Talleres Nova: three e-bike models from three scarce components
Talleres Nova assembles three e-bike models, A, B and C, from motors, batteries and controllers. Each model A needs 1 motor, 2 batteries and 1 controller; each B needs 2 motors, 1 battery and 3 controllers; each C needs 1 motor, 3 batteries and 2 controllers. For the next weekly batch the warehouse holds 160 motors, 250 batteries and 250 controllers. The production manager wants to use all three stocks exactly. A provisional plan is 30 units of A, 40 of B and 50 of C. Later a supplier delays 10 motors but can deliver 20 extra controllers, so the available vector becomes 150 motors, 250 batteries and 270 controllers. The manager still wants to exhaust every component and cannot build a negative number of bicycles.
1. Write the first stock problem in matrix form, compute the determinant of the coefficient matrix and solve the system. Does the first stock vector determine a unique production plan?
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Let x, y, z be the numbers of models A, B and C. The resource equations are: x + 2y + z = 160 2x + y + 3z = 250 x + 3y + 2z = 250 In matrix form A·(x,y,z)^T = b, where A = [[1,2,1],[2,1,3],[1,3,2]] and b = (160,250,250)^T. Then det A = 1(1·2 − 3·3) − 2(2·2 − 3·1) + 1(2·3 − 1·1) = (2 − 9) − 2(4 − 3) + (6 − 1) = −7 − 2 + 5 = −4. Since det A ≠ 0, the system has a unique solution for every right-hand side. Solving: subtract equation 1 from equation 3 to get y + z = 90. Subtract twice equation 1 from equation 2: −3y + z = −70, so z = 3y − 70. Then y + (3y − 70) = 90, giving 4y = 160, y = 40. Hence z = 50 and x = 160 − 80 − 50 = 30. The plan is 30 A, 40 B and 50 C, and it uses all the first stock.
2. With the revised stock vector (150,250,270), solve the system and decide whether a feasible production plan exists. Explain what the negative value means in the workshop.
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The new system is: x + 2y + z = 150 2x + y + 3z = 250 x + 3y + 2z = 270 Subtracting equation 1 from equation 3 gives y + z = 120. Subtracting twice equation 1 from equation 2 gives −3y + z = −50, so z = 3y − 50. Substituting into y + z = 120: y + 3y − 50 = 120, so 4y = 170, y = 42.5. Then z = 77.5 and x = 150 − 2(42.5) − 77.5 = 150 − 85 − 77.5 = −12.5. The algebraic solution is unique but asks for −12.5 units of model A. Since production quantities cannot be negative, there is no feasible plan that uses exactly all revised stocks. The negative value means the revised motor stock is too small relative to the extra controllers: to absorb all controllers with the same component technology, the workshop would have to 'unbuild' model A. The manager must leave some components unused or change the product mix, not merely solve the equations.