Two machines, two products: is the overtime offer worth ₹1,500?
Vikas runs a fabrication unit in Nashik that makes two products, P and Q, on two machines. Each unit of P needs 2 hours on machine M1 and 1 hour on M2; each unit of Q needs 1 hour on M1 and 3 hours on M2. For the coming week M1 is available for 100 hours and M2 for 150 hours, and Vikas intends to keep both machines fully loaded. Contribution is ₹250 per unit of P and ₹180 per unit of Q. His accountant writes the weekly plan as a matrix equation AX = B, finds A⁻¹, solves it, and reports the plan together with the total contribution. Separately, a machine-hire firm offers Vikas 20 extra hours on M2 for a fixed charge of ₹1,500, while another supplier offers extra M1 hours at ₹40 per hour. Vikas wants the arithmetic checked, the two offers judged, and a clear statement of which capacity he should be buying.
1. Set up the production plan as a matrix equation and solve it using the inverse of the coefficient matrix. Show the check and state why the answer is feasible.
Read the worked answer
Let x = units of P and y = units of Q. Both machines fully loaded gives
2x + y = 100 x + 3y = 150
In matrix form AX = B with A = [[2, 1], [1, 3]], X = [x, y]ᵀ, B = [100, 150]ᵀ.
|A| = 2(3) − 1(1) = 5 ≠ 0, so A⁻¹ exists. adj A = [[3, −1], [−1, 2]], hence A⁻¹ = (1/5)[[3, −1], [−1, 2]].
X = A⁻¹B = (1/5)[3(100) − 1(150); −1(100) + 2(150)] = (1/5)[150; 200] = [30; 40].
Check: 2(30) + 40 = 100 ✓ and 30 + 3(40) = 150 ✓.
Plan: 30 units of P and 40 units of Q. Both values are non-negative integers, so the plan is physically feasible and uses both machines exactly - the assumed equalities are validated, not contradicted.
2. Compute the contribution of the plan. Then evaluate the offer of 20 extra M2 hours for ₹1,500 by re-solving the system and comparing the contribution gain with the charge.
Read the worked answer
Base contribution = 30(250) + 40(180) = 7,500 + 7,200 = ₹14,700.
With M2 capacity 170 hours the system becomes 2x + y = 100 x + 3y = 170
From the first equation y = 100 − 2x. Substituting: x + 300 − 6x = 170 ⇒ −5x = −130 ⇒ x = 26, y = 100 − 52 = 48.
New contribution = 26(250) + 48(180) = 6,500 + 8,640 = ₹15,140.
Gain = 15,140 − 14,700 = ₹440, against a charge of ₹1,500. The hire costs ₹75 per extra hour (1500 ÷ 20) while each extra M2 hour is worth only ₹22 (440 ÷ 20). Reject the offer - the extra capacity is real but far too cheaply valued to justify the fixed charge.
3. Using the same contributions, work out the marginal value of one extra M1 hour. Given M1 hours at ₹40 each and the M2 figures from the previous part, which capacity should Vikas buy, and what caution applies to this conclusion?
Read the worked answer
Raise M1 to 101 hours: 2x + y = 101 x + 3y = 150
y = 101 − 2x ⇒ x + 303 − 6x = 150 ⇒ −5x = −153 ⇒ x = 30.6, y = 39.8.
Change in contribution = (30.6 − 30)(250) + (39.8 − 40)(180) = 0.6(250) − 0.2(180) = 150 − 36 = ₹114 per extra M1 hour.
Comparison:
- M1 hour: worth ₹114, costs ₹40 → accept.
- M2 hour: worth ₹22, costs ₹75 → reject.
Interpretation. M1 is the true bottleneck. Expanding M1 by one hour releases P output (+0.6) and forces Q down (−0.2), yet the net effect is strongly positive because P carries the higher contribution per machine-hour on M1. Expanding M2 adds Q but also cuts P, netting only ₹22.
Caution. These are local marginal values. They hold only while the same two constraints stay active and both x and y remain positive; a large purchase of M1 hours would eventually make the M2 constraint slack or make y = 0, after which the ₹114 rate collapses. Vikas should re-solve for any purchase larger than a few hours rather than multiplying ₹114 by the number of hours.