Least Cost Method

Least Cost Method

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Least cost Method is one of the method to obtain feasible solution of Transportation Problem. We already understand the North West Corner Method for transportation problem, to obtain feasible solution.

To understand Least Cost Method we will go through the numerical provided as follows (the same numerical we used in North West Corner Method) :

Numerical
A mobile phone manufacturing company has three branches located in three different regions, say Jaipur, Udaipur and Mumbai. The company has to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the Least Cost method to find a basic feasible solution (BFS).

Note that all the explanation is provided in “CYAN” colour. You have to write in examination the only thing which are given in this regular colour under each steps(if any), else you can directly solve matrix of the problem as explained here.

north-west-corner-numerical-question-image

Solution:

Step 1: Balance the problem

Balance the problem meaning we need to check that if;

Σ Supply=Σ Demand\color{#32c5d4} \Sigma \text { Supply} = \Sigma \text { Demand}

If this holds true, then we will consider the given problem as a balanced problem.

Now, what if it’s not balanced?

i.e., Σ SupplyΣ Demand\color{#32c5d4} \text {i.e., } \Sigma \text { Supply} \not = \Sigma \text { Demand}

If such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced.

You can watch a video on this type of numerical, which is known as Unbalanced Transportation Problems.

\to The given transportation problem is balanced.

Step 2: Select the lowest cost from the entire matrix and allocate the minimum of supply or demand.

We are using Least Cost Method here, so we will be indentifying the lowest cell value in this entire matrix.

Here, in this matrix we have 1 (For cell: Jaipur-Delhi) as the lowest value.

lcm-2

So, moving with that cell, and allocating the minimum of demand or supply, i.e. allocating 40 here (as supply value is 40, whereas demand is of 60).

Checking out the first row and not the last column, because we are allocating 40 in the cell for supply, as it is minimum.

Subtracting allocated value (i.e. 40) from corresponding supply and demand.

Step 3: Remove the row or column whose supply or demand is fulfilled and prepare a new matrix

As we have fulfilled the demand or supply for that row or column respectively, remove that row or column and prepare a new matrix, as shown below:

lcm-3

Step 4: Repeat the procedure until all the allocations are over

Repeat the same procedure of allocation of the smallest value in the new generated matrix and check out demand or supply based on the smallest value (of demand or supply) as shown below, until all allocations are over.

lcm-4.1

Tie in the following step!

lcm-4.2

You may find tie in selecting cell here in the above matrix, as we have minimun cell value 3 for two cells in the matrix above.

So whenever you find this kind of situation while evaluting matrix by Least Cost Method, you will have multiple solution to the problem, called alternate solutions.

The alternate solution of the this problem is provided at the end of this note-blog. In that we will be selecting the intersecting cell of Udaipur - Kanpur.

lcm-4.3

Step 5: After all the allocations are over, write the allocations and calculate the transportation cost

Once all allocations are over, prepare the table with all allocations marked and calculate the transportation cost as follows:

lcm-5

 Transportation cost=(1×40)+(3×40)+(3×20)+(6×30)+(2×40)=Rs 480\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 40) + (3 \times 20) + (6 \times 30) + (2 \times 40) \\ &= \text {Rs } 480 \end{aligned}

Alternate Solution

As we have tie in selecting minimum value at this step.

We can select any of this cell as mentioned earlier.
We have already seen above the solution by selecting minimum value as of Udaipur-Delhi intersecting cell, we will select now the another one and continuing the allocation using Least Cost Method as follows:

lcm-6-4.2

lcm-6-4.3

lcm-6-5

So, we will be having alternate solution as mentioned above, and alternate transportation cost as follows:

 Transportation cost=(1×40)+(3×60)+(6×10)+(2×40)+(8×20)=Rs 520\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 60) + (6 \times 10) + (2 \times 40) + (8 \times 20) \\ &= \text {Rs } 520 \end{aligned}

\to Thus, we have optimum Transportation cost as Rs. 480.

Find solution of same numerical by:

  1. Vogel's Approximation Method (VAM)

  2. North West Corner Method

Least cost Method is one of the method to obtain feasible solution of Transportation Problem. We already understand the North West Corner Method for transportation problem, to obtain feasible solution.

To understand Least Cost Method we will go through the numerical provided as follows (the same numerical we used in North West Corner Method) :

Numerical
A mobile phone manufacturing company has three branches located in three different regions, say Jaipur, Udaipur and Mumbai. The company has to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the Least Cost method to find a basic feasible solution (BFS).

Note that all the explanation is provided in “CYAN” colour. You have to write in examination the only thing which are given in this regular colour under each steps(if any), else you can directly solve matrix of the problem as explained here.

north-west-corner-numerical-question-image

Solution:

Step 1: Balance the problem

Balance the problem meaning we need to check that if;

Σ Supply=Σ Demand\color{#32c5d4} \Sigma \text { Supply} = \Sigma \text { Demand}

If this holds true, then we will consider the given problem as a balanced problem.

Now, what if it’s not balanced?

i.e., Σ SupplyΣ Demand\color{#32c5d4} \text {i.e., } \Sigma \text { Supply} \not = \Sigma \text { Demand}

If such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced.

You can watch a video on this type of numerical, which is known as Unbalanced Transportation Problems.

\to The given transportation problem is balanced.

Step 2: Select the lowest cost from the entire matrix and allocate the minimum of supply or demand.

We are using Least Cost Method here, so we will be indentifying the lowest cell value in this entire matrix.

Here, in this matrix we have 1 (For cell: Jaipur-Delhi) as the lowest value.

lcm-2

So, moving with that cell, and allocating the minimum of demand or supply, i.e. allocating 40 here (as supply value is 40, whereas demand is of 60).

Checking out the first row and not the last column, because we are allocating 40 in the cell for supply, as it is minimum.

Subtracting allocated value (i.e. 40) from corresponding supply and demand.

Step 3: Remove the row or column whose supply or demand is fulfilled and prepare a new matrix

As we have fulfilled the demand or supply for that row or column respectively, remove that row or column and prepare a new matrix, as shown below:

lcm-3

Step 4: Repeat the procedure until all the allocations are over

Repeat the same procedure of allocation of the smallest value in the new generated matrix and check out demand or supply based on the smallest value (of demand or supply) as shown below, until all allocations are over.

lcm-4.1

Tie in the following step!

lcm-4.2

You may find tie in selecting cell here in the above matrix, as we have minimun cell value 3 for two cells in the matrix above.

So whenever you find this kind of situation while evaluting matrix by Least Cost Method, you will have multiple solution to the problem, called alternate solutions.

The alternate solution of the this problem is provided at the end of this note-blog. In that we will be selecting the intersecting cell of Udaipur - Kanpur.

lcm-4.3

Step 5: After all the allocations are over, write the allocations and calculate the transportation cost

Once all allocations are over, prepare the table with all allocations marked and calculate the transportation cost as follows:

lcm-5

 Transportation cost=(1×40)+(3×40)+(3×20)+(6×30)+(2×40)=Rs 480\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 40) + (3 \times 20) + (6 \times 30) + (2 \times 40) \\ &= \text {Rs } 480 \end{aligned}

Alternate Solution

As we have tie in selecting minimum value at this step.

We can select any of this cell as mentioned earlier.
We have already seen above the solution by selecting minimum value as of Udaipur-Delhi intersecting cell, we will select now the another one and continuing the allocation using Least Cost Method as follows:

lcm-6-4.2

lcm-6-4.3

lcm-6-5

So, we will be having alternate solution as mentioned above, and alternate transportation cost as follows:

 Transportation cost=(1×40)+(3×60)+(6×10)+(2×40)+(8×20)=Rs 520\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 60) + (6 \times 10) + (2 \times 40) + (8 \times 20) \\ &= \text {Rs } 520 \end{aligned}

\to Thus, we have optimum Transportation cost as Rs. 480.

Find solution of same numerical by:

  1. Vogel's Approximation Method (VAM)

  2. North West Corner Method

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