JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III OPERATIONS RESEARCH Previous Question Paper (August - 2022 - R18)

To find the cheapest mixture, we formulate this as a Linear Programming Problem (LPP). Let x be the amount of Alloy A (in units of 10 grams) and y be the amount of Alloy B (in units of 10 grams).

The composition per 10 grams is:

Let C be the cost of 10g of Alloy A. Then the cost of 10g of Alloy B is 1.5C. The objective is to minimize the total cost Z = C(x + 1.5y).

Constraints:

1. 2x + y ≥ 10 (Copper)
2. x + y ≥ 8 (Zinc)
3. x + y ≥ 12 (Lead)

Since the Lead requirement (x + y ≥ 12) is stricter than the Zinc requirement (x + y ≥ 8), the Zinc constraint is redundant. We focus on:

Subject to:

Evaluating the corner points of the feasible region:

1. Intersection of 2x + y = 10 and x + y = 12: Subtracting the equations gives x = -2, which is outside the feasible region.
2. Intersection of 2x + y = 10 and x = 0: y = 10. Point (0, 10). Z = 0 + 1.5(10) = 15.
3. Intersection of x + y = 12 and y = 0: x = 12. Point (12, 0). Z = 12 + 1.5(0) = 12.

Comparing the values, the minimum cost occurs at x = 12 and y = 0. Thus, 120 grams of Alloy A and 0 grams of Alloy B should be used.

To solve the given linear programming problem, we first analyze the constraints. Observe the third and fourth constraints:

Notice that the third constraint is exactly twice the fourth constraint (2 × (a + 3b - 2c + 4d) = 2a + 6b - 4c + 8d). Since these two equations are linearly dependent, they represent the same hyperplane. We can simplify the system by using only one of these equations: a + 3b - 2c + 4d = 0. Given the non-negativity constraints (a, b, c, d ≥ 0), the equation a + 3b + 4d = 2c must hold. This implies c = 0.5a + 1.5b + 2d.

Substitute this expression for c into the objective function and the remaining inequality constraints:

Objective Function: Z = a + 1.5b + 5(0.5a + 1.5b + 2d) + 2d = a + 1.5b + 2.5a + 7.5b + 10d + 2d = 3.5a + 9b + 12d.

Constraint 1: 3a + 2b + (0.5a + 1.5b + 2d) + 4d ≤ 63 ⇒ 3.5a + 3.5b + 6d ≤ 63.

Constraint 2: 2a + b + 5(0.5a + 1.5b + 2d) + d ≤ 42 ⇒ 2a + b + 2.5a + 7.5b + 10d + d ≤ 42 ⇒ 4.5a + 8.5b + 11d ≤ 42.

To maximize Z = 3.5a + 9b + 12d subject to the reduced constraints, we observe that the coefficients of b and d in the objective function are significantly higher relative to their consumption of the resources in the constraints compared to a. Setting a = 0, we have:

1. 3.5b + 6d ≤ 63
2. 8.5b + 11d ≤ 42

Testing the boundary of the second constraint (as it is more restrictive for b): If d = 0, 8.5b = 42 ⇒ b ≈ 4.94. If b = 0, 11d = 42 ⇒ d ≈ 3.81. Testing the intersection of the two constraints: 3.5b + 6d = 63 and 8.5b + 11d = 42. Solving this system yields negative values for b or d, which violates non-negativity. Thus, the optimal solution lies on the axes. Comparing Z values: at (b=4.94, d=0), Z = 9(4.94) = 44.46; at (b=0, d=3.81), Z = 12(3.81) = 45.72. The maximum value is approximately 45.72 at a=0, b=0, c=7.62, d=3.81.