Presentation on theme: "1 Chapter 14 Making the most of things: Linear Programming."— Presentation transcript:
1 Chapter 14 Making the most of things: Linear Programming
2 Procedure 1. Define carefully the DECISION VARIABLES. 2. Define the OBJECTIVE FUNCTION 3. Write down in mathematical notation the CONSTRAINTS of the problem. 4. Solve the problem. Where there are only 2 variables it is possible to solve the model using a graphical technique.
3 Example A small company assembles two types of cooker, the Ambassador and the Baron. Each Ambassador which is produced earns £40 contribution, while each Baron cooker earns £50. The company's management need to decide the number of cookers of each type which should be assembled next week in order to maximise contribution. An Ambassador cooker takes 4 hours to assemble and a Baron takes 6 hours. Both cookers need to be assembled by a skilled worker, but currently the company only employ two of these workers, both of whom work a 48 hour week. The Ambassador cooker has two built in clocks, while the Baron has only one clock. Because of a strike at the suppliers, only 20 of these clocks can be obtained per week. For marketing reasons, the company has decided that at least 6 Baron cookers should be produced per week. How many cookers of each type should be produced?
4 The objective function Let A = number of Ambassador cookers to make per week B = number of Baron cookers to make per week This is a maximization problem The objective function is:- Max Profit = 40A + 50B
5 The constraints Subject to 4A + 6B 96 Time constraint (2 times 48) 2A + B 20Clocks constraint B 6Marketing constraint
6 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B
7 The marketing constraint The line B = 6 is a horizontal line and the region B 6 is above the line
8 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B Marketing constraint: B 6
9 The clock constraint The straight line 2A + B = 20 can be plotted by noting that when A = 0, B = 20 And when B = 0, A = 10 These two points can now be plotted and a line drawn between them. To find the region that satisfies the constraint test a point on one side of the line (the origin is a good point). Since 0,0 satisfies the constraint the region is below the line
10 05 152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B Marketing constraint: B 6 Clocks constraint: 2A + B 20
11 The time constraint Again the points where the line crosses the axes can be found, i.e. A = 0, B = 16 B = 0, A = 24 The region is again below the line
12 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B B 6 2A + B 20 Time constraint 4A + 6B 48
13 The feasible region The feasible region is that region that satisfies all constraints. It is shown shaded in the last graph The optimal solution cannot be outside this region. How do we find the optimal combination of the number of Ambassador and Baron cookers to make?
14 What combination of A and B will give a profit of say £500 i.e. 40A + 50B = 500 This is another straight line so A = 0, B = 10 B = 0, A = 12.5 Plot these two points on the graph.
15 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B B 6 2A + B 20 4A + 6B 48 40A + 50B = 500
16 The line lies partly inside and partly outside the feasible region. All points on the line that are within the feasible region will give a profit of £500. For example A = 5 and B = 6 is a feasible combination and gives a profit of £500
17 Can we get better than £500? Try £750 40A + 50B = 750 A = 0, B = 15 B = 0, A = 18.75
18 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B 40A + 50B = 750
19 Try £1000 40A + 50B = 1000 A = 0, B= 20 B= 0, A =25
20 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B 40A + 50B = 1000
21 The line is completely outside the feasible region so none of the points on the line would be acceptable. Notice that all the profit lines we have drawn are parallel. They are called isoprofit lines. If you move the line representing £750 outwards it leaves the feasible region at a corner point. This is the optimal solution.
22 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B (3, 14)
23 The optimal solution The optimal solution is when A = 3 and B =14 This gives a profit of 40 3 + 50 14 = £820
24 LP Theorems 1.The optimal solution of a linear programming problem will always lie at a corner point of the feasible region 2.The optimal solution can involve fractional values of the decision variables
25 Finding the optimal solution Either draw in the isoprofit line and move it to the furthest point of the feasible region Or evaluate each corner point and chose the largest value In the next graph the 4 corner points have been labelled p, q, r and s
26 0510152520 0 5 10 15 20 Number of Ambassadors A Number of Barons B B 6 2A + B 20 Time constraint 4A + 6B 48 p q r s
27 Using corner points Again you can see that point q is the optimum
28 Minimisation problems In some situations you may want to minimise costs. The procedure is exactly the same except the you chose the corner point that is closest to the origin when you plot your isocost line Or you choose the corner point that gives you the minimum cost (2 nd method)