Transportation & Transshipment Problems Explained with Worked Examples

Transshipment Model or Transportation Model

The transportation problem is a subclass of linear programming where the objective is to transport specified quantities of a single homogeneous product from multiple origins to various destinations in such a way that the total transportation cost is minimized.

Transportation models deal with finding the optimal allocation of resources, ensuring that products produced at different factories or plants (supply origins) are distributed efficiently to warehouses, distribution centers, or retail stores (demand destinations).

The primary objective is to:

• Fully satisfy the demand at each destination.
• Stay within the supply limits at each origin.
• Minimize the overall cost of transportation.

This model becomes especially relevant whenever goods move from manufacturers to final consumers through various intermediaries such as wholesalers, distributors, or retailers. By minimizing transportation costs, businesses can maximize profitability.

A special case of this model is the transshipment model, where goods pass through intermediate points before reaching the final destination. For example, products may first move from manufacturing plants to regional distribution centers, and then from these centers to retail outlets.

Constraints in a Transportation/Transshipment Problem

1. The total amount leaving each source must not exceed its supply.
2. The total amount entering each destination must not exceed its demand.
3. In transshipment problems, the total amount entering an intermediate node must equal the total amount leaving it.

Initial Solution Methods for Transportation Problems

Since transportation problems are usually solved in two phases finding an initial feasible solution and then optimizing it several methods exist to generate the initial allocation.

1. North-West Corner Method (NWCM)

The North-West Corner Rule is a simple method for finding an initial feasible solution.

Steps:

• Start with the top-left (north-west) cell of the transportation tables.
• Allocate as much as possible: min(supply, demand).
• Adjust the supply and demand accordingly.
• If supply is exhausted, move down to the next row; if demand is satisfied, move right to the next column.
• If supply equals demand in a cell, you can proceed either to the next row or next column.
• Repeat until all supplies and demands are allocated.

2. Minimum Cost Method (Least Cost Method / Matrix Minimum Method)

This method selects allocations based on the lowest transportation cost available.

Steps:

• Identify the cell with the minimum unit transportation cost.
• If multiple cells have the same cost, choose the one in the row/column with the lowest index.
• Allocate as much as possible to this cell without violating supply/demand constraints.
• If demand is satisfied, cross out the column; if supply is exhausted, cross out the row.
• Repeat until all allocations are made.

3. Vogel’s Approximation Method (VAM)

The VAM is a more refined method that often yields a better starting solution than NWCM or the Minimum Cost Method.

Steps:

• For each row, calculate the penalty: difference between the lowest and next-lowest transportation costs. Record it at the row’s side.
• For each column, calculate the penalty in the same way and record it below the column.
• Identify the row or column with the highest penalty.
• In that row or column, allocate as much as possible to the cell with the minimum transportation cost.
• Adjust supply and demand. If either is satisfied, cross out the respective row/column.
• In case of ties, give preference to the uppermost row or leftmost column.
• Repeat the process until all requirements are satisfied.

The initial solution obtained from these methods can then be optimized using techniques such as the MODI method or stepping-stone method to ensure the total cost is truly minimized.

Worked Numerical Example (Transportation Problem)

Goal: Minimize total transportation cost while meeting all demand and not exceeding any supply.

Cost (₹ per unit)

	D1	D2	D3
S1	4	8	8
S2	16	24	16
S3	8	16	24

Supply: S1=20, S2=30, S3=25 (Total 75)
Demand: D1=10, D2=35, D3=30 (Total 75)

Step 1: Initial Feasible Solution via VAM

We compute row/column penalties (difference between the two lowest costs) and allocate where the penalty is highest, in the lowest-cost cell of that row/column.

Iteration 1

• Row penalties: S1=4 (8−4), S2=0 (16−16), S3=8 (16−8)
• Column penalties: D1=4 (8−4), D2=8 (16−8), D3=8 (16−8)
• Max penalty tie → prefer row. Pick S3; lowest cost in S3 is D1 (8).
  Allocate min(S3=25, D1=10) = 10 to (S3, D1).
  Update: S3→15 left, D1 satisfied (0).

Iteration 2 (remaining columns D2, D3)

• Row penalties: S1=0 (8−8), S2=8 (24−16), S3=8 (24−16)
• Column penalties: D2=8 (16−8), D3=8 (16−16=0 → next pair gives 8)
• Max penalty tie → prefer row. Pick S2; lowest cost is D3 (16).
  Allocate min(S2=30, D3=30) = 30 to (S2, D3).
  Update: S2 satisfied (0), D3 satisfied (0).

Iteration 3 (only D2 open)

• Rows left: S1 (20), S3 (15); Column left: D2 (35)
• Allocate at lowest cost cells to fill D2:
• (S3, D2) gets min(15, 35)=15 → D2 now 20 left, S3 done
• (S1, D2) gets min(20, 20)=20 → D2 done, S1 done

Initial solution (also optimal here):

• (S3→D1) = 10
• (S2→D3) = 30
• (S3→D2) = 15
• (S1→D2) = 20

Step 2: Total Cost & Optimality Check

Total cost =

• S3→D1: 10×8 = 80
• S2→D3: 30×16 = 480
• S3→D2: 15×16 = 240
• S1→D2: 20×8 = 160
  Grand Total = ₹960

Using the MODI (u–v) test shows all opportunity costs (Δᵢⱼ) ≥ 0, so this is optimal.

Mini Example: How a Transshipment Node is Modeled

Suppose shipments must pass through a Distribution Center (DC) before reaching stores.

Costs (₹/unit)

• Plants → DC:
• P1→DC = 5, P2→DC = 7
• DC → Stores:
• DC→R1 = 6, DC→R2 = 4

Supplies: P1=30, P2=20 (Total 50)
Demands: R1=25, R2=25 (Total 50)
DC: Intermediate node (no net supply or demand)

How to set it up:

1. Duplicate the DC into two nodes for a transportation tableau: DC-in (as a destination) and DC-out (as a source).
2. Add a zero-cost, unlimited-capacity link between DC-in and DC-out to enforce flow conservation (inflow = outflow).
3. Build a standard transportation table:
• Rows (sources): P1, P2, DC-out (supply of DC-out equals total flow arriving to DC-in).
• Columns (destinations): DC-in, R1, R2 (demand at DC-in equals total flow sent from plants).
4. Include costs P→DC in column DC-in, and DC→R in row DC-out. The DC-out→DC-in cell carries 0 cost (linking the duplicated node).
5. Solve as a normal transportation problem (VAM → MODI). The duplication ensures everything entering DC equals everything leaving DC.

This trick lets you keep the convenient transportation tableau while correctly modeling transshipment (intermediate) flow balance.

Frequently Asked Questions (FAQs)

What is the main difference between a transportation model and a transshipment model?

A transportation model directly moves goods from supply points to demand points, while a transshipment model allows goods to pass through intermediate points (like warehouses or distribution centers) before reaching their final destination.

Why is Vogel’s Approximation Method preferred over the North-West Corner Method?

VAM generally provides a better initial feasible solution because it considers penalties based on cost differences, while NWCM ignores transportation costs at the beginning.

What are common optimization methods used after finding the initial solution?

The most widely used optimization techniques are the Modified Distribution Method (MODI) and the Stepping-Stone Method.

Can transportation models handle multiple products?

The basic model assumes a single homogeneous product. For multiple products, extended or multi-commodity transportation models are used.

Where are transportation and transshipment models applied in real life?

They are used in logistics, supply chain management, distribution planning, airline scheduling, and even energy and utility distribution.

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