The use of linear programming for the allocation of scarce resources LEARNING NOTE 11.1 The use of linear programming for the allocation of scarce resources

LN11.1 (1a) Linear programming

LN11.1 (1b) Example (cont.)

LN11.1 (2) Materials constraint (8Y + 4Z  3,440 (When Y= 0, Z = 860; when Z= 0, Y = 430

LN11.1 (3) Labour constraint 6Y + 8Z  2,880 (when Z = 0, Y = 480; when Y = 0, Z = 360)

LN11.1 (4) Machine capacity constraint 4Y + 6Z  2,760 (when Z = 0, Y = 690; when y = 0, Z = 460)

LN11.1 (5) Sales limitation Y  420

LN11.1 (6) Optimum solution Feasible production combination = Area ABCDE

LN11.1 (7a) Optimum solution 1. The optimum output can be determined by solving the simultaneous equations that intersect at point C: 8Y + 4Z = 3 440 6Y + 8Z = 2 880 so that Y = 400 and Z = 60

LN11.1 (7b) Optimum solution 2. The materials and labour constraints are binding and therefore have opportunity costs. The marginal contribution from obtaining one extra unit of materials can be calculated by solving the following equations: 8Y + 4Z = 3 441 (revised materials constraint) 6Y + 8Z = 2 880 (unchanged labour constraint) Y = 400.2 units, Z = 59.85 units Therefore the planned output of Y would be increased by 0.2 units and Z reduced by 0.15 units and contribution will increase by £0.40 (the opportunity cost).

LN11.1 (7c) Optimum solution 3. The marginal contribution from obtaining one extra labour hour can be found in a similar way: 8Y + 4Z = 3 400 (unchanged materials constraint) 6Y + 8Z = 2 881 (revised labour constraint) Y = 399.9 and Z = 60.2 Marginal contribution = £1.80