Exact resource planning for a three-pack cooperative
A village industrial cooperative in Thanjavur makes three gift packs, P1, P2 and P3, from labour hours, packing film and fragrance sachets. Each unit of P1 needs 2 labour hours, 1 film sheet and 1 sachet; each P2 needs 1 labour hour, 2 film sheets and 1 sachet; each P3 needs 1 labour hour, 1 film sheet and 2 sachets. On Monday the cooperative has available 100 labour hours, 120 film sheets and 140 sachets. The production clerk proposes making 15 units of P1, 25 of P2 and 55 of P3, arguing that because the three resource totals are close to the pack counts, the plan must be feasible. The accountant wants a unique production plan that uses every resource exactly, and also wants to know how the plan changes if 8 extra labour hours become available and nothing else changes. No pack can be negative.
1. Write the resource constraints as a linear system, compute the determinant of the coefficient matrix, and solve for the exact-usage plan. State what the determinant tells the cooperative.
Read the worked answer
Let x, y and z be the numbers of P1, P2 and P3. The constraints are:
2x + y + z = 100
x + 2y + z = 120
x + y + 2z = 140
The coefficient matrix is A = [[2,1,1],[1,2,1],[1,1,2]]. Expanding along the first row:
det A = 2(4 - 1) - 1(2 - 1) + 1(1 - 2) = 6 - 1 - 1 = 4.
Since det A is not zero, A is invertible and the exact-usage plan is unique. Adding the three equations gives 4(x + y + z) = 360, so x + y + z = 90. From the first equation, x + (x + y + z) = 100, so x = 10. From the second, y = 30, and from the third, z = 50. Check: 2(10) + 30 + 50 = 100; 10 + 2(30) + 50 = 120; 10 + 30 + 2(50) = 140. The exact-usage plan is (10, 30, 50).
2. Test the clerk's proposed plan (15, 25, 55). Which resources are overused or unused? Is the plan feasible?
Read the worked answer
Substitute x = 15, y = 25, z = 55 into the three resource expressions.
Labour: 2(15) + 25 + 55 = 30 + 25 + 55 = 110. This exceeds 100 by 10 labour hours.
Film: 15 + 2(25) + 55 = 15 + 50 + 55 = 120. This exactly uses the 120 film sheets.
Sachets: 15 + 25 + 2(55) = 15 + 25 + 110 = 150. This exceeds 140 by 10 sachets.
The plan is infeasible because it requires 10 more labour hours and 10 more sachets than are available, even though film is exactly exhausted. The clerk viewed only the totals and ignored the different per-pack coefficients.
3. Find the exact-usage plan if labour availability rises by 8 hours to 108 while film remains 120 and sachets remain 140. Compare with the original plan and explain the direction of change.
Read the worked answer
The new right-hand side is (108, 120, 140). Adding the equations gives 4(x + y + z) = 368, so x + y + z = 92. Then:
x = 108 - 92 = 16,
y = 120 - 92 = 28,
z = 140 - 92 = 48.
Check: 2(16) + 28 + 48 = 108; 16 + 2(28) + 48 = 120; 16 + 28 + 2(48) = 140.
The new plan is (16, 28, 48), compared with (10, 30, 50). The extra labour is absorbed mainly by increasing P1, which uses 2 labour hours but only 1 film sheet and 1 sachet per unit. P2 and P3 fall because film and sachets did not increase, and those packs use more of those resources.