Coir Unit Product Mix: Solving a Three-Product Plan with the Matrix Inverse
A coir products unit near Pollachi makes three items from the same raw fibre: pith blocks (A), yarn (B) and garden mats (C). Each unit of A needs 2 labour-hours, 1 machine-hour and 3 kg of fibre; each unit of B needs 1, 2 and 1; each unit of C needs 3, 1 and 2. For the coming month the unit has 130 labour-hours, 100 machine-hours and 130 kg of fibre, and the current plan is said to use all three resources fully. The accountant writes the plan as a matrix equation and solves it by the inverse method. Meanwhile a neighbouring unit offers to rent out up to 8 spare machine-hours at Rs 250 per hour. Contribution per unit is Rs 180 for B, Rs 120 for A and Rs 140 for C, but a buyer contract caps B at 32 units for the month. The owner wants to know whether the rental offer is worth taking, and how much of it can actually be used.
1. Write the resource constraints as a matrix equation and solve for the monthly production of A, B and C using the matrix inverse. Show the determinant, the adjoint and the check against the labour constraint.
Read the worked answer
Let x1, x2, x3 be units of A, B, C.
Labour: 2x1 + x2 + 3x3 = 130 Machine: x1 + 2x2 + x3 = 100 Fibre: 3x1 + x2 + 2x3 = 130
M = [[2,1,3],[1,2,1],[3,1,2]], b = (130,100,130).
det M = 2(4-1) - 1(2-3) + 3(1-6) = 6 + 1 - 15 = -8. Non-zero, so a unique plan exists.
Cofactors give the adjoint matrix [[3,1,-5],[1,-5,1],[-5,1,3]] (it is symmetric), hence M inverse = (-1/8) times that matrix, i.e. row entries (-0.375, -0.125, 0.625), (-0.125, 0.625, -0.125), (0.625, -0.125, -0.375).
x1 = -0.375(130) - 0.125(100) + 0.625(130) = -48.75 - 12.5 + 81.25 = 20 x2 = -0.125(130) + 0.625(100) - 0.125(130) = -16.25 + 62.5 - 16.25 = 30 x3 = 0.625(130) - 0.125(100) - 0.375(130) = 81.25 - 12.5 - 48.75 = 20
Plan: 20 units of A, 30 of B, 20 of C. Check labour: 2(20) + 30 + 3(20) = 130, correct; machine 20 + 60 + 20 = 100; fibre 60 + 30 + 40 = 130.
2. Use the inverse to find the change in the product mix when one additional machine-hour becomes available. Interpret each of the three entries in business language.
Read the worked answer
The effect of a small change in availability is dx = M inverse times the change in b. One extra machine-hour means the change vector is (0, 1, 0), so dx is simply the second column of the inverse:
dx = (-0.125, +0.625, -0.125).
Reading this economically: one extra machine-hour is absorbed by raising B by 5/8 of a unit, while A and C each fall by 1/8 of a unit. The reason is technical dependence: B is the most machine-intensive product (2 machine-hours per unit), so extra machine capacity is best used by shifting fibre and labour out of A and C into B. For 8 extra hours the naive answer would be (20 - 1, 30 + 5, 20 - 1) = (19, 35, 19), which already breaks the B cap of 32 units. The cap binds when B rises by 2 units, i.e. after 2 divided by 0.625 = 3.2 extra machine-hours. So at most about 3.2 of the 8 hours can actually be used (mix roughly 19.6, 32, 19.6); the remaining 4.8 hours have no use.
3. Using the contribution figures, advise the owner on how many of the 8 rented machine-hours to take at Rs 250 each, and whether the rental price is justified at all.
Read the worked answer
Value one extra machine-hour at contribution, using the mix change from the inverse:
Gain: 0.625 units of B x Rs 180 = Rs 112.50 Loss: 0.125 units of A x Rs 120 = Rs 15.00 Loss: 0.125 units of C x Rs 140 = Rs 17.50 Net contribution from one extra machine-hour = 112.50 - 15.00 - 17.50 = Rs 80.
Renting costs Rs 250 per hour, so each hour rented destroys Rs 170 of contribution. Recommendation: rent none of the 8 hours. Two separate reasons point the same way, which makes the advice robust: (i) the marginal contribution of machine time (Rs 80) is far below its rental price (Rs 250), and (ii) even if the price were acceptable, the B cap of 32 units means only 3.2 hours could be absorbed before the mix change becomes infeasible. The owner should instead ask whether the B cap can be renegotiated, since B is the product that converts machine time into contribution most effectively.