i (i=1,2,3,4) produced on machine X per weekyi = amount of product i (i=2,3,4) produced on machine Y per weekwhere xi >= 0 i=1,2,3,4 and yi >= 0 i=2,3,4Note here that as product 1 must be processed on both machines X and Y we do not define y1.Constraintsfloor space0.1x1 + 0.15(x2 + y2) + 0.5(x3 + y3) + 0.05(x4 + y4) <= 50customer requirementsx2 + y2 = 2(x3 + y3)Note here that as this is only an approximate (±5% say) constraint we might do better to express this constraint as0.95[2(x3 + y3)] <= x2 + y2 <= 1.05[2(x3 + y3)]available time10x1 + 12x2 + 13x3 + 8x4 <= 0.95(35)(60) (machine X)27x1 + 19y2 + 33y3 + 23y4 <= 0.93(35)(60) (machine Y)Objectivemaximise profit, i.e.maximise 10x1 + 12(x2 + y2) + 17(x3 + y3) + 8(x4 + y4)
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