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11.7 Least Cost Method

The Least Cost Method (LCM) is a more cost-conscious approach to generating an initial feasible solution for a balanced transportation model. Unlike the North-West Corner Method, LCM prioritizes allocations based on minimizing transportation costs from the outset.

Steps for Applying the Least Cost Method:

  1. Identify the Cell with the Lowest Transportation Cost:
    Scan the entire transportation tableau and locate the cell with the minimum unit cost.
  2. Allocate the Maximum Possible Quantity to This Cell:
    Assign as much as possible to the selected cell, based on the available supply and demand.
  3. Adjust Supply and Demand:
    Subtract the allocated quantity from the corresponding row (supply) and column (demand).
  4. Eliminate Satisfied Rows or Columns:
    If a row or column is fully satisfied (i.e., reduced to zero), it is crossed out or marked as complete.
  5. Repeat the Process:
    Continue selecting the next lowest-cost cell from the remaining tableau and repeat the allocation process until all supply and demand constraints are met.

This method typically results in a lower total transportation cost than the North-West Corner Method, as it incorporates cost efficiency into the allocation process from the beginning.

Example: Mega Farms Inc.

Mega Farms Inc. operates three strawberry farms (S1, S2, and S3) which supply four regional markets (D1, D2, D3, and D4). The supply capacities and market demands are as follows:

  • Supply Capacities:
    • S1: 60 cases
    • S2: 80 cases
    • S3: 100 cases
  • Market Demands:
    • D1: 50 cases
    • D2: 70 cases
    • D3: 80 cases
    • D4: 40 cases

The transportation cost per case (in $) from each farm to each market is provided in the following tableau:

D1 D2 D3 D4 Supply
S1 $6 $2 $14 $8 60 Cases
S2 $4 $12 $10 $18 80 Cases
S3 $16 $6 $6 $4 100 Cases
Demand 50 Cases 70 Cases 80 Cases 40 Cases

Mega Farms Inc. wants to determine how many cases should be sent from which farm to which market so that the total cost of transportation is minimized.

 

Determining the Total Transportation Cost Using the Least Cost Method

The model tableau is shown below:

D1 D2 D3 D4 Supply
S1 $6 $2 $14 $8 60
S2 $4 $12 $10 $18 80
S3 $16 $6 $6 $4 100
Demand 50 70 80 40

Step 1: Select the Cell with the Lowest Cost

  • The cell with the Lowest Cost is (S1, D2) with a cost of $2
  • Allocate all 60 units from S1
  • Update:
    • S1 supply exhausted
    • D2 remaining demand = [latex]70 - 60 = 10[/latex]
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 60 0
S2 $4 $12 $10 $18 80
S3 $16 $6 $6 $4 100
Demand 50 70 10 80 40

Step 2: Next Lowest Cost

  • There is a tie for the next lowest cost at $4, (S2, D1) and (S3, D4)
    • Randomly select (S2, D1)
  • Allocate 50 units (the full demand of D1)
  • Update:
    • S2 remaining supply = [latex]80 - 50 = 30[/latex]
    • D1 demand fulfilled
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 0
S2 (50) $4 $12 $10 $18 80 30
S3 $16 $6 $6 $4 100
Demand 50 0 10 80 40

Step 3: Next Lowest Cost

  • Move to the Next Lowest Cost, (S3, D4), Cost = $4
  • Allocate 40 units (total demand of D4)
  • Update:
    • S3 remaining supply = [latex]100 - 40 = 60[/latex]
    • D4 demand fulfilled
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 0
S2 (50) $4 $12 $10 $18 30
S3 $16 $6 $6 (40) $4 100 60
Demand 0 10 80 40 0

Step 4: Next Lowest Cost

  • There is another tie at $6 (S3, D2) and (S3, D3)
    • Randomly select (S3, D3)
  • Allocate 60 units exhausting S3
  • Update:
    • S3 supply exhausted
    • D3 remaining demand = [latex]80 - 60 = 20[/latex]
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 0
S2 (50) $4 $12 $10 $18 30
S3 $16 $6 (60) $6 (40) $4 60 0
Demand 0 10 80 20 0

Step 5: Next Lowest Cost

  • The Next Lowest Cost is (S2, D3), Cost = $10
  • Allocate 20 units (remaining demand from D3)
  • Update:
    • S2 remaining supply = [latex]30 - 20 = 10[/latex]
    • D3 demand fulfilled
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 0
S2 (50) $4 $12 (20) $10 $18 30 10
S3 $16 $6 (60) $6 (40) $4 0
Demand 0 10 20 0 0

Step 6: Final Allocation

  • The Final Allocation will be to (S2, D2), Cost = $12
  • Allocate the remaining 10 units
  • Update:
    • S2 exhausted
    • D2 demand fulfilled
D1 D2 D3 D4 Supply
S1 $6 (60) $2 $14 $8 0
S2 (50) $4 (10) $12 (20) $10 $18 10 0
S3 $16 $6 (60) $6 (40) $4 0
Demand 0 10 0 0 0

After completing the allocation process using the Least Cost Method, all supply capacities have been fully utilized, and all market demands have been satisfied. This confirms that a feasible solution has been reached.

To compute the total transportation cost, we multiply the quantity allocated to each cell by its corresponding unit transportation cost and sum the results:

[latex]\begin{align*} \text{Total Cost}&= (\small\text{S2/D1} \times \small{4}) + (\small\text{S1/D2} \times 2) + (\small\text{S2/D2} \times 12)+ (\small\text{S2/D3} \times 10) + (\small\text{S3/D3} \times 6) + (\small\text{S3/D4} \times 4) \\ \text{Total Cost}&= (50 \times 4) + (60 \times 2) + (10 \times 12) + (20 \times 10) + (60 \times 6) + (40 \times 4)\\ \text{Total Cost}&= 200 + 120 + 120 + 200 + 360 + 160 \\ \text{Total Cost}&= 1160 \end{align*}[/latex]

Thus, the initial basic feasible solution obtained using the Least Cost Method results in a total transportation cost of $1160.

This solution is significantly more cost-efficient than the one obtained using the North-West Corner Method ($1760), as it incorporates transportation costs into the allocation process. However, while this method often yields a near-optimal solution, it does not guarantee optimality.

Video: Transportation Problem - Least Cost Initial Allocation

Video: "Transportation : Least Cost Initial Allocation" by Maths Resource [5:51] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

To explore the possibility of further cost reduction, we now turn to Vogel’s Approximation Method (VAM)—a more refined approach that considers both cost and opportunity cost in its allocation strategy.

 

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