Questions & explanations
1. In a transportation LPP, two godowns supply 100 and 50 units to three shops demanding 60, 50, 40 units. If x and y are units from godown A to shops X and Y, what is the minimum cost if cost per unit from A to X,Y,Z is 6,3,2.5 and from B to X,Y,Z is 4,2,3?
- Rs 405
- Rs 415
- Rs 420
- Rs 510
Answer: Rs 510
Let x, y be units from A to X, Y. Then A to Z = 100-x-y, B to X = 60-x, B to Y = 50-y, B to Z = x+y-60. Non-negativity: x≥0, y≥0, x≤60, y≤50, x+y≥60, x+y≤100. Cost Z = 6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60) = 2.5x+1.5y+410. Feasible vertices: (10,50) and (60,0). Z(10,50)=510, Z(60,0)=560. Minimum cost is Rs 510.
2. The feasible region of the system x + y ≤ 4, 2x + y ≤ 6, x ≥ 0, y ≥ 0 is:
- a quadrilateral with vertices (0,0), (3,0), (2,2), (0,4)
- a triangle with vertices (0,0), (3,0), (0,4)
- a quadrilateral with vertices (0,0), (4,0), (2,2), (0,6)
- a triangle with vertices (0,0), (4,0), (0,6)
Answer: a quadrilateral with vertices (0,0), (3,0), (2,2), (0,4)
The feasible region is the intersection of all half-planes. Graph each inequality: x+y≤4 gives half-plane below line through (4,0) and (0,4); 2x+y≤6 gives half-plane below line through (3,0) and (0,6); x≥0 and y≥0 restrict to first quadrant. The overlap is a quadrilateral with vertices (0,0), (3,0), (2,2), (0,4).
3. For the LPP: maximize Z = kx + 2y subject to 2x + y ≤ 8, x + 2y ≤ 10, x, y ≥ 0, the vertices are (0,0), (4,0), (2,4), (0,5). For what value of k does the LPP have multiple optimal solutions?
- k = 1
- k = 2
- k = 3
- k = 4
Answer: k = 1
For alternate optima, two vertices must give the same Z and that value must be the maximum. Equate Z at (2,4) and (0,5): 2k+8 = 10 ⇒ k = 1. At k=1, Z values: (0,0)=0, (4,0)=4, (2,4)=10, (0,5)=10. The maximum is 10, achieved at both vertices, so multiple optimal solutions exist.
4. For the transportation problem with godowns A(100), B(50) and shops X(60), Y(50), Z(40), the minimum transport cost (in ₹) is:
- ₹410
- ₹470
- ₹455
- ₹440
Answer: ₹440
Let x = A→X, y = A→Y. Then Z = 2.5x + 1.5y + 410. Constraints: x≥0, y≥0, x≤60, y≤50, x+y≥60, x+y≤100. Feasible region corners: (0,60), (0,50), (10,50), (60,0), (60,40). Evaluate Z: (0,60)→500, (0,50)→485, (10,50)→440, (60,0)→560, (60,40)→560. Minimum is 440 at (10,50).
5. A company makes two products P1 and P2. Resources: labour 24 h, machine 20 h, finishing 12 h. P1 needs 2,2,1 h; P2 needs 3,1,2 h. Profit: ₹40 and ₹30 per unit. Which constraint is redundant?
- 2x + y ≤ 20
- 2x + 3y ≤ 24
- x + 2y ≤ 12
- x ≥ 0, y ≥ 0
Answer: 2x + 3y ≤ 24
The feasible region is bounded by x+2y≤12, 2x+y≤20, and non-negativity. The labour constraint 2x+3y≤24 is redundant because its boundary line lies outside the region formed by the other constraints; removing it does not change the feasible region.
6. A cottage industry makes pedestal lamps (x) and wooden shades (y). Each lamp needs 2 h on sprayer and 3 h with operator; each shade needs 1 h on sprayer and 2 h with operator. Sprayer available 12 h/day, operator 20 h/day. Profit is ₹25 per lamp and ₹15 per shade. What is the maximum daily profit?
- ₹150
- ₹160
- ₹170
- ₹180
Answer: ₹160
Formulate: maximize Z = 25x + 15y subject to 2x + y ≤ 12, 3x + 2y ≤ 20, x, y ≥ 0. Vertices: (0,0), (6,0), (4,4), (0,10). Z values: 0, 150, 160, 150. Maximum is 160 at (4,4). The feasible region is bounded, so optimum exists at a corner.
7. A furniture dealer makes tables (₹50 profit) and chairs (₹15 profit). Space: total items ≤ 60. Money: 250 per table, 50 per chair, total ≤ ₹5000. How many tables and chairs maximize profit?
- 20 tables, 0 chairs
- 10 tables, 50 chairs
- 0 tables, 60 chairs
- 10 tables, 40 chairs
Answer: 10 tables, 50 chairs
Let x = tables, y = chairs. Maximize Z = 50x + 15y subject to x + y ≤ 60, 250x + 50y ≤ 5000 (simplifies to 5x + y ≤ 100), x, y ≥ 0. Corner points: (0,0): Z=0; (20,0): Z=1000; (10,50): Z=1250; (0,60): Z=900. Maximum Z = 1250 at (10,50).
8. A company makes two products A and B. Machine 1 (40 h) needs 2 h for A, 1 h for B; Machine 2 (40 h) needs 1 h for A, 2 h for B; Machine 3 (40 h) needs 1 h for A, 1 h for B. Profit is ₹6 per A and ₹4 per B. What is the maximum profit?
- ₹160
- ₹120
- ₹140
- ₹133.33
Answer: ₹133.33
Constraints: 2x+y≤40, x+2y≤40, x+y≤40, x,y≥0. The third constraint is redundant. Intersection of first two gives (40/3,40/3). Profit = 6*(40/3)+4*(40/3)=400/3≈133.33. Check corners: (20,0)=120, (0,20)=80, (0,0)=0. Maximum is 400/3.
9. Two godowns A (100 quintals) and B (50 quintals) supply three shops X (60), Y (50), Z (40). Transport cost per quintal: A→X ₹6, A→Y ₹3, A→Z ₹2.5; B→X ₹4, B→Y ₹2, B→Z ₹3. Let x = A→X, y = A→Y. What is the total cost function?
- Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60)
- Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(60 - x - y)
- Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(100 - x - y)
- Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x - y + 60)
Answer: Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60)
From A to Z: 100 - x - y. From B to X: 60 - x. From B to Y: 50 - y. From B to Z: total demand 40 minus (100 - x - y) from A = x + y - 60. Summing costs gives Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60).
10. A furniture dealer sells tables (₹50 profit) and chairs (₹15 profit). Space allows at most 60 pieces. He has ₹5000 to invest; each table costs ₹250, each chair costs ₹50. Which LPP formulation maximizes profit?
- Max Z = 250x + 50y, subject to x + y ≤ 60, 50x + 15y ≤ 5000, x ≥ 0, y ≥ 0
- Max Z = 50x + 15y, subject to x + y ≤ 60, 250x + 50y ≥ 5000, x ≥ 0, y ≥ 0
- Max Z = 50x + 15y, subject to x + y ≤ 60, 250x + 50y ≤ 5000, x ≥ 0, y ≥ 0
- Max Z = 50x + 15y, subject to x + y ≥ 60, 250x + 50y ≤ 5000, x ≥ 0, y ≥ 0
Answer: Max Z = 50x + 15y, subject to x + y ≤ 60, 250x + 50y ≤ 5000, x ≥ 0, y ≥ 0
Profit per table is ₹50 and per chair is ₹15, so objective is Z = 50x + 15y. Space constraint: total pieces ≤ 60 gives x + y ≤ 60. Investment constraint: total cost ≤ ₹5000 gives 250x + 50y ≤ 5000. Non-negativity: x ≥ 0, y ≥ 0.
11. Consider the LPP: Maximise Z = x + y subject to x + y ≥ 3, x ≥ 0, y ≥ 0. Which of the following is true?
- The maximum value is 3
- The maximum value is 0
- The maximum value is 1.5
- No maximum exists
Answer: No maximum exists
The feasible region is unbounded in the direction of increasing x and y. The corner point (3,0) gives Z=3, but the open half-plane x+y > 3 intersects the feasible region (e.g., (4,0) gives Z=4). Hence, no finite maximum exists.
12. Maximise Z = 2x + 3y subject to 2x + 3y ≤ 12, x + y ≤ 5, x ≥ 0, y ≥ 0. Which of the following is true?
- The LPP has a unique optimal solution
- The LPP has multiple optimal solutions
- The LPP has no feasible solution
- The LPP has unbounded solution
Answer: The LPP has multiple optimal solutions
The objective function Z = 2x+3y is parallel to the constraint 2x+3y ≤ 12. The corner points (0,4) and (3,2) both give Z=12, and every point on the line segment joining them is optimal. Hence multiple optimal solutions exist.