Solve question (i) using graphical method, to minimise the weekly total cost.

Linear programming is a powerful mathematical technique used in operations research, management science, and economics to optimize a linear objective function subject to linear equality and inequality constraints. It’s widely applied in resource allocation, production planning, and transportation problems. Minimizing total cost is a common application, where businesses aim to find the most cost-effective way to achieve a desired output. The graphical method is particularly useful for problems with two decision variables, allowing for a visual representation of the feasible region and the optimal solution. This response will demonstrate the graphical method to minimize weekly total cost, assuming a hypothetical scenario for illustrative purposes. Problem Formulation Let's assume a company produces two products, A and B. The weekly total cost is to be minimized. We need to define the variables and formulate the problem. Decision Variables: x = Number of units of product A produced weekly y = Number of units of product B produced weekly Objective Function: Minimize Total Cost (Z) = 10x + 15y (Assuming product A costs $10/unit and product B costs $15/unit) Constraints: 2x + y ≤ 40 (Resource constraint 1 - e.g., labor hours) x + 2y ≤ 30 (Resource constraint 2 - e.g., raw materials) x ≥ 0, y ≥ 0 (Non-negativity constraints) Graphical Solution The graphical method involves the following steps: Step 1: Plot the Constraints Convert each inequality constraint into an equation and plot the corresponding line on a graph with x and y axes. 2x + y = 40 => y = 40 - 2x x + 2y = 30 => y = 15 - 0.5x x = 0 (y-axis) y = 0 (x-axis) Step 2: Identify the Feasible Region The feasible region is the area on the graph that satisfies all the constraints simultaneously. For each inequality, determine which side of the line represents the feasible region. This is done by testing a point (e.g., (0,0)) in the inequality. 2x + y ≤ 40: 2(0) + 0 ≤ 40 (True) - Feasible region is below the line. x + 2y ≤ 30: 0 + 2(0) ≤ 30 (True) - Feasible region is below the line. x ≥ 0, y ≥ 0 - Feasible region is in the first quadrant. The feasible region is the polygon formed by the intersection of these areas. Step 3: Identify the Corner Points The corner points (vertices) of the feasible region are the points where the constraint lines intersect. These points are potential optimal solutions. In this case, the corner points are: (0, 0) (20, 0) (0, 15) Intersection of 2x + y = 40 and x + 2y = 30 Step 4: Solve for the Intersection Point To find the intersection point, solve the system of equations: 2x + y = 40 x + 2y = 30 Multiply the first equation by 2: 4x + 2y = 80. Subtract the second equation from this: 3x = 50 => x = 50/3 ≈ 16.67. Substitute x back into the second equation: (50/3) + 2y = 30 => 2y = 30 - (50/3) = 40/3 => y = 20/3 ≈ 6.67. Therefore, the intersection point is approximately (16.67, 6.67). Step 5: Evaluate the Objective Function at Each Corner Point Substitute the coordinates of each corner point into the objective function Z = 10x + 15y. Corner Point (x, y) Z = 10x + 15y (0, 0) Z = 10(0) + 15(0) = 0 (20, 0) Z = 10(20) + 15(0) = 200 (0, 15) Z = 10(0) + 15(15) = 225 (16.67, 6.67) Z = 10(16.67) + 15(6.67) ≈ 166.7 + 100.05 ≈ 266.75 Step 6: Determine the Optimal Solution The optimal solution is the corner point that yields the minimum value of the objective function. In this case, the minimum cost is $0 at (0,0). However, this implies producing zero units of both products, which might not be a realistic scenario. If we consider only positive production levels, the minimum cost is $200 at (20,0). The graphical method provides a clear visual approach to solving linear programming problems. In this example, minimizing the weekly total cost involved formulating the problem, plotting constraints, identifying the feasible region, and evaluating the objective function at corner points. The optimal solution, assuming positive production, is to produce 20 units of product A and 0 units of product B, resulting in a minimum weekly cost of $200. This method is effective for problems with two variables but becomes more complex with increasing dimensionality, necessitating the use of more advanced techniques like the simplex method.

Question 3 | UPSC Mains MANAGEMENT-PAPER-II 2011

Linear programming is a powerful mathematical technique used in operations research, management science, and economics to optimize a linear objective function subject to linear equality and inequality constraints. It’s widely applied in resource allocation, production planning, and transportation problems. Minimizing total cost is a common application, where businesses aim to find the most cost-effective way to achieve a desired output. The graphical method is particularly useful for problems with two decision variables, allowing for a visual representation of the feasible region and the optimal solution. This response will demonstrate the graphical method to minimize weekly total cost, assuming a hypothetical scenario for illustrative purposes.

Problem Formulation

Let's assume a company produces two products, A and B. The weekly total cost is to be minimized. We need to define the variables and formulate the problem.

Decision Variables:

x = Number of units of product A produced weekly
y = Number of units of product B produced weekly

Objective Function:

Minimize Total Cost (Z) = 10x + 15y (Assuming product A costs $10/unit and product B costs $15/unit)

Constraints:

2x + y ≤ 40 (Resource constraint 1 - e.g., labor hours)
x + 2y ≤ 30 (Resource constraint 2 - e.g., raw materials)
x ≥ 0, y ≥ 0 (Non-negativity constraints)

Graphical Solution

The graphical method involves the following steps:

Step 1: Plot the Constraints

Convert each inequality constraint into an equation and plot the corresponding line on a graph with x and y axes.

2x + y = 40 => y = 40 - 2x
x + 2y = 30 => y = 15 - 0.5x
x = 0 (y-axis)
y = 0 (x-axis)

Step 2: Identify the Feasible Region

The feasible region is the area on the graph that satisfies all the constraints simultaneously. For each inequality, determine which side of the line represents the feasible region. This is done by testing a point (e.g., (0,0)) in the inequality.

2x + y ≤ 40: 2(0) + 0 ≤ 40 (True) - Feasible region is below the line.
x + 2y ≤ 30: 0 + 2(0) ≤ 30 (True) - Feasible region is below the line.
x ≥ 0, y ≥ 0 - Feasible region is in the first quadrant.

The feasible region is the polygon formed by the intersection of these areas.

Step 3: Identify the Corner Points

The corner points (vertices) of the feasible region are the points where the constraint lines intersect. These points are potential optimal solutions.

In this case, the corner points are:

(0, 0)
(20, 0)
(0, 15)
Intersection of 2x + y = 40 and x + 2y = 30

Step 4: Solve for the Intersection Point

To find the intersection point, solve the system of equations:

2x + y = 40
x + 2y = 30

Multiply the first equation by 2: 4x + 2y = 80. Subtract the second equation from this: 3x = 50 => x = 50/3 ≈ 16.67. Substitute x back into the second equation: (50/3) + 2y = 30 => 2y = 30 - (50/3) = 40/3 => y = 20/3 ≈ 6.67. Therefore, the intersection point is approximately (16.67, 6.67).

Step 5: Evaluate the Objective Function at Each Corner Point

Substitute the coordinates of each corner point into the objective function Z = 10x + 15y.

Corner Point (x, y)	Z = 10x + 15y
(0, 0)	Z = 10(0) + 15(0) = 0
(20, 0)	Z = 10(20) + 15(0) = 200
(0, 15)	Z = 10(0) + 15(15) = 225
(16.67, 6.67)	Z = 10(16.67) + 15(6.67) ≈ 166.7 + 100.05 ≈ 266.75

Step 6: Determine the Optimal Solution

The optimal solution is the corner point that yields the minimum value of the objective function. In this case, the minimum cost is $0 at (0,0). However, this implies producing zero units of both products, which might not be a realistic scenario. If we consider only positive production levels, the minimum cost is $200 at (20,0).

Solve question (i) using graphical method, to minimise the weekly total cost.

Model Answer

Introduction