Assignment vs Transportation Problem: Differences + Example

Last updated: June 6, 2026 • Sources • Methodology

Assignment vs Transportation Problem

Assignment is a one-to-one matching model (each row/column used exactly once). Transportation ships quantities from sources to destinations to satisfy supply and demand (many-to-many). Assignment is a special case of transportation (supplies/demands all equal 1), but it is typically solved with the Hungarian algorithm.

For solved assignment practice, see assignment examples.

On this page

Quick takeaway
Models side-by-side
When to use which
Worked examples (solved)
Questions people ask

Quick takeaway: Use assignment for one-to-one pairings. Use transportation for quantity shipping with general supplies/demands. Assignment can be written as transportation by setting all supplies/demands to 1, but transportation cannot generally be reduced to assignment without losing meaning.

Models side-by-side

Assignment (balanced n×n)

minimize   Σ_i Σ_j c(i,j) x(i,j)
subject to Σ_j x(i,j) = 1
           Σ_i x(i,j) = 1
           x(i,j) ∈ {0,1}

Transportation

minimize   Σ_i Σ_j c(i,j) x(i,j)
subject to Σ_j x(i,j) = supply(i)
           Σ_i x(i,j) = demand(j)
           x(i,j) ≥ 0

If an assignment matrix is rectangular, use dummy rows/columns: unbalanced assignment.

When to use which

Use assignment when

each worker must get exactly one job and each job must be filled once,
the decision is a pairing (binary), not a shipped quantity,
cost depends only on the pair (i,j).

Use transportation when

you ship divisible quantities,
supplies/demands are general numbers,
a source can serve multiple destinations and vice versa.

Worked examples (solved)

Worked example
Example 1: Assignment as transportation (supplies/demands = 1)

Assignment costs:

      J1  J2  J3
W1     4   1   3
W2     2   0   5
W3     3   2   2

Transportation conversion

Set supplies and demands to 1:

Supply(W1)=Supply(W2)=Supply(W3)=1
Demand(J1)=Demand(J2)=Demand(J3)=1

Solved assignment

One optimal assignment is W1→J2 (1), W2→J1 (2), W3→J3 (2), total 5. (Manual steps: assignment examples.)

Worked example
Example 2: Transportation problem solved + optimality check (MODI)

Two sources ship to three destinations. Total supply = total demand = 50.

Costs per unit:
          D1  D2  D3   Supply
S1         2   4   5     20
S2         3   1   7     30
Demand    10  25  15

Step 1 — Start with a feasible plan (least-cost style)

A good feasible shipment plan is:

x(S2,D2)=25 (use the cheapest lane cost=1)
x(S1,D1)=5 and x(S2,D1)=5 (to meet D1=10 while respecting supplies)
x(S1,D3)=15 (use remaining S1 supply to meet D3)

Shipment table (nonlisted cells are 0):

          D1   D2   D3   Supply
S1         5    0   15     20
S2         5   25    0     30
Demand    10   25   15

Total cost

S1→D1: 5×2 = 10
S1→D3: 15×5 = 75
S2→D1: 5×3 = 15
S2→D2: 25×1 = 25

Total = 10 + 75 + 15 + 25 = 125.

Step 2 — MODI optimality check (u–v potentials)

For a basic feasible solution, compute potentials u_i (rows) and v_j (columns) so that for each basic cell: u_i + v_j = c(i,j). We have basic cells: (S1,D1), (S1,D3), (S2,D1), (S2,D2).

Set u(S1)=0. Then:

From (S1,D1): v(D1)=2
From (S1,D3): v(D3)=5
From (S2,D1): u(S2)+2=3 → u(S2)=1
From (S2,D2): 1+v(D2)=1 → v(D2)=0

Now compute reduced costs for nonbasic cells: Δ(i,j) = c(i,j) - (u_i + v_j). If all Δ ≥ 0, the solution is optimal (for minimization).

Nonbasic (S1,D2): Δ = 4 – (0+0) = 4 ≥ 0
Nonbasic (S2,D3): Δ = 7 – (1+5) = 1 ≥ 0

All reduced costs are nonnegative → the solution is optimal.

This “optimality check” is a lecture-friendly reason transportation is not solved by Hungarian: the structure uses flow/potentials logic, not one-to-one matching zeros.

Questions people ask

Questions people ask
Is assignment always a special case of transportation?

Yes, mathematically: if you set every supply and every demand equal to 1 and restrict flows to 0/1, the transportation constraints force exactly one unit shipped from each row and into each column—matching the assignment structure.

Assignment is still treated separately because (1) it has a strong matching structure, and (2) Hungarian/Kuhn–Munkres exploits that structure efficiently.

Questions people ask
Why doesn’t Hungarian solve transportation problems?

Transportation decisions are quantities, not one-to-one pairings. A single source can ship to multiple destinations. The constraint geometry and the algorithmic structure are different: transportation is typically solved by transportation simplex / min-cost flow methods, with optimality checks based on potentials and reduced costs (like the MODI method).

Hungarian is a matching algorithm built around creating/selecting independent zeros (tight reduced costs) in a square matrix. That is not the natural structure of quantity-shipping flow problems.

Questions people ask
What if my assignment is not square?

That’s the unbalanced assignment case. Add dummy rows/columns and interpret dummies as “idle” or “unfilled.” See unbalanced assignment.

Disclaimer

Educational content only. Real logistics and staffing problems often include constraints not captured by basic assignment or transportation models (time windows, multi-period planning, capacity rules, fairness, routing). Validate assumptions and solution meaning before operational use. Read the full disclaimer.

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