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(i) Variables : 

Let x₁ and x₂ denote the number of ordinary and auto-cut voltage stabilized.

(ii) Objective function: 

Profit on x₁ units of ordinary stabilizers = 100x₁

Profit on x₂ units of auto-cut stabilized = 150x₂ 

Total profit = 100x₁ + 150x₂

Let Z = 100x₁ + 150x₂, which is the objective function. 

Since the profit is to be maximized. We have to 

Maximize, Z = 100x₁ + 15x₂

(iii) Constraints: 

The assembling and testing time required for x₁ units of ordinary stabilizers = 0.8x₁ and for x₂ units of auto-cut stabilizers = 1.2x₂ 

Since the manufacturing capacity is 720 hours per week. 

We get 0.8x₁ + 1.2x₂ ≤ 720 

Maximum weekly sale of ordinary stabilizer is 600 

i.e., x₁ ≤ 600 

Maximum weekly sales of auto-cut stabilizer is 400 

i.e., x₂ ≤ 400

(iv) Non-negative restrictions: 

Since the number of both the types of stabilizer is nonnegative, we get x₁, x₂ ≥ 0. 

Thus, the mathematical formulation of the LPP is, 

Maximize Z = 100x₁ + 150x₂ 

Subject to the constraints 

0.8x₁ + 1.2x₂ ≤ 720, x₁ ≤ 600, x₂ ≤ 400, x₁, x₂ ≥ 0

(i) Variables: 

Let x₁ and x₂ denote the two types products A and B respectively.

(ii) Objective function: 

Profit on x₁ units of type A product = 30x₁ 

Profit on x₂ units of type B product = 40x₂ 

Total profit = 30x₁ + 40x₂ 

Let Z = 30x₁ + 40x₂, which is the objective function. 

Since the profit is to be maximized, we have to maximize Z = 30x₁ + 40x₂ 

(iii) Constraints: 

60x₁ + 120x₂ ≤ 12,000 

8x₁ + 5x₂ ≤ 600 

3x₁ + 4x₂ ≤ 500

(iv) Non-negative constraints: 

Since the number of products on type A and type B are non-negative, we have x₁, x₂ ≥ 0 

Thus, the mathematical formulation of the LPP is 

Maximize Z = 30x₁ + 40x₂ 

Subject to the constraints, 

60x₁ + 120x₂ ≤ 12,000 

8x₁ + 5x₂ ≤ 600 

3x₁ + 4x₂ ≤ 500 

x₁, x₂ ≥ 0

(i) Variables: 

Let x₁ and x₂ denotes the number of pens in type A and type B.

(ii) Objective function: 

Profit on x₁ pens in type A is = 5x₁

Profit on x₂ pens in type B is = 3x₂ 

Total profit = 5x₁ + 3x₂ 

Let Z = 5x₁ + 3x₂, which is the objective function. 

Since the B total profit is to be maximized, we have to maximize Z = 5x₁ + 3x₂

(iii) Constraints: 

Raw materials required for each pen A is twice as that of pen B. 

i.e., for pen A raw material required is 2x₁ and for B is x₂. 

Raw material is sufficient only for 1000 pens per day 

∴ 2x₁ + x₂ ≤ 1000 

Pen A requires 400 clips per day 

∴ x₁ ≤ 400 

Pen B requires 700 clips per day 

∴ x₂ ≤ 700

(iv) Non-negative restriction:

Since the number of pens is non-negative, we have x₁ > 0, x₂ > 0. 

Thus, the mathematical formulation of the LPP is 

Maximize Z = 5x₁ + 3x₂ 

Subj ect to the constrains 

2x₁ + x₂ ≤ 1000, x₁ ≤ 400, x₂ ≤ 700, x₁, x₂ ≥ 0

(i) Maximin principle

Max (10, 5, 3) = 10. Since the maximum pay-off is 10 units, the marketing manager has to choose Egg shampoo by Maximin rule.

(ii) Minimax principle

Min (30, 40, 55) = 30. Since the minimum pay-off is 30 units, the marketing manager has to choose Egg shampoo by minimax rule.

(c) occupied and unoccupied cells

(a) Total supply ≠ Total demand

The correct answer is : (c) 1 or 0

(b) Minimise the total cost

1. units supplied, units demanded 

2. Dummy rows, dummy columns 

3. Vogel’s approximation method 

4. Degeneracy 

5. Network flow problems 

6. zero

(d) All of the above

The correct answer is : (c) (a) or (b)

Z_A = 20(0) + 70(8) 

= 560; and Z_B = 20(12) + 70(5) 

= 590 (max) 

∴ Optimal soplution is at B(12,5); is 

x = 12 and y = 5.

’’Whenever there is a situation of conflict and competition between two or more opposing teams, We refer to the situation as a game”.

While playing a game, the pure strategy of a player is the predetermined decision to adopt a specified course of action irrespective of the course of action of the opponent.

The strategy of player is the predetermined rule by which he chooses his course of action while playing the game.

1. there are finite numbers of players 

2. Each player have finite number of courses of action OR The game is said to be played, one of the player should adopt one of course action while playing the game.

Here α – Maximin and β – Minimax are equal and are equal to ν = 2.

It is the gain or loss of the players while playing a game.

The game is not fair, (the game is said to be fair if ν = 0).

For some LPP the optimum value of Z may be infinity, then LPP is said to have unbounded solution.

The assignment problem is a special case of the transportation problem. The differences are given below.

Suppose that we have ‘m’ jobs to be performed on ‘n’ machines. The cost of assigning each job to each machine is C_ij . (i = 1, 2,…, n and j = 1, 2,…. n).Our objective is to assign different jobs to different machines (one job per machine) to minimize the overall cost. This is known as the assignment problem.

(a) top left corner

(a) exactly one

(c) Smallest two costs

Feasible Solution: A feasible solution to a transportation problem is a set of non-negative values x_ij (i = 1, 2,.., m, j = 1, 2, …n) that satisfies the constraints. 

Non-degenerate basic feasible Solution: If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocation in independent positions, it is called a Non-degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

(c) less than m + n – 1

(b) the balance between total activities and total resources

(a) Least cost method

(b) Travelling salesman problem

(c) Multiple optimal solutions exist