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Every question from the 2025-26 Final and Re-Examination β all seven units, 30 parts β screenshotted from the original paper and worked in full, with the method, the arithmetic, and the way to lay the answer out on the page.
Each question shows the original screenshot from the paper, then a full solution, then a marks breakdown taken from the official synoptic answer key. Every solution here has been checked against that key β and where the key contains an arithmetic slip or an internal contradiction, it is flagged rather than copied.
The Units 4β7 questions (game theory, queuing, simulation, decision theory, AHP) were added for the Term End Examination, together with Re-Exam Q7A and Q7B. Three of the official keys for those parts contain errors β Re-Exam Q2, Q6B and Q7A β each flagged in a red card where it occurs.
Final Q1A β Primal to Dual 2.5 marks Β· Unit 1
Step 1 β Convert to canonical form. The primal is a Max problem, so every constraint must become β€.
| Original | Action | Canonical | Dual variable |
|---|---|---|---|
| Xβ + 2Xβ + Xβ β€ 8 | Already β€ | Xβ + 2Xβ + Xβ β€ 8 | Yβ |
| 2Xβ + Xβ + 3Xβ β₯ 10 | Multiply by β1 | β2Xβ β Xβ β 3Xβ β€ β10 | Yβ |
| Xβ + Xβ + Xβ = 6 | Split: the β€ half | Xβ + Xβ + Xβ β€ 6 | Yβ |
| Split: the β₯ half, Γ(β1) | βXβ β Xβ β Xβ β€ β6 | Yβ |
Three printed constraints, but the equality splits into two β so the dual has four variables. Counting them before you start is the single best guard against losing this mark.
Step 2 β Set up the dual. Max primal β Min dual. The primal's RHS values (8, β10, 6, β6) become the dual's objective coefficients:
Step 3 β One dual constraint per primal variable, reading down each column of the canonical form. Max primal βΉ dual constraints are β₯, with the primal objective coefficients as RHS.
| Primal variable | Column coefficients (Yβ, Yβ, Yβ, Yβ) | Dual constraint |
|---|---|---|
| Xβ | 1, β2, 1, β1 | Yβ β 2Yβ + Yβ β Yβ β₯ 2 |
| Xβ | 2, β1, 1, β1 | 2Yβ β Yβ + Yβ β Yβ β₯ 3 |
| Xβ | 1, β3, 1, β1 | Yβ β 3Yβ + Yβ β Yβ β₯ 4 |
Final answer β write exactly this:
β Matches the synoptic key exactly Key p. 1.
Re-Exam Q1A β Primal to Dual (Min version) 2.5 marks Β· Unit 1
Step 1 β Canonical form. This one is a Min problem, so every constraint must become β₯ β the opposite direction to the Final's version.
| Original | Action | Canonical | Dual variable |
|---|---|---|---|
| 3Xβ + 2Xβ β€ 12 | Multiply by β1 | β3Xβ β 2Xβ β₯ β12 | Yβ |
| Xβ + Xβ β₯ 4 | Already β₯ | Xβ + Xβ β₯ 4 | Yβ |
| Xβ β Xβ = 3 | Split: the β₯ half | Xβ β Xβ β₯ 3 | Yβ |
| Split: the β€ half, Γ(β1) | βXβ + Xβ β₯ β3 | Yβ |
Step 2 β Dual objective. Min primal β Max dual, RHS values become the coefficients:
Step 3 β Dual constraints, one per primal variable, direction β€:
| Primal variable | Column coefficients (Yβ, Yβ, Yβ, Yβ) | Dual constraint |
|---|---|---|
| Xβ | β3, 1, 1, β1 | β3Yβ + Yβ + Yβ β Yβ β€ 5 |
| Xβ | β2, 1, β1, 1 | β2Yβ + Yβ β Yβ + Yβ β€ 2 |
The printed key gives the first dual constraint as β3Yβ + 2Yβ + Yβ β Yβ β€ 5 Key p. 1. But the coefficient of Xβ in the primal constraint Xβ + Xβ β₯ 4 is 1, not 2 β so the dual coefficient on Yβ must be 1. The version above is the correct derivation. If you write +Yβ you are right; the key slipped.
Final Q1B β Assignment: balanced? variables? constraints? 2.5 marks Β· Unit 3
Identify the type first. There is no supply or demand row β every worker takes exactly one task. This is an assignment problem, m = 3 workers, n = 3 tasks.
| Sub-question | Answer | Justification to write |
|---|---|---|
| 1. Is the problem balanced? | Yes | For an assignment problem, balanced means number of rows = number of columns. Here 3 = 3, so it is balanced and no dummy is needed. |
| 2. Number of constraints (excluding NNC) | 6 | m + n = 3 + 3 = 6 β one constraint per worker (each does exactly one task) and one per task (each is done by exactly one worker). |
| 3. Number of variables | 9 | m Γ n = 3 Γ 3 = 9 β one binary variable Xα΅’β±Ό per cell of the cost matrix. |
β Matches the key: "Yes Problem is balanced as Number of rows = number of column; m+n = 6; mn = 9" Key p. 1.
Re-Exam Q1B β Transportation, and this one is unbalanced 2.5 marks Β· Unit 3
Identify the type. A supply column and a demand row are given βΉ transportation problem.
Balance test β always compute both totals explicitly:
| Sub-question | Answer | Justification to write |
|---|---|---|
| 1. Balanced? If not, how to balance? | No. Add a dummy row (source) with supply = 10 and all costs = 0. | Balanced requires Ξ£ supply = Ξ£ demand. Demand is short by 10 on the supply side, so a dummy source supplies the missing 10 units at zero cost β those units are never actually shipped. |
| 2. Number of constraints (excluding NNC) | 6 | m + n = 3 + 3 = 6, counted on the original matrix. |
| 3. Number of variables | 9 | m Γ n = 3 Γ 3 = 9. |
β Matches the key Key p. 1, which also counts m + n and m Γ n on the original 3 Γ 3 matrix rather than the balanced 4 Γ 3 one.
Final Q3B β Formulate an LPP (chairs and tables) 5 marks Β· Unit 1
Step 0 β build the resource table. Do this before writing anything else:
| Resource | Per chair (Xβ) | Per table (Xβ) | Available |
|---|---|---|---|
| Carpentry (hours) | 4 | 3 | 120 |
| Finishing (hours) | 2 | 3 | 90 |
| Profit (βΉ/unit) | 200 | 300 | β maximise |
The question says "Formulate" β so stop here. No graph, no simplex.
| Component | Marks |
|---|---|
| Correctly defining variables | 1 |
| Objective function formed properly | 1 |
| Two constraints formulated accurately | 2 |
| Non-negativity stated | 1 |
Read that breakdown carefully. Defining the variables and stating non-negativity are worth 2 of the 5 marks β two lines that take ten seconds. Never skip them.
Re-Exam Q3B β Formulate an LPP (two products, two raw materials) 5 marks Β· Unit 1
| Resource | Per P1 (Xβ) | Per P2 (Xβ) | Available |
|---|---|---|---|
| Raw material A | 5 | 3 | 150 |
| Raw material B | 3 | 6 | 180 |
| Profit (βΉ/unit) | 40 | 50 | β maximise |
β Matches the key exactly Key p. 3, including the same 1 / 1 / 2 / 1 mark split.
Final Q4A β Interpret the sensitivity report 5 marks Β· Unit 1
(1) Which constraints are binding? Give the evidence.
All three are binding. The test has two halves and you should quote both:
| Constraint | Slack / Surplus | Dual value | Verdict |
|---|---|---|---|
| Constraint 1 | 0 | 14.69 | Binding |
| Constraint 2 | 0 | 2.19 | Binding |
| Constraint 3 | 0 | 6.88 | Binding |
Zero slack means the resource is fully consumed; a non-zero shadow price means more of it would still be worth something. Both conditions hold for all three, so every resource is a genuine bottleneck.
(2) Interpret the shadow price 14.69 for Constraint 1 in economic terms.
(3) If the RHS of Constraint 2 increases by 10 units, what is the change in objective value? Is it valid?
Because the change is within range, the shadow price is still the right multiplier and the same variables remain basic.
(4) All variables have reduced cost = 0. What does that imply?
All three variables are basic β they are in the optimal solution at positive levels (Xβ = 36.25, Xβ = 35, Xβ = 21.25). A reduced cost of zero means no improvement to that variable's objective coefficient is needed to justify producing it; it is already worth producing.
A reduced cost of 0 on a non-basic variable would signal alternate optimal solutions. That is not the case here, because all three variables have non-zero values and so are basic. Saying that sentence out loud earns the mark Key p. 4.
(5) If the coefficient of Xβ rises from βΉ45 to βΉ55, does the basis stay optimal?
But the objective value does change. Add this line to separate a good answer from a full one:
Re-Exam Q4B β Sensitivity report with a non-binding constraint 5 marks Β· Unit 1
(1) Which constraints are binding?
| Constraint | Slack / Surplus | Dual value | Verdict |
|---|---|---|---|
| Material | 0 | 22 | Binding |
| Labour | 0 | 6 | Binding |
| Machine Hours | 40 | 0 | NOT binding |
Material and Labour are binding β slack = 0 and shadow price > 0. Machine Hours is not: it has 40 units of unused capacity and a shadow price of 0, so it is not limiting the solution at all.
(2) Interpret the shadow price of 22 for Material.
(3) If Machine Hours RHS increases by 20 units, will profit change?
No β profit is unchanged. Two reasons, both worth stating: the shadow price is 0, so ΞZ = 0 Γ 20 = 0; and there are already 40 unused machine hours. Adding more of a resource you are not fully using cannot help. Machine Hours is a non-binding constraint.
(4) The reduced cost of Xβ is 4 β interpret it.
Xβ has value 0: it is not produced in the optimal solution. The reduced cost of 4 can be stated two equivalent ways, and either earns the mark:
- Forcing one unit of Xβ into the solution would reduce total profit by βΉ4.
- Equivalently, Xβ's profit coefficient would have to rise by βΉ4 β from βΉ30 to βΉ34 β before it becomes worth producing.
Notice the internal check: Xβ's upper bound in the table is 34 = 30 + 4. The report is consistent with itself, and quoting that agreement is a strong touch.
(5) If the profit coefficient of Xβ increases by βΉ20, does the solution stay optimal?
Quantities stay Xβ = 30, Xβ = 40, Xβ = 0, but profit rises:
Final Q4B β Formulate an ILP (Maverick Analytics) 5 marks Β· Unit 2
Step 1 β spot the variable type. "Which projects should be undertaken" βΉ each project is a yes/no decision βΉ binary variables, not quantities.
Step 2 β objective function. Maximise total expected profit:
Step 3 β resource constraints. One per quarter, read straight down the table columns:
Step 4 β the logical constraint. This is the sentence buried at the end of the stem: "Projects C and E cannot be undertaken together."
Step 5 β the binary condition. Without this line the model is an LP, not an ILP:
| Component | Marks |
|---|---|
| Identification of decision variables | 1 |
| Formulation of objective function | 1 |
| Formulation of constraints | 2 |
| Binary condition + overall clarity of model structure | 1 |
The binary condition is a full mark on its own. Write it as its own labelled line.
Re-Exam Q4A β Formulate an ILP (Orion Tech Services) 5 marks Β· Unit 2
The stem says "Projects B and D cannot be executed together", but the table lists projects P, Q, R, S, T β there is no B or D. This is leftover wording from the Final paper's version (which did have projects AβE). The defensible reading is that B and D mean the second and fourth projects in the table, i.e. Q and S. State that assumption in one line β the paper's own instruction 7 says "Assume suitable data if necessary", so you are explicitly permitted to.
The synoptic key prints its Q4A answer as an image with no extractable text Key p. 3, but its published mark split is identical to the Final's: variables 1, objective 1, constraints 2, binary condition 1.
Final Q5B β Formulate an assignment problem (flight scheduling) 5 marks Β· Unit 3
Step 1 β work out when each outbound flight lands. Flight time is 2 hours 15 minutes.
| Flight (Hyd β Kol) | Departs | Arrives Kolkata |
|---|---|---|
| 101 | 8:00 AM | 10:15 AM |
| 102 | 1:00 PM | 3:15 PM |
| 103 | 5:30 PM | 7:45 PM |
| 104 | 9:00 PM | 11:15 PM |
| 105 | 11:30 PM | 1:45 AM (next day) |
Step 2 β build the idle-time (cost) matrix. Idle time = return departure β arrival, rolling to the next day whenever the return leaves before the aircraft lands.
| Arrives β / Returns β | 201 (10:30) | 202 (14:30) | 203 (19:45) | 204 (23:15) |
|---|---|---|---|---|
| 101 (10:15) | 0.25 | 4.25 | 9.5 | 13 |
| 102 (15:15) | 19.25 | 23.25 | 4.5 | 8 |
| 103 (19:45) | 14.75 | 18.75 | 24 | 3.5 |
| 104 (23:15) | 11.25 | 15.25 | 20.5 | 24 |
| 105 (01:45) | 8.75 | 12.75 | 18 | 21.5 |
Sample working: flight 102 lands at 15:15; return 203 leaves at 19:45; idle = 19:45 β 15:15 = 4 h 30 m = 4.5. Flight 104 lands 23:15; return 202 leaves 14:30 the next day; idle = 15 h 15 m = 15.25.
Step 3 β balance the problem. There are 5 outbound flights but only 4 returns β the stem tells you flight 105 has no direct return option. So:
Step 4 β write the assignment LPP.
The key prints two idle-time tables that disagree with each other Key p. 5. Checked against the timetable, the correct values are the ones above; the key's errors are:
- Cell 102 β 203: key says 4.25. Arrival 15:15, departure 19:45 βΉ 4.5.
- Cell 104 β 202: key's two tables say 14.25 and 3.25. Arrival 23:15, departure 14:30 next day βΉ 15.25.
Re-Exam Q5B β Assignment with a turnaround requirement 5 marks Β· Unit 3
Step 1 β arrival times. Flight time 2.5 hours.
| Flight (Del β Che) | Departs | Arrives Chennai |
|---|---|---|
| 201 | 7:00 AM | 9:30 AM |
| 202 | 11:30 AM | 2:00 PM |
| 203 | 3:00 PM | 5:30 PM |
| 204 | 6:30 PM | 9:00 PM |
Step 2 β idle time, net of the 30-minute turnaround. This is what makes this version harder than the Final's. A pairing is only feasible if the gap is more than the 30-minute turnaround; otherwise the aircraft must wait until the same flight the next day. The cost recorded is the idle time beyond the required turnaround, i.e. gap β 0.5.
| Arrives β / Returns β | 301 (9:30) | 302 (12:30) | 303 (18:00) | 304 (21:30) |
|---|---|---|---|---|
| 201 (9:30) | 23.5 | 2.5 | 8 | 11.5 |
| 202 (14:00) | 19 | 22 | 3.5 | 7 |
| 203 (17:30) | 15.5 | 18.5 | 24 | 3.5 |
| 204 (21:00) | 12 | 15 | 20.5 | 24 |
Sample working: 202 lands 14:00, return 303 departs 18:00 βΉ gap 4 h, minus the 0.5 h turnaround = 3.5. Where the gap is only 0.5 h β 201β301 and 203β303 and 204β304 β the turnaround exactly consumes it, so the key rolls the aircraft to the same service next day, giving 24. β Every cell above matches the key Key p. 5.
Step 3 β the formulation. Here it is 4 Γ 4, already balanced β no dummy needed.
Final Q7A β Identify the special case from the final tableau 5 marks Β· Unit 1
(1) The special case is DEGENERACY.
(2) Justification β three observations, each pointing at a specific part of the tableau:
| Where to look | What you see | What it proves |
|---|---|---|
| Quantity (RHS) column | slack 2 is a basic variable with Quantity = 0 |
A basic variable at zero level is the definition of a degenerate basic feasible solution. |
| Cβ±Ό β Zβ±Ό row | All values β€ 0 (0, β2, β4, β10, 0, 0, 0) | For a maximisation problem this is the optimality condition β so the tableau is already optimal. Hence degeneracy at optimality. |
| Cβ±Ό β Zβ±Ό, non-basic columns | Xβ = β2, Xβ = β4, Sβ = β10 β all strictly negative | Rules out multiple optimal solutions, which would need a non-basic variable at exactly 0. This is the distinction the question is testing. |
(3) One important implication for the simplex method:
β Matches the key's three-part structure β identification, justification, implication β almost word for word Key p. 7.
Under time pressure it is easy to see "a zero in the Cβ±Ό β Zβ±Ό row" and answer multiple optima. But those zeros sit in the basic columns (Xβ, Sβ, Sβ, Sβ), where a zero is expected and means nothing. The zero that matters for multiple optima must be on a non-basic variable. Here the zero that matters is in the Quantity column instead β different row, different diagnosis.
Final Q7B β Transportation format with a profit twist 5 marks Β· Unit 3
Step 1 β spot the twist. This is not a plain cost-minimisation. You are given plant costs, transportation costs and selling prices β so the cell value must be a profit, and the objective becomes Max.
Step 2 β build the profit matrix. Compute cell by cell:
| Cell | Working | Profit |
|---|---|---|
| P1 β A | 25 β 14 β 4 | 7 |
| P2 β A | 25 β 12 β 5 | 8 |
| P3 β A | 25 β 10 β 6 | 9 |
| P1 β B | 28 β 14 β 6 | 8 |
| P2 β B | 28 β 12 β 8 | 8 |
| P3 β B | 28 β 10 β 5 | 13 |
| P1 β C | 24 β 14 β 5 | 5 |
| P2 β C | 24 β 12 β 7 | 5 |
| P3 β C | 24 β 10 β 6 | 8 |
| P1 β D | 30 β 14 β 7 | 9 |
| P2 β D | 30 β 12 β 6 | 12 |
| P3 β D | 30 β 10 β 8 | 12 |
Step 3 β lay it out in transportation format.
| Plant | A | B | C | D | Supply |
|---|---|---|---|---|---|
| P1 | 7 | 8 | 5 | 9 | 100 |
| P2 | 8 | 8 | 5 | 12 | 150 |
| P3 | 9 | 13 | 8 | 12 | 200 |
| Demand | 120 | 150 | 100 | 80 | 450 / 450 |
β This profit matrix matches the synoptic key cell for cell Key p. 8.
Step 4 β balance check and the counts the question asks for.
Step 5 β the LPP.
Seven constraints β exactly the m + n you computed. Because supply and demand are perfectly balanced, the supply constraints may be written as = rather than β€.
Re-Exam Q7A β Transportation to maximise profit, unbalanced 5 marks Β· Unit 3
- Step 1 β Total cost per unit at each plant P1 = 12 + 7 = βΉ19, P2 = 10 + 8 = βΉ18, P3 = 8 + 7 = βΉ15, P4 = 7 + 5 = βΉ12
- Step 2 β Profit per unit = sales price β plant cost β transport
Plant β / Warehouse β A (βΉ30) B (βΉ32) C (βΉ28) D (βΉ34) E (βΉ30) Capacity P1 (cost βΉ19) 7 5 7 5 9 100 P2 (cost βΉ18) 5 5 3 9 7 200 P3 (cost βΉ15) 11 12 10 17 7 120 P4 (cost βΉ12) 15 15 8 18 15 80 Dummy 0 0 0 0 0 10 Demand 80 120 150 70 90 510 Sample: P3 β D = 34 β 15 β 2 = 17. P4 β A = 30 β 12 β 3 = 15.
- Step 3 β Balanced? Ξ£ capacity = 100 + 200 + 120 + 80 = 500; Ξ£ demand = 80 + 120 + 150 + 70 + 90 = 510 Not balanced β demand exceeds supply by 10 βΉ add a dummy plant with capacity 10 and profit 0
- Step 4 β LPP form Xα΅’β±Ό = units from plant i (P1βP4, dummy) to warehouse j (AβE) Max Z = 7XβA + 5XβB + 7XβC + 5XβD + 9XβE + 5XβA + β¦ + 15XβE Ξ£β±Ό Xα΅’β±Ό = capacity of i (5 rows); Ξ£α΅’ Xα΅’β±Ό = demand of j (5 columns); Xα΅’β±Ό β₯ 0 Variables 5 Γ 5 = 25 (20 before balancing); constraints 5 + 5 = 10 (9 before)
- Going further β the optimal plan Maximum profit βΉ4790: P1βC 100; P2βA 70; P2βC 40; P2βE 90; P3βB 50; P3βD 70; P4βA 10; P4βB 70; DummyβC 10
The key's Q7A table is the plain cost table 6, 4, 6, 4, 8 / 6, 6, 4, 10, 8 / β¦ with a dummy row of 10 Key p. 7 β that is the transportation problem from Assignment 6, not this question's profit data. Its balancing idea (dummy supply of 10) is right; the matrix to write is the profit matrix above.
Re-Exam Q7B β Identify the special case (multiple optima) 5 marks Β· Unit 1
(1) The special case is MULTIPLE (ALTERNATIVE) OPTIMAL SOLUTIONS.
| Where to look | What you see | What it proves |
|---|---|---|
| Cβ±Ό β Zβ±Ό row | 0, 0, β1, 0 β none positive | Max problem βΉ optimal at Z = 10 |
| Non-basic Xβ | Xβ is not in the basis, yet its Cβ±Ό β Zβ±Ό = 0 | Entering Xβ changes the solution but not Z βΉ another optimum exists |
| Quantity column | Xβ = 5, slack 2 = 3 β both positive | Not degenerate; no artificial βΉ not infeasible |
(2) The other optimum: enter Xβ. Ratios 5 Γ· 0.5 = 10 (Xβ row) and 3 Γ· 0.5 = 6 (slack-2 row) βΉ slack 2 leaves at ΞΈ = 6, giving Xβ = 6 and Xβ = 5 β 0.5 Γ 6 = 2. Both (5, 0) and (2, 6) give Z = 2Xβ + Xβ = 10, and so does every point on the segment between them.
(3) Implication for the decision-maker: there is a free choice among several plans with the same profit, so management can use a secondary criterion β easier scheduling, fewer products, more balanced resource use. β Matches the key Key p. 7. (The tableau's last column is labelled "S1" twice; the second is Sβ.)
Final Q1C β AHP: which city for the headquarters? 2.5 marks Β· Unit 7
- Step 1 β Complete with reciprocals and add the columns (0.5 mark)
Ahmedabad Hyderabad Delhi Ahmedabad 1 5 7 Hyderabad 0.2 1 3 Delhi 0.143 0.333 1 Sum 1.343 6.333 11 - Step 2 β Normalise each column and average each row (2 marks)
Ahmedabad Hyderabad Delhi Priority Ahmedabad 0.745 0.789 0.636 0.724 Hyderabad 0.149 0.158 0.273 0.193 Delhi 0.106 0.053 0.091 0.083 - Step 3 β Interpret
(1) Hyderabad is rated 1/5 against Ahmedabad (Ahmedabad strongly preferred over it) and 3 against Delhi (Hyderabad moderately preferred over Delhi). (2) Ahmedabad is the most strategically preferred city (0.724); Delhi the least (0.083). β Matches the key Key p. 1. Consistency ratio 0.057 < 0.10.
Re-Exam Q1C β AHP from verbal judgements 2.5 marks Β· Unit 7
- Step 1 β Translate the judgements with the printed scale
Judgement Rating Mirror cell FizzUp moderately preferable to CoolMist 3 CoolMist vs FizzUp = 1/3 FizzUp equally-to-moderately preferable to Zestora 2 Zestora vs FizzUp = 1/2 CoolMist strongly preferable to Zestora 5 Zestora vs CoolMist = 1/5 - Step 2 β Pairwise matrix and column sums (0.5 mark)
FizzUp CoolMist Zestora FizzUp 1 3 2 CoolMist 0.333 1 5 Zestora 0.5 0.2 1 Sum 1.833 4.2 8 - Step 3 β Normalise and average (2 marks)
FizzUp CoolMist Zestora Priority FizzUp 0.545 0.714 0.25 0.503 CoolMist 0.182 0.238 0.625 0.348 Zestora 0.273 0.048 0.125 0.148 - Step 4 β Interpret
FizzUp has the highest flavour priority (0.503), then CoolMist (0.348), then Zestora (0.149). β Matches the key Key p. 1. One line worth adding: the consistency ratio is 0.41 β well above 0.10 β because FizzUp β» CoolMist (3) and CoolMist β» Zestora (5) imply FizzUp β»β» Zestora, yet the panel said only 2. The panel's judgements should be revisited.
Final Q1D β Payoff matrix, Maximin and Maximax 2.5 marks Β· Unit 7
- Step 1 β Payoff matrix (0.5 mark)
Strategy High Moderate Low Row max Row min Mass production (Aβ) 120 70 β40 120 β40 Custom manufacturing (Aβ) 100 85 20 100 20 Limited edition (Aβ) 80 60 40 80 40 - Step 2 β Criteria (2 marks)
Maximin = max(β40, 20, 40) = 40 βΉ Limited edition (Aβ)
Maximax = max(120, 100, 80) = 120 βΉ Mass production (Aβ)
A pessimist protects against the β40 loss and picks the limited edition; an optimist chases the 120 and picks mass production. β Key p. 2.
Re-Exam Q1D β The Laplace principle 2.5 marks Β· Unit 7
- Step 1 β Payoff matrix
Strategy High (Sβ) Moderate (Sβ) Low (Sβ) Average Online campaign (Aβ) 90 60 30 60 In-store promotions (Aβ) 70 50 40 53.33 Mixed approach (Aβ) 80 65 35 60 - Step 2 β How Laplace works
With no probabilities available, Laplace treats every state as equally likely (1/3 each), so the expected payoff of each strategy is the simple average of its row; choose the highest.
- Step 3 β Answer
Online campaign and Mixed approach tie at 60 β either is preferred under Laplace
State the tie β the key accepts both Key p. 2. A tie-breaker if asked: the mixed approach has the better worst case (35 vs 30).
Final Q2 β Simulate a single-counter queue 10 marks Β· Unit 6
- Step 1 β Random-number ranges (00β99)
Inter-arrival (min) Probability Cumulative Random-number range 1 0.2 0.2 00 β 19 2 0.3 0.5 20 β 49 3 0.3 0.8 50 β 79 4 0.1 0.9 80 β 89 5 0.1 1 90 β 99 Service (min) Probability Cumulative Random-number range 1 0.2 0.2 00 β 19 3 0.5 0.7 20 β 69 5 0.3 1 70 β 99 - Step 2 β Simulation table (customer 1 arrives at 0)
Cust RN (arrival) Inter-arrival Arrival time RN (service) Service time Service starts Waiting Service ends Time in system Server idle 1 β 0 0 28 3 0 0 3 3 0 2 61 3 3 01 1 3 0 4 1 0 3 55 3 6 61 3 6 0 9 3 2 4 01 1 7 85 5 9 2 14 7 0 5 33 2 9 67 3 14 5 17 8 0 6 19 1 10 53 3 17 7 20 10 0 7 25 2 12 62 3 20 8 23 11 0 8 79 3 15 79 5 23 8 28 13 0 9 93 5 20 66 3 28 8 31 11 0 10 18 1 21 63 3 31 10 34 13 0 11 49 2 23 33 3 34 11 37 14 0 12 92 5 28 77 5 37 9 42 14 0 - Step 3 β The five answers (1 mark each)
1. Service start of customer 11 = minute 34
2. Waiting time of customer 5 = 5 minutes
3. Arrival time of customer 4 = minute 7
4. Time in system of customer 9 = 11 minutes
5. Service end of customer 7 = minute 23
β All five match the key Key p. 2. Method in detail: Unit 6 Β· Simulating a Queue.
Re-Exam Q2 β Simulate production against a 152-scooter lorry 10 marks Β· Unit 6
- Step 1 β Random-number ranges
Production Probability Cumulative Random-number range 146 0.04 0.04 00 β 03 147 0.09 0.13 04 β 12 148 0.12 0.25 13 β 24 149 0.14 0.39 25 β 38 150 0.11 0.5 39 β 49 151 0.1 0.6 50 β 59 152 0.2 0.8 60 β 79 153 0.12 0.92 80 β 91 154 0.08 1 92 β 99 - Step 2 β Simulate 15 days, carrying unshipped scooters forward
Day RN Production Waiting from yesterday Available Shipped Left waiting Empty spaces 1 80 153 0 153 152 1 0 2 81 153 1 154 152 2 0 3 76 152 2 154 152 2 0 4 75 152 2 154 152 2 0 5 64 152 2 154 152 2 0 6 43 150 2 152 152 0 0 7 18 148 0 148 148 0 4 8 26 149 0 149 149 0 3 9 10 147 0 147 147 0 5 10 12 147 0 147 147 0 5 11 65 152 0 152 152 0 0 12 68 152 0 152 152 0 0 13 69 152 0 152 152 0 0 14 61 152 0 152 152 0 0 15 57 151 0 151 151 0 1 - Step 3 β Averages Average scooters waiting = 9/15 = 0.6 per day Average empty spaces = 18/15 = 1.2 per day
The key carries the backlog forward on day 2 (2 waiting) and on day 6 (0 empty spaces), but shows 0 waiting on days 3β5, where exactly 152 are made and nothing clears the 2 already waiting Key p. 2. Applied consistently, waiting totals 9 (average 0.6), not 3 (0.2). Its empty-space figure of 18/15 = 1.2 is correct.
Final Q3A β SpeedEx vs FastMove payoff matrix 5 marks Β· Unit 4
- Step 1 β Assumptions (1 mark)
Two-person zero-sum game; SpeedEx is the row player, so each cell is SpeedEx's profit and FastMove's is its negative. Equal strategies βΉ random award βΉ 0.
- Step 2 β Payoff matrix (2 + 1 marks)
SpeedEx β / FastMove β C (Conservative) A (Aggressive) Row min C (Conservative) 0 β8 β8 A (Aggressive) +8 0 0 Column max 8 0 - Step 3 β Interpretation (1 mark)
Maximin = Minimax = 0 βΉ saddle point at (A, A), value V = 0
Each firm's best strategy is aggressive β it is never worse than the alternative whatever the rival does β and with V = 0 the game is fair. β Same matrix as the key Key p. 3.
Re-Exam Q3A β NetLink vs AirWave payoff matrix 5 marks Β· Unit 4
- Step 1 β Assumptions (1 mark)
Two-person zero-sum game; NetLink is the row player, so each cell is NetLink's profit and AirWave's is its negative. Equal strategies βΉ random award βΉ 0.
- Step 2 β Payoff matrix (2 + 1 marks)
NetLink β / AirWave β P (Premium) D (Discount) Row min P (Premium) 0 β7 β7 D (Discount) +7 0 0 Column max 7 0 - Step 3 β Interpretation (1 mark)
Maximin = Minimax = 0 βΉ saddle point at (D, D), value V = 0
Each firm's best strategy is discount β it is never worse than the alternative whatever the rival does β and with V = 0 the game is fair. β Same matrix as the key Key p. 3.
Final Q5A β Broker's desk: traffic intensity and queue parameters 5 marks Β· Unit 5
- Step 1 β Rates Ξ» = 20 per hour; service 2 min βΉ ΞΌ = 30 per hour
- Step 2 β (1) Traffic intensity Ο = 20/30 = 0.667
- Step 3 β (2) Queue parameters Ls = 20/(30 β 20) = 2 Lq = 400/(30 Γ 10) = 1.33 Ws = 1/10 hr = 6 min Wq = 20/300 hr = 4 min
- Step 4 β (3) Ξ» = 25 per hour
Ο = 0.833; Ws = 1/5 hr = 12 min; Wq = 25/150 hr = 10 min
Arrivals up 25% βΉ queue wait up 150% (4 β 10 min) and time in system doubles (6 β 12 min): the desk is approaching saturation. β Key p. 5.
Re-Exam Q5A β Mall billing counter and peak-hour stability 5 marks Β· Unit 5
- Step 1 β Rates Ξ» = 10 per hour; 1 customer per minute βΉ ΞΌ = 60 per hour
- Step 2 β (1) P(counter busy) Ο = 10/60 = 0.167
- Step 3 β (2) Queue parameters Ls = 10/50 = 0.2 Lq = 100/(60 Γ 50) = 0.033 Ws = 1/50 hr = 1.2 min Wq = 10/3000 hr = 0.2 min
- Step 4 β (3) Peak Ξ» = 20 per hour
Ο = 0.333 < 1 βΉ still stable
Ws = 1/40 hr = 1.5 min; Wq = 20/2400 hr = 0.5 min
The counter stays stable (Ο well below 1), but the queue wait rises 2.5Γ. Only if Ξ» approached 60 per hour would the queue grow without limit. β Key p. 5 (it rounds Ws to 1.48 min; 1.5 is exact).
Final Q6A β Decision tree and EMV 5 marks Β· Unit 7
- Step 1 β Decision tree
Decision node β , chance nodes β with EMVs; the kept branch is solid
- Step 2 β EMVs (backward pass) S1 = 20(0.1) + 60(0.5) + 60(0.4) = 56 S2 = β10(0.1) + 70(0.5) + 100(0.4) = 74 S3 = β140(0.1) + 30(0.5) + 210(0.4) = 85
- Step 3 β Decision and justification
Construct new facilities (S3), EMV 85
Highest EMV. It carries the only large loss (β140) but with just a 10% chance; with high demand at 40% the 210 dominates. β Key p. 6.
Re-Exam Q6A β BrightStar Publishing: decision tree with an agency report 5 marks Β· Unit 7
- (1) The decision tree (3 marks)
Three first-stage choices; the agency branch leads to a report (F/U) and then the same decision again ($'000)
- (2) Without the agency (2 marks)
EV(d1 publish) = 0.20(β100) + 0.30(50) + 0.50(150) = β20 + 15 + 75 = 70
EV(d2 sell) = 100 in every state = 100
Recommended: sell the rights to the competitor β expected value $100,000
β Key p. 6.
- Going further β is the agency worth $5,000? Favourable: d1 = 0.09(β100) + 0.26(50) + 0.65(150) = 101.5 > 100 βΉ publish Unfavourable: d1 = 0.45(β100) + 0.39(50) + 0.16(150) = β1.5 < 100 βΉ sell EV with agency = 0.69(101.5) + 0.31(100) = 101.04; EVSI = 1.04 ($1,040) Net of the $5,000 fee = 96.04 < 100 βΉ do not hire the agency
Final Q6B β Hurwicz with Ξ± = 0.6 5 marks Β· Unit 7
- Step 1 β Row max, row min, weighted value
Strategy Max Min 0.6Β·Max + 0.4Β·Min Social media (M1) 120 20 80 Email (M2) 100 30 72 Influencer (M3) 150 10 94 - Step 2 β Decision
Influencer partnerships (M3), Hurwicz value 94
β Key p. 7.
Re-Exam Q6B β Hurwicz with Ξ± = 0.7 (the key is wrong) 5 marks Β· Unit 7
- Step 1 β Row max, row min, weighted value
Option Max Min 0.7Β·Max + 0.3Β·Min Upgrading (U1) 180 40 126 + 12 = 138 Training (U2) 140 60 98 + 18 = 116 Outsourcing (U3) 160 50 112 + 15 = 127 - Step 2 β Decision Upgrade the machinery (U1), Hurwicz value 138
The key lists Max/Min as 120/20, 100/30, 150/10 β the Final paper's Q6B data β and concludes Outsourcing with 108 Key p. 6. With this paper's own table the answer is U1, 138. If you meet the key's numbers in revision, recognise them as a copy error.
| Slot | What it always is | Unit | Marks |
|---|---|---|---|
| Q1A | Primal β Dual conversion, always with one = and one wrong-way constraint | 1 | 2.5 |
| Q1B | Cost matrix β balanced? Β· m + n Β· m Γ n | 3 | 2.5 |
| Q1C | AHP priorities from a 3 Γ 3 pairwise matrix | 7 | 2.5 |
| Q1D | Build a payoff matrix; Maximin/Maximax or Laplace | 7 | 2.5 |
| Q2 | Monte Carlo simulation β a queue or a production/lorry problem | 6 | 10 |
| Q3A | Formulate a 2 Γ 2 zero-sum payoff matrix | 4 | 5 |
| Q3B | Formulate an LPP from a two-product, two-resource story | 1 | 5 |
| Q4 (one part) | Interpret a sensitivity report β 5 sub-questions | 1 | 5 |
| Q4 (other part) | Formulate an ILP β project selection + one logical constraint | 2 | 5 |
| Q5A | M/M/1 queue β Ο, all parameters, effect of more arrivals | 5 | 5 |
| Q5B | Flight-pairing assignment problem | 3 | 5 |
| Q6A | Decision tree with EMV (the Re-Exam adds survey information) | 7 | 5 |
| Q6B | Hurwicz criterion with a given Ξ± | 7 | 5 |
| Q7A / Q7B | Special case from a final tableau Β· transportation with a profit twist | 1, 3 | 5 + 5 |
Both papers had exactly this shape. Q1 (10 marks, compulsory) mixes four units; then you choose 4 of Q2βQ7. Units 1β3 can supply all four optional answers, but the safest plan for the Term End is to be ready for Q2 simulation (10 marks, pure procedure) and Q6 decision theory as well β both are mechanical once practised, and together with Q1 they are 30 marks.