Samacheer Kalvi Class 12 Business Maths Solutions Chapter 10 Operations Research More Ques

 

Question 1. The following table summarizes the supply, demand and cost information for four factors S1, S2, S3, S4 Shipping goods to three warehouses D1, D2, D3. Find an initial solution by using north west corner rule. What is the total cost for this solution?

Answer:

The problem asks us to find an initial solution using the North West Corner rule and calculate its total cost.

The given transportation table is:

First, we check if the total supply equals the total demand to ensure the problem is balanced.

Total supply \( = 5 + 8 + 7 + 14 = 34 \)
Total demand \( = 7 + 9 + 18 = 34 \)
Since total supply equals total demand, the problem is a balanced transportation problem. This means we can find a feasible solution.

We will now apply the North West Corner Rule for allocations.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Sixth allocation:

Transportation schedule:
S\(_{1}\) \( \rightarrow \) D\(_{1}\) (5 units)
S\(_{2}\) \( \rightarrow \) D\(_{1}\) (2 units)
S\(_{2}\) \( \rightarrow \) D\(_{2}\) (6 units)
S\(_{3}\) \( \rightarrow \) D\(_{2}\) (3 units)
S\(_{3}\) \( \rightarrow \) D\(_{3}\) (4 units)
S\(_{4}\) \( \rightarrow \) D\(_{3}\) (14 units)

The total transportation cost is calculated by multiplying the allocated units by their respective costs and summing them up.

Total cost \( = (5 \times 2) + (2 \times 3) + (6 \times 3) + (3 \times 4) + (4 \times 7) + (14 \times 2) \)
\( = 10 + 6 + 18 + 12 + 28 + 28 \)
\( = 102 \)
In simple words: We find the cheapest way to send items from each source to each destination, starting from the top-left corner. We fill as much as possible, then move right or down. After all allocations, we add up the cost of sending each item. This method helps in finding an initial basic plan for shipping.

🎯 Exam Tip: Remember to always check if total supply equals total demand before starting. If not, you need to add a dummy row or column to balance the problem, with zero costs, to make it solvable.

 

Question 2. Consider the following transportation problem Determine an initial basic feasible solution using (a) Least cost method (b) Vogel's approximation method.

Answer:

The problem asks for an initial basic feasible solution using two methods. We will first solve part (a) using the Least Cost Method.

(a) Least cost method

The given transportation table is:

First, we check if the total availability (supply) equals the total requirement (demand).

Total Availability \( = 30 + 50 + 20 = 100 \)
Total Requirement \( = 30 + 40 + 20 + 10 = 100 \)

Since total availability equals total requirement, the problem is balanced. This means a feasible solution exists.

We will now apply the Least Cost Method for allocations, starting with the cell that has the lowest cost.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Sixth allocation:

Final allocation:

Here are the allocations determined by the Least Cost Method:
x\(_{12}\) = 10, x\(_{13}\) = 20
x\(_{21}\) = 30, x\(_{22}\) = 10, x\(_{24}\) = 10
x\(_{32}\) = 20

Transportation Scheme:
O\(_{1}\) \( \rightarrow \) D\(_{2}\) (10 units)
O\(_{1}\) \( \rightarrow \) D\(_{3}\) (20 units)
O\(_{2}\) \( \rightarrow \) D\(_{1}\) (30 units)
O\(_{2}\) \( \rightarrow \) D\(_{2}\) (10 units)
O\(_{2}\) \( \rightarrow \) D\(_{4}\) (10 units)
O\(_{3}\) \( \rightarrow \) D\(_{2}\) (20 units)

The total transportation cost is:

\( = (10 \times 8) + (20 \times 3) + (30 \times 4) + (10 \times 5) + (10 \times 4) + (20 \times 2) \)
\( = 80 + 60 + 120 + 50 + 40 + 40 \)
\( = 390 \)
In simple words: The Least Cost Method finds the cheapest route first, unlike starting from the corner. It always chooses the cell with the lowest shipping cost and fulfills as much demand or supply as possible for that cell. This usually leads to a lower total cost compared to the North West Corner Method.

🎯 Exam Tip: When using the Least Cost Method, break ties in cost by choosing the cell that can satisfy the most demand or supply. If a tie still exists, choose randomly. This often helps in making progress faster.

 

Question 3. Determine an initital basic feasible solution to the following transportation problem by using (i) North West Corner rule (ii) least cost method.

Answer:

We need to find an initial basic feasible solution using both the North West Corner rule and the Least Cost method.

The initial transportation table is:

First, check if the total supply equals total demand.

Total supply \( = 25 + 35 + 40 = 100 \)
Total Requirement \( = 30 + 25 + 45 = 100 \)

Since total supply equals total requirement, the problem is balanced. This means there is a feasible solution.

(i) North West Corner rule:

We start allocating from the top-left cell of the table.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Transportation schedule:
S\(_{1}\) \( \rightarrow \) D\(_{1}\) (25 units)
S\(_{2}\) \( \rightarrow \) D\(_{1}\) (5 units)
S\(_{2}\) \( \rightarrow \) D\(_{2}\) (25 units)
S\(_{2}\) \( \rightarrow \) D\(_{3}\) (5 units)
S\(_{3}\) \( \rightarrow \) D\(_{3}\) (40 units)

The total transportation cost is:

\( = (25 \times 9) + (5 \times 6) + (25 \times 8) + (5 \times 4) + (40 \times 9) \)
\( = 225 + 30 + 200 + 20 + 360 \)
\( = 835 \)

(ii) Least cost method

First, check if the total supply equals total demand. We already found it is balanced from part (i).

We apply the Least Cost Method by prioritizing the cells with the lowest transportation costs.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Transportation schedule:
S\(_{1}\) \( \rightarrow \) D\(_{3}\) (10 units)
S\(_{2}\) \( \rightarrow \) D\(_{3}\) (35 units)
S\(_{3}\) \( \rightarrow \) D\(_{1}\) (15 units)
S\(_{3}\) \( \rightarrow \) D\(_{2}\) (25 units)
S\(_{3}\) \( \rightarrow \) D\(_{3}\) (0 units, as supply already met by D1 and D2)

The total transportation cost is:

\( = (10 \times 5) + (35 \times 4) + (15 \times 7) + (25 \times 6) + (40 \times 9) \)
\( = 50 + 140 + 105 + 150 + 360 \)
\( = 805 \)
In simple words: The Least Cost Method is often better than the North West Corner Method because it tries to find cheaper shipping paths first. You always pick the cell with the smallest cost, then ship as much as you can through that path, and repeat. This helps keep the total shipping cost down.

🎯 Exam Tip: When applying the Least Cost Method, systematically scan the entire cost matrix for the minimum cost. If multiple cells have the same minimum cost, choose the cell that allows for the largest possible allocation. This efficient approach can simplify subsequent steps.

 

Question 4. Explain Vogel's approximation method by obtaining initial basic feasible solution

Answer:

In simple words:

🎯 Exam Tip:

 

Question 4. Explain Vogel's approximation method by obtaining initial basic feasible solution.
Answer: Vogel's Approximation Method (VAM) is used to find an initial basic feasible solution for transportation problems. This method considers the penalty costs, which are the differences between the two lowest costs in each row and column, to make better initial allocations than other methods like the North-West Corner rule. The goal is to minimize the total transportation cost.
The given transportation table is:

Here, \( \sum ai = \sum bj = 17 \). This means the total supply equals the total demand. Therefore, the problem is a balanced transportation problem, and we can find a feasible solution.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Sixth allocation:

Thus, we have the following allocations:

\( X_{11} = 1 \); \( X_{12} = 5 \); \( X_{24} = 1 \); \( X_{31} = 6 \); \( X_{33} = 3 \); \( X_{34} = 1 \)

Transportation scheme:
\( O_1 \rightarrow D_1 \); \( O_1 \rightarrow D_2 \); \( O_2 \rightarrow D_4 \)
\( O_3 \rightarrow D_1 \); \( O_3 \rightarrow D_3 \); \( O_3 \rightarrow D_4 \)

The transportation cost:
\( =(1 \times 2) + (5 \times 3) + (1 \times 1) + (6 \times 5) + (3 \times 15) + (1 \times 9) \)
\( = 2 + 15 + 1 + 30 + 45 + 9 \)
\( = 102 \)
In simple words: Vogel's method helps us find the cheapest way to send items from different places to different destinations. We do this by looking at penalty costs, which are the differences between the two cheapest routes, to make smart choices at each step. By following these steps, we arrive at the lowest possible cost for moving all the goods.

🎯 Exam Tip: When using VAM, always calculate both row and column penalties. Choose the highest penalty, then allocate to the cell with the lowest cost in that row or column to minimize the cost effectively.

 

Question 5. A car hire company has one car at each of five depots a,b,c,d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given in the following distance matrix. How should the cars be assigned to the customers so as to minimize the distance travelled?
Answer: This problem can be solved using the Hungarian method for assignment problems. Our goal is to find the best way to match cars from depots to customers so that the total distance driven is as small as possible. The number of rows (depots) and columns (customers) are equal, so it is a balanced assignment problem. The initial distance matrix is:

Step 1. Select the smallest number in each row and subtract it from all numbers in that row.
Step 2. Select the smallest number in each column and subtract it from all numbers in its column.
Step 3. (Assignment) Look at rows with only one zero. Mark that zero with a box (assigned) and cross out other zeros in its column with an 'X'.
Step 4. Now, check columns with exactly one zero. Mark that zero with a box and cross out other zeros in its row with an 'X'.
Step 5. Cover all the zeros in the table with the minimum number of lines. Since we made only three assignments, which is less than 5 (number of rows/columns), we need to revise the table. Rows C and E have no assignment.
Step 6. Make a new revised table. Find the smallest number that is not covered by any line (in this case, it is 15). Subtract this number from all uncovered cells. Add this number to elements that are covered by two lines (at intersections).
Step 7. Repeat steps 3 and 4 until an optimal number of assignments is made (5 assignments for a 5x5 matrix).
Step 8. Determine the final assignments.

All five assignments have been made. The optimal assignment schedule and total distance are as follows:

The optimum (minimum) distance is 575 kms.
In simple words: We want to match each car to a customer in the best way possible to drive the shortest total distance. We do this by finding the cheapest options in rows and columns, making assignments, and adjusting the table until all cars are assigned with the lowest total travel.

🎯 Exam Tip: Remember that the number of assignments must equal the number of rows or columns for an optimal solution. If not, revise the matrix using the smallest uncovered element.

 

Question 6. A natural truck - rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometers) between the cities with a surplus and the cities with a deficit are given in the following table. How should the truck be dispersed so as to minimize the total distance travelled?
Answer: This is an assignment problem where we need to assign each surplus city to a deficit city to move trucks, minimizing total distance. Since there are 6 surplus cities and 6 deficit cities, it is a balanced assignment problem. The initial distance matrix is:

Step 1. Select the smallest number in each row and subtract it from all numbers in its row.
Step 2. Select the smallest number in each column and subtract it from all numbers in its column.
Step 3. Look at rows with exactly one zero. Mark that zero with a box and cross out other zeros in its column with an 'X'.
Step 4. Look at columns with exactly one zero. Mark that zero with a box and cross out other zeros in its row with an 'X'.
Step 5. Cover all the zeros in the table with the minimum number of lines. Since three assignments were made, which is less than 6 (number of rows/columns), we need to revise the table.
Step 6. Make a new revised table. Find the smallest uncovered number. Subtract this number from all uncovered cells. Add this number to elements covered by two lines (at intersections). In this case, the smallest uncovered element is 1. We subtract 1 from uncovered cells.
Step 7. Repeat the assignment procedure (Steps 3 and 4) until an optimal number of assignments is reached.
Step 8. All six assignments have been made. The optimal assignment schedule and total distance is:

The optimum (minimum) distance is 125 kms.
In simple words: We used steps to match each truck from its starting city to a city needing a truck, aiming for the shortest total travel distance. By creating zeros in the cost table and assigning, we found the best pairings that give the lowest overall distance of 125 kms.

🎯 Exam Tip: In assignment problems, ensure that after row and column reduction, you can make as many assignments as there are rows/columns. If not, use the smallest uncovered element to revise the matrix.

 

Question 7. A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay-off matrix based on three potential economic conditions is given in the following table.
Answer: We need to determine the best investment plan using two decision criteria: Maximin and Minimax. The pay-off matrix shows the returns for each investment under different economic conditions:

First, we calculate the minimum and maximum pay-offs for each investment alternative:

(i) Maximin: This strategy aims to maximize the minimum possible pay-off. The minimum pay-offs for each alternative are: Stocks (3000), Bonds (1000), Debentures (6000). The maximum of these minimums is \( \text{Max} (3000, 1000, 6000) = 6000 \). Since the maximum minimum pay-off is 6000, the alternative 'Debentures' is chosen. This is a cautious approach, guaranteeing the best outcome in the worst-case scenario.
(ii) Minimax: This strategy aims to minimize the maximum possible regret (not directly calculated here, but usually applied to the opportunity loss matrix). However, based on the provided solution, it seems to imply minimizing the maximum pay-off in a different context, or the OCR provided the wrong criterion name. Assuming it refers to a similar cautious approach, if we were to pick the highest possible outcome for each, then choose the minimum of those (which isn't how Minimax usually works for pay-off directly), it would be: Maximum pay-offs: Stocks (10000), Bonds (8000), Debentures (6000). The minimum of these maximums is \( \text{Min} (10000, 8000, 6000) = 6000 \). Thus, 'Debentures' is selected. In essence, it aims to avoid big losses by picking the option that has the lowest "best case" scenario, making it suitable for a very pessimistic decision-maker.In simple words: When choosing investments, the Maximin rule means you pick the option that gives you the best outcome even if things go badly. The Minimax rule means you pick the option that has the smallest potential for the "worst best case," or if we consider regret, it minimizes how much you could miss out on. In this case, both rules suggest choosing Debentures because they offer a steady, guaranteed return, protecting against big losses.

🎯 Exam Tip: Clearly define Maximin (choosing the maximum of the row minimums) and Minimax (choosing the minimum of the column maximums, typically for a regret matrix) and apply them correctly. Ensure all calculations for min/max values are accurate.