Unit 03 β€” Transportation & Assignment

Recognising the two problem types, checking whether they are balanced, counting variables and constraints, and converting the cost matrix into a full LPP.

SESSIONS: 4 TEXTBOOK: ANDERSON & SWEENEY β€” CH. 6, SEC. 6.1 pp. 259–265, SEC. 6.2 pp. 274–278 SCOPE: FORMULATION & BALANCE β€” SEE SCOPE NOTE Full PPT: Assignment class notes 6 Note 2 β€” Transportation 23
⚠ Scope note β€” read this before you revise

The course session plan lists the full Unit 3 as including North-West Corner Rule, Least Cost Method, VAM, the Modified Distribution (MODI) method, degeneracy, maximisation problems and the Hungarian Method Policy p. 7.

Your notes do not cover any of those solution algorithms. Every page of the Unit 3 material β€” the handwritten Assignment Problems set pp. 1–6 and the transportation section of Note 2 pp. 12–18 β€” stops at formulation: identify the type, check balance, count variables and constraints, write the LPP.

This page therefore covers formulation thoroughly and does not invent algorithm content you were not taught. That is also what the papers asked: both 2025-26 Q1B items and both Q5B/Q7B items are pure formulation and balance questions. If your class later covers NWCR/VAM/MODI/Hungarian, tell me and I will add them.

β—† What the papers ask from this unit
  • Q1B (2.5 marks), both papers β€” a cost matrix, then three fixed sub-questions: Is the problem balanced? Number of constraints (excluding NNC)? Number of variables? Final p. 1 Re-Exam p. 1. This is the most formulaic 2.5 marks on the paper. Solved β†’
  • Q5B (5 marks), both papers β€” flight scheduling: "formulate an assignment problem to minimise aircraft overnight idle time". Solved β†’
  • Q7B (5 marks), Final β€” "formulate in transportation format and identify the number of constraints and variables", with a profit twist. Solved β†’

1. Telling the Two Problems Apart Core syllabus concept

1Understand the Concept

Both problems arrive as a cost matrix. One sentence separates them Class notes p. 3:

If demand and supply are given, it is a transportation problem. If instead every row must be matched one-to-one with exactly one column, it is an assignment problem.
Transportation problemAssignment problem
Rows / columnsSources (plants, warehouses) Γ— Destinations (markets, regions)Workers / machines / flights Γ— Tasks / jobs / return flights
Extra dataA supply column and a demand rowNone β€” just the cost matrix
Balanced whenTotal supply = total demandNumber of rows (m) = number of columns (n)
Xα΅’β±Ό meansQuantity shipped from i to j (any non-negative number)1 if row i is assigned to column j, else 0
Row constraintΞ£β±Ό Xα΅’β±Ό ≀ supplyα΅’Ξ£β±Ό Xα΅’β±Ό = 1 (each worker gets exactly one task)
Column constraintΞ£α΅’ Xα΅’β±Ό = demandβ±ΌΞ£α΅’ Xα΅’β±Ό = 1 (each task gets exactly one worker)
Sources
Class Assignment vs transportation, cost matrix pp. 1–4 Β· Note 2 pp. 12–14

2. Balance, Variables and Constraints Core syllabus concept

1The three counting rules

Q1B in both papers is exactly these three rules applied to an m Γ— n matrix. Memorise them as a block:

Number of VARIABLES = m Γ— n (one Xα΅’β±Ό per cell of the matrix) Number of CONSTRAINTS = m + n (one per row + one per column, excluding NNC) BALANCED? Assignment β†’ m = n Transportation β†’ Ξ£ supply = Ξ£ demand

2How to balance an unbalanced problem

Whichever side is less, add a corresponding DUMMY row or column, and set every cost in the dummy to zero Class notes p. 1.

  • Assignment: if rows β‰  columns, add a dummy row (or column) of zeros so the matrix becomes square.
  • Transportation: if Ξ£ supply β‰  Ξ£ demand, add a dummy source (if supply is short) or dummy destination (if demand is short), with zero costs and the missing quantity as its supply/demand p. 3.
Why zero costs: a dummy represents goods that are never actually shipped, or a task never actually performed. Charging it nothing keeps the objective value honest.

3Worked β€” the exact Q1B pattern, both versions

Final Exam Q1B β€” assignment, 3 workers Γ— 3 tasks QP p. 1
T1T2T3
W1927
W2643
W3581

Answer: (1) Yes, balanced β€” number of rows = number of columns = 3. (2) Constraints m + n = 3 + 3 = 6. (3) Variables m Γ— n = 3 Γ— 3 = 9. Matches the synoptic key exactly Key p. 1.

Re-Exam Q1B β€” transportation, and this one is not balanced QP p. 1
P1P2P3Supply
W192730
W264325
W358120
Demand203035

Answer: Ξ£ supply = 30 + 25 + 20 = 75; Ξ£ demand = 20 + 30 + 35 = 85. (1) No, not balanced β€” demand exceeds supply by 10, so add a dummy row (source) with supply 10 and zero costs. (2) Constraints m + n = 3 + 3 = 6. (3) Variables m Γ— n = 3 Γ— 3 = 9. Matches the key Key p. 1.

⚠ The counting subtlety

The synoptic key counts m + n and m Γ— n on the original matrix, not on the balanced one. Answer the counting parts from the matrix as given, and mention the dummy separately when asked how to balance it. Also note the phrase "excluding NNC" in the question β€” the non-negativity restriction is not counted as a constraint.

Practice H1–H2 β€” one unbalanced assignment, one unbalanced transportation β†’

Sources
Class Balance rule, dummy rows, counting pp. 1, 3 Β· Note 2 pp. 12–13
Exam Final Q1B p. 1 + key p. 1 Β· Re-Exam Q1B p. 1 + key p. 1

3. Converting the Matrix into an LPP Core syllabus concept

1The template β€” assignment problem

4 employees Γ— 4 tasks, the class example Class notes pp. 1–2
EmployeesTask 1Task 2Task 3Task 4
A1012815
B8101211
C1215913
D1191410
Solution

Step 1 β€” housekeeping (write these four lines first, they carry marks):

Balanced problem : Yes (m = n = 4) No. of variables : m Γ— n = 4 Γ— 4 = 16 No. of constraints: m + n = 4 + 4 = 8 Decision variable: Xα΅’β±Ό where i β†’ row (employee A to D), j β†’ column (task 1 to 4) Xα΅’β±Ό = 1 if employee i is assigned task j, else 0

Step 2 β€” objective function. Read the matrix left to right, row by row:

Min Z = 10X₁₁ + 12X₁₂ + 8X₁₃ + 15X₁₄ + 8X₂₁ + 10Xβ‚‚β‚‚ + 12X₂₃ + 11Xβ‚‚β‚„ + 12X₃₁ + 15X₃₂ + 9X₃₃ + 13X₃₄ + 11X₄₁ + 9Xβ‚„β‚‚ + 14X₄₃ + 10Xβ‚„β‚„

Step 3 β€” constraints. Each employee gets exactly one task; each task goes to exactly one employee.

Row constraints (one task per employee): X₁₁ + X₁₂ + X₁₃ + X₁₄ = 1 X₂₁ + Xβ‚‚β‚‚ + X₂₃ + Xβ‚‚β‚„ = 1 X₃₁ + X₃₂ + X₃₃ + X₃₄ = 1 X₄₁ + Xβ‚„β‚‚ + X₄₃ + Xβ‚„β‚„ = 1 Column constraints (one employee per task): X₁₁ + X₂₁ + X₃₁ + X₄₁ = 1 X₁₂ + Xβ‚‚β‚‚ + X₃₂ + Xβ‚„β‚‚ = 1 X₁₃ + X₂₃ + X₃₃ + X₄₃ = 1 X₁₄ + Xβ‚‚β‚„ + X₃₄ + Xβ‚„β‚„ = 1 NNC: Xα΅’β±Ό β‰₯ 0 (and binary, 0 or 1)

Eight constraints β€” exactly the m + n you counted in Step 1. That internal check is worth doing out loud in the answer.

2The template β€” transportation problem

3 origins Γ— 4 destinations, the class example Class notes pp. 3–4
D1D2D3D4Supply
O₁641514
Oβ‚‚892716
O₃43625
Demand61015435 / 35
Solution

Step 1 β€” housekeeping.

Balanced : Yes β€” Ξ£ supply = 14 + 16 + 5 = 35 = Ξ£ demand = 6 + 10 + 15 + 4 = 35 Decision variable: Xα΅’β±Ό = quantity shipped from origin i (1–3) to destination j (1–4) No. of variables : m Γ— n = 3 Γ— 4 = 12 No. of constraints: m + n = 3 + 4 = 7

Step 2 β€” objective function.

Min Z = 6X₁₁ + 4X₁₂ + 1X₁₃ + 5X₁₄ + 8X₂₁ + 9Xβ‚‚β‚‚ + 2X₂₃ + 7Xβ‚‚β‚„ + 4X₃₁ + 3X₃₂ + 6X₃₃ + 2X₃₄

Step 3 β€” constraints. Note the asymmetry: supply is a ceiling (≀), demand must be met exactly (=).

Row (supply) constraints: X₁₁ + X₁₂ + X₁₃ + X₁₄ ≀ 14 X₂₁ + Xβ‚‚β‚‚ + X₂₃ + Xβ‚‚β‚„ ≀ 16 X₃₁ + X₃₂ + X₃₃ + X₃₄ ≀ 5 Column (demand) constraints: X₁₁ + X₂₁ + X₃₁ = 6 X₁₂ + Xβ‚‚β‚‚ + X₃₂ = 10 X₁₃ + X₂₃ + X₃₃ = 15 X₁₄ + Xβ‚‚β‚„ + X₃₄ = 4 NNC: Xα΅’β±Ό β‰₯ 0

Seven constraints = m + n. When the problem is perfectly balanced you may write the supply constraints as = too; the class notes use ≀, which is the safer general form.

⚠ Unit 3 β€” where marks are lost
  • Writing Xα΅’β±Ό without saying what i and j range over. "i β†’ row 1 to 3, j β†’ column 1 to 4" is a marked line.
  • Mixing up the row and column constraint directions. Supply ≀, demand =. For assignment, both are = 1.
  • Transposing the cost matrix in the objective. X₁₃ is row 1, column 3 β€” read row-wise and it is impossible to slip.
  • Answering "how many constraints" with m Γ— n. Variables are m Γ— n; constraints are m + n. They are swapped surprisingly often under time pressure.

Practice H3–H5 β€” full LPP conversion, profit variant, forbidden pairing β†’

Sources
Class Full LPP conversion for both problem types pp. 1–4 Β· Note 2 pp. 12–18
Course Session plan for Unit 3 Policy p. 7
Exam Final Q5B, Q7B pp. 4, 6 Β· Re-Exam Q5B p. 3