Decision Variables

When formulating a transportation problem, we define variables for goods transported from each source to each destination:

• For two sources and two destinations: Let $x$ be goods from source 1 to destination 1, and $y$ be goods from source 1 to destination 2. Then automatically, the remaining goods complete the supply to destinations.

• For two sources and three destinations: Let $x$, $y$, and $z$ be goods from source 1 to destinations 1, 2, and 3 respectively. The remaining quantities come from source 2.

Objective Function

The objective is always to minimize total transportation cost, expressed as: $\text{Minimize } z = \sum (\text{cost per unit} \times \text{units transported})$

Constraints

Transportation problems have three types of constraints:

1. Supply constraints: Total goods shipped from each source cannot exceed available supply
2. Demand constraints: Total goods received at each destination must meet the requirement
3. Non-negativity: All variables must be greater than or equal to zero

A problem is balanced when total supply equals total demand.

Step 1: Define Variables

Let $x$ = units transported from source 1 to destination 1 Let $y$ = units transported from source 1 to destination 2

Then automatically:

• Units from source 2 to destination 1 = demand₁ $- x$
• Units from source 2 to destination 2 = demand₂ $- y$

Step 2: Write the Objective Function

Multiply each variable by its respective transportation cost and sum all terms.

Step 3: Write Constraints

• From source 1: $x + y \leq \text{supply}_1$
• From source 2: $(\text{demand}_1 - x) + (\text{demand}_2 - y) \leq \text{supply}_2$
• At destination 1: $x \leq \text{demand}_1$
• At destination 2: $y \leq \text{demand}_2$
• Non-negativity: $x \geq 0$, $y \geq 0$

Step 4: Solve Graphically

Plot all constraints on a coordinate plane, identify the feasible region (bounded polygon), evaluate the objective function at each corner point, and select the minimum value.

Problem: Mariam has two warehouses—in Kigoma (80 tonnes) and Mtwara (60 tonnes). She needs to transport 60 tonnes to Singida and 40 tonnes to Moshi. The transportation costs per tonne are:

Find the minimum transportation cost.

Solution

Step 1: Define variables

• Let $x$ = tonnes from Kigoma to Singida
• Let $y$ = tonnes from Kigoma to Moshi

Then automatically:

• From Mtwara to Singida: $60 - x$
• From Mtwara to Moshi: $40 - y$

Step 2: Objective function

$\text{Minimize } z = 180000x + 120000y + 240000(60-x) + 100000(40-y)$ $z = 18400000 - 60000x + 20000y$

Step 3: Constraints

• From Kigoma: $x + y \leq 80$
• From Mtwara: $(60-x) + (40-y) \leq 60 \Rightarrow x + y \geq 40$
• At Singida: $x \leq 60$
• At Moshi: $y \leq 40$
• Non-negativity: $x \geq 0$, $y \geq 0$

Step 4: Graphical solution

Evaluating at corner points:

Minimum cost: 14,800,000 TSh at point D(60, 0)

Interpretation:

• 60 tonnes from Kigoma to Singida
• 0 tonnes from Kigoma to Moshi
• 0 tonnes from Mtwara to Singida
• 40 tonnes from Mtwara to Moshi

Problem: A supplier has stores at Mbagala (40 sacks) and Ubungo (36 sacks). Customers at Mbewi, Manzese, and Kinyerezi require 32, 24, and 20 sacks respectively. Transportation costs per sack:

Find the minimum cost and determine if the problem is balanced.

Solution

Step 1: Define variables

• Let $x$ = sacks from Mbagala to Mbewi
• Let $y$ = sacks from Mbagala to Manzese
• Let $z$ = sacks from Mbagala to Kinyerezi

Then:

• From Ubungo to Mbewi: $32 - x$
• From Ubungo to Manzese: $24 - y$
• From Ubungo to Kinyerezi: $20 - z$

Since total from Mbagala = 40: $x + y + z = 40$, so $z = 40 - (x + y)$

Step 2: Objective function

$w = 160x + 100y + 150z + 100(32-x) + 120(24-y) + 100(20-z)$ $w = 60x - 20y + 50z + 8080$

Substituting $z = 40 - (x + y)$:

$w = 10x - 70y + 10080$

Step 3: Constraints

• $x \leq 32$, $y \leq 24$
• $x + y \geq 20$ (from $z \geq 0$)
• $x + y \leq 40$ (from $z \leq 20$)
• $x \geq 0$, $y \geq 0$

Step 4: Graphical solution

Evaluating at corner points:

Minimum cost: 8,400 TSh at point B(0, 24)

Transportation plan:

• From Mbagala: 0 to Mbewi, 24 to Manzese, 16 to Kinyerezi
• From Ubungo: 32 to Mbewi, 0 to Manzese, 4 to Kinyerezi

Balance check: Total supply = 40 + 36 = 76 sacks; Total demand = 32 + 24 + 20 = 76 sacks. The problem is balanced.

1. Identify sources and destinations with their supplies and demands
2. Define decision variables for one source to all destinations
3. Express remaining variables in terms of demand minus already allocated
4. Formulate objective function by multiplying variables by unit costs
5. Write supply and demand constraints
6. Add non-negativity constraints
7. Graph constraints and identify feasible region
8. Evaluate objective function at each corner point
9. Select minimum cost solution

A Tanzanian maize mill owner with storage facilities in Morogoro and Iringa can use transportation linear programming to decide how much maize to transport to their distribution centers in Dar es Salaam, Dodoma, and Arusha. By formulating the problem with two sources and three destinations, they minimize transportation costs in Tanzanian shillings while meeting the demand at each center—directly applying this method to reduce operational costs in their business.