Assignment 3 β Sensitivity Analysis
Four QM for Windows ranging reports read line by line: binding constraints, shadow prices, reduced costs, the range of optimality, and the effect of changing a right-hand side β plus the standard form of each LPP.
- Binding or not? Slack/surplus = 0 βΉ binding (fully used). Slack > 0 βΉ unused by exactly that amount; its dual value is 0.
- RHS change? Check the new RHS against that constraint's lower/upper bound. Inside βΉ the same basis; ΞZ = dual value Γ ΞRHS. Outside βΉ the basis changes and the problem must be re-solved.
- Objective coefficient change? Check the new coefficient against that variable's bounds. Inside βΉ the same optimal quantities; Z changes by Ξc Γ (variable's value). Outside βΉ re-solve.
- Reduced cost of a variable at 0 = how much its coefficient must improve before it becomes worth producing; equivalently, the amount Z worsens per unit if you force it in.
Every interpretation, with the class example: Unit 1 Β· Sensitivity Analysis.
- Four-part answer template: rule β number from the table β arithmetic β one-line business meaning. Every sub-question, every time.
- "Solution doesn't change" β "Z doesn't change". A coefficient moved inside its range keeps the quantities but changes Z β always recompute Z.
- Min problems: QM shows the dual value of a β₯ constraint as negative (e.g. β2.5). Read it as "each extra unit required raises the cost by 2.5".
- Standard form (asked for Q2βQ4) = add a slack to every β€ and subtract a surplus from every β₯, so every constraint becomes an equation; for the simplex, also add an artificial variable to every β₯.
Q1 β Reading the full report (Max Z = 7Xβ + 5Xβ + 6Xβ)
The QM for Windows ranging output for Max Z = 7Xβ + 5Xβ + 6Xβ (the constraints are those of Assignment 2 Q2: 3Xβ + 2Xβ β€ 12, Xβ + 3Xβ β€ 9, 2Xβ + Xβ β€ 8, Xβ + Xβ + 2Xβ β€ 10, 2Xβ + Xβ + Xβ β€ 11):
| Variable | Value | Reduced cost | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Xβ | 2.1 | 0 | 7 | 1.17 | 9.5 |
| Xβ | 2.85 | 0 | 5 | 3.33 | 16.67 |
| Xβ | 2.3 | 0 | 6 | 0.17 | 23.5 |
| Constraint | Dual value | Slack / surplus | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Constraint 1 | 1.75 | 0 | 12 | 5 | 15 |
| Constraint 2 | 1.75 | 0 | 9 | 1.33 | 9.82 |
| Constraint 3 | 0.75 | 0 | 8 | 1.67 | 9.29 |
| Constraint 4 | 0 | 0.45 (printed "45") | 10 | 9.55 | Infinity |
| Constraint 5 | 0 | 1.65 | 11 | 9.35 | Infinity |
(1) Are all constraints exhausted? If not, which are unused and by how much? (2) What do the upper and lower bounds in the variable table mean? (3) Economic interpretation of the dual value / shadow price, with an example.
Assignment 3 p. 1
- (1) Not all constraints are exhausted
Constraint Slack Status 1, 2, 3 0 Binding β fully used; dual values 1.75, 1.75, 0.75 4 0.45 Unused by 0.45: Xβ + Xβ + 2Xβ = 2.1 + 2.85 + 4.6 = 9.55 of 10 5 1.65 Unused by 1.65: 2(2.1) + 2.85 + 2.3 = 9.35 of 11 The report prints Constraint 4's slack as "45" β the column is too narrow for the leading point. It is 0.45, which matches its lower bound of 9.55 (= 10 β 0.45).
- (2) The variable bounds are the range of optimality
They are the range over which each objective coefficient can move without changing the optimal quantities. Example: Xβ's profit of 7 may lie anywhere from 1.17 to 9.5 and the best plan stays Xβ = 2.1, Xβ = 2.85, Xβ = 2.3 β only Z changes. Push it outside the range and a different corner becomes optimal, so the problem must be re-solved.
- (3) Shadow price
The shadow price is the increase in Z from one more unit of a constraint's resource, valid while the RHS stays inside its range. Constraint 1's 1.75 means a 13th unit of resource 1 adds 1.75 to profit: raising it from 12 to 14 gives Z = 42.75 + 2 Γ 1.75 = 46.25 (14 lies inside 5β15). It is also the most the firm should pay for an extra unit. Constraints 4 and 5 have shadow price 0 β they already have spare capacity, so more of them is worth nothing.
Q2 β Min Z = 5Xβ + 4Xβ + 6Xβ
| Variable | Value | Reduced cost | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Xβ | 7 | 0 | 5 | 0 | 8 |
| Xβ | 0 | 1.5 | 4 | 2.5 | Infinity |
| Xβ | 0 | 6 | 6 | 0 | Infinity |
| Constraint | Dual value | Slack / surplus | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Constraint 1 | 0 | 3 | 10 | 7 | Infinity |
| Constraint 2 | β2.5 | 0 | 14 | 0 | 16 |
| Constraint 3 | 0 | 1 | 8 | 7 | Infinity |
Convert to standard form, and answer: (1) which constraints are unutilised and how much is consumed; (2) the meaning of Xβ's reduced cost of 6; (3) does raising Xβ's cost from 6 to 9 change the optimum; (4) does reducing C2 from 14 to 8 change it?
Assignment 3 p. 1
- Standard form
Min Z = 5Xβ + 4Xβ + 6Xβ + 0Sβ + 0Sβ + 0Sβ
Xβ + Xβ + Xβ + Sβ = 10
2Xβ + Xβ β Sβ = 14
Xβ + 2Xβ + Xβ + Sβ = 8
Xβ, Xβ, Xβ, Sβ, Sβ, Sβ β₯ 0
(For the simplex table add an artificial Aβ to C2 with cost +M.) The optimum is Xβ = 7, Xβ = Xβ = 0, Z = 35.
- (1) Unutilised constraints C1: uses 7 of 10 βΉ 3 units unused (slack 3) C3: uses 7 of 8 βΉ 1 unit unused (slack 1) C2: 2(7) = 14, exactly the requirement βΉ binding
- (2) Reduced cost of 6 on Xβ
Xβ is not used. Forcing one unit of Xβ into the solution would raise the cost by 6; put the other way, Xβ's cost would have to fall by 6 (from 6 to 0) before it would be worth using.
- (3) Cost of Xβ from 6 to 9
Xβ's range of optimality: 0 to Infinity; 9 lies inside
βΉ No change: Xβ stays at 0 and Z stays 35
Making an unused input dearer can only make it less attractive.
- (4) C2 from 14 to 8
C2's RHS range: 0 to 16; 8 lies inside βΉ the same basis (only Xβ is used)
But the values change: 2Xβ = 8 βΉ Xβ = 4
ΞZ = dual Γ ΞRHS = 2.5 Γ (β6) = β15 βΉ Z = 35 β 15 = 20
So the basis does not change but the optimal quantity and cost do β say both.
Q3 β Max Z = 5Xβ + 5Xβ + 3Xβ
| Variable | Value | Reduced cost | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Xβ | 4.5 | 0 | 5 | 5 | 9 |
| Xβ | 1.5 | 0 | 5 | 3.67 | 5 |
| Xβ | 4 | 0 | 3 | 0 | 5 |
| Constraint | Dual value | Slack / surplus | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Constraint 1 | 1 | 0 | 12 | 10 | 15 |
| Constraint 2 | 3 | 0 | 10 | 6 | 12 |
| Constraint 3 | 0 | 0 | 15 | 9 | 18 |
| Constraint 4 | 0 | 6 | 15 | 9 | Infinity |
| Constraint 5 | 0 | 2 | 6 | 4 | Infinity |
Standard form, then: (1) impact of raising C4 from 15 to 25; (2) economic implication of raising C2 from 10 to 12; (3) does the solution change if C1 falls from 12 to 9; (4) if Xβ's profit rises from 5 to 8, must the problem be re-solved, and what happens to Z?
- Note on C3
The handout prints C3 as Xβ + Xβ β€ 15, but the report only fits 3Xβ + Xβ β€ 15 (the Assignment 2 Q5 constraint): 3(4.5) + 1.5 = 15 gives slack 0, as printed, whereas Xβ + Xβ = 6 would leave slack 9. Answer from the report. The solution is Xβ = 4.5, Xβ = 1.5, Xβ = 4, Z = 42 β and it is one of several optima (Assignment 2 Q5).
- Standard form Max Z = 5Xβ + 5Xβ + 3Xβ + 0Sβ + β¦ + 0Sβ 2Xβ + 2Xβ + Sβ = 12 Xβ + Xβ + Xβ + Sβ = 10 3Xβ + Xβ + Sβ = 15 Xβ + 3Xβ + Sβ = 15 Xβ + Sβ = 6 all variables β₯ 0
- (1) C4 from 15 to 25 C4 is non-binding: slack 6, dual value 0, range 9 to Infinity βΉ No impact on the solution or on Z; the slack simply grows to 16
- (2) C2 from 10 to 12
Dual value 3; range 6 to 12 β 12 is still inside (at the limit)
ΞZ = 3 Γ 2 = +6 βΉ Z = 48
Each extra unit of resource 2 is worth 3 in profit, so the firm should pay at most 3 per unit for more of it. Beyond 12 the shadow price no longer holds.
- (3) C1 from 12 to 9
C1's range is 10 to 15; 9 is BELOW the lower bound
βΉ The current basis is no longer optimal β the problem must be re-solved
Re-solving gives Xβ = 4.5, Xβ = 0, Xβ = 5.5, Z = 39: Xβ drops out of the solution, which is exactly the basis change the range warned about.
- (4) Xβ's profit from 5 to 8 Xβ's range of optimality is 5 to 9; 8 lies inside βΉ No need to re-solve: Xβ = 4.5, Xβ = 1.5, Xβ = 4 stay optimal ΞZ = 3 Γ 4.5 = 13.5 βΉ Z = 42 + 13.5 = 55.5
Q4 β Min Z = 5Xβ + 5Xβ + 3Xβ β binding constraints
| Variable | Value | Reduced cost | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Xβ | 2 | 0 | 5 | 5 | 10 |
| Xβ | 8 | 0 | 5 | 2.5 | 5 |
| Xβ | 4 | 0 | 3 | 0 | Infinity |
| Constraint | Dual value | Slack / surplus | Original value | Lower bound | Upper bound |
|---|---|---|---|---|---|
| Constraint 1 | β5 | 0 | 10 | 7.5 | 10 |
| Constraint 2 | 0 | 0 | 2 | βInfinity | 2 |
| Constraint 3 | 0 | 5 | 3 | βInfinity | 8 |
| Constraint 4 | β3 | 0 | 4 | 0 | Infinity |
| Constraint 5 | 0 | 0 | 12 | 12 | 17 |
Standard form, then: (1) which constraints are binding; (2) is C5 fully utilised; (3) how much of C3 is utilised; (4) the final answer and Z.
Assignment 3 p. 2
- Standard form Min Z = 5Xβ + 5Xβ + 3Xβ + 0Sβ + β¦ + 0Sβ Xβ + Xβ β Sβ = 10 Xβ β Sβ = 2 Xβ β Sβ = 3 Xβ β Sβ = 4 2Xβ + Xβ β Sβ = 12 all variables β₯ 0 (artificials Aββ¦Aβ are added for the simplex)
- (1) Binding constraints
C1: Xβ + Xβ = 10 β surplus 0 (dual β5)
C2: Xβ = 2 β surplus 0
C4: Xβ = 4 β surplus 0 (dual β3)
C5: 2(2) + 8 = 12 β surplus 0
Binding: C1, C2, C4, C5. Non-binding: C3
C2 and C5 are binding with a dual value of 0 β a sign of degeneracy (the same LP is Assignment 2 Q6).
- (2) C5
Yes β 2Xβ + Xβ = 2(2) + 8 = 12, exactly the requirement, surplus 0. But its dual value is 0, so tightening it slightly would not raise the cost.
- (3) C3 Xβ = 8 against a requirement of at least 3 βΉ exceeded by 5 (surplus 5)
- (4) Final answer
Xβ = 2, Xβ = 8, Xβ = 4, Min Z = 5(2) + 5(8) + 3(4) = 62
Because 5Xβ + 5Xβ is parallel to C1, any Xβ from 2 to 7 with Xβ = 10 β Xβ also gives 62.