Reactor Capacity, Shadow Prices, and a New Product
A specialty chemicals plant makes two products, A and B, on a shared reactor and purification train. Monthly capacity is 100 reactor-hours and 80 purification-hours. Product A earns a $40 contribution per tonne and uses 2 reactor-hours and 1 purification-hour; B earns $30 and uses 1 and 2. The planning model is max 40A+30B subject to 2A+B≤100, A+2B≤80, A,B≥0. The solver returns A=40, B=20, contribution $2,200, with dual values $16.67 per reactor-hour and $6.67 per purification-hour. The plant manager proposes (i) outsourcing up to 10 extra reactor-hours at $12 per hour and (ii) launching product C, which would consume 1.5 reactor-hours and 1 purification-hour per tonne and earn $25 per tonne. The operations director notes that both dual values are positive and asks for a fast yes/no on both proposals, adding that if 10 hours help, 30 hours should help three times as much.
1. Interpret the two dual values and decide whether proposal (i), buying 10 extra reactor-hours at $12 each, creates value. State the assumption about the range over which the dual is valid.
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The dual value of a capacity constraint is the local marginal contribution of one more unit of that capacity, holding the current basis. Reactor hours are worth $16.67/hour and purification hours $6.67/hour.
For 10 extra reactor-hours:
- Marginal benefit = 10 × 16.67 = $166.70.
- Cost = 10 × 12 = $120.
- Net gain = $46.70.
Verification: solve 2A+B=110 and A+2B=80. This gives A=46.67, B=16.67. Contribution = 40×46.67 + 30×16.67 = $2,366.80, an increase of $166.80 over $2,200. Both variables remain positive, so the same basis and duals remain valid. The purchase is worthwhile if the hours are truly incremental, quality is unaffected, and no other bottleneck binds. If tonnes must be integer, re-solve the integer model; the LP gain is only an approximation.
2. Evaluate product C using the dual values. What is its reduced cost, and what would have to change for C to be worthwhile?
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Product C consumes 1.5 reactor-hours and 1 purification-hour. Its opportunity cost at current duals is:
1.5 × 16.67 + 1 × 6.67 = 25.00 + 6.67 = $31.67 per tonne.
Its contribution is $25 per tonne, so the reduced cost is 25.00 − 31.67 = −$6.67 per tonne. At the current plan, introducing C destroys $6.67 of contribution for every tonne produced because it displaces more valuable use of the constrained resources.
Break-even contribution is $31.67 per tonne. C becomes attractive if its price rises by more than $6.67, if its resource use falls, or if other changes alter the dual values. If management forces a small positive quantity of C, the profit loss is about $6.67 per tonne only while the duals remain stable; after a basis change, re-solve because the marginal costs change. Do not justify C by saying 'the duals are positive'-positive duals mean resources are scarce, not that any user of them is profitable.
3. The director asks whether 30 extra reactor-hours can be valued at three times the 10-hour gain. Explain the range-of-validity issue and verify the 30-hour case.
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No. A shadow price is valid only while the same constraints remain binding and the same basis remains optimal. With reactor capacity R and purification capacity 80, the intersection of the two binding constraints is:
A = (2R − 80)/3, B = (160 − R)/3.
B remains nonnegative only when R ≤ 160. Thus the reactor dual of $16.67 applies for R between 40 and 160, not indefinitely.
At R=130 (30 extra hours), A=60 and B=10, both nonnegative, so the basis is unchanged. Extra contribution = 30 × 16.67 = $500. Cost = 30 × 12 = $360. Net gain = $140. The 30-hour purchase is still worthwhile.
At R=170, however, B would be negative, so the basis changes. Purification becomes the only binding constraint: A=80, B=0, and reactor capacity has slack. The marginal value of an extra reactor-hour beyond R=160 is $0, not $16.67. The director's linear extrapolation fails because the bottleneck shifts.