QMT PAST YEAR QUESTION PDF

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TOPIC : LINEAR PROGRAMMINGJUN 2019 (PA) a) True. (1 mark) 2.(3 marks) Corner points Z = 30X 1 + 20X 2 (0 , 0) 0 (0 , 800) 16, (350 , 0) 10, (200 , 600) 18, (350 , 300) 16,Optimal solution: X 1 = 200, X 2 = 600, Z = RM18,000. ( marks)JUN 2019 (PB)a) X 1 = number of simple canoes to be manufactured X ...

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TOPIC : LINEAR PROGRAMMING JUN 2019 (PA) 1. a) True.

(1 mark)

2.

(3 marks) Corner points (0 , 0) (0 , 800) (350 , 0) (200 , 600) (350 , 300)

Z = 30X1 + 20X2 0 16,000 10,500 18,000 16,500

Optimal solution: X1 = 200, X2 = 600, Z = RM18,000. marks)

(2

JUN 2019 (PB) a) X1 = number of simple canoes to be manufactured X2 = number of luxury canoes to be manufactured Maximize Z = 70X1 + 100X2 Subject to: 4.5X1 + 5X2 ≤ 1800 2X1 + X2 ≤ 600 2X1 + 4X2 ≤ 1200 X2 ≤ 2/3 (X1 + X2) X1 , X2 ≥ 0

(5 marks)

b) i) Cj 0 500 300 Zj C j - Zj

50 - 50

110,000

(3 marks) ii) X 1 = 100 X 2 = 200 Profit = RM110,000 (2 marks) iii) It is not worth to increase the number of special circuits since the shadow price is equal to RM0. (Y 1 = 0) (2 marks)

iv) Y2 = RM250. v) There is no alternative optimal solution since no non-basic variables equal to zero in Cj - Zj row. marks) DEC 2018 (PA)

(1 mark) (2

1. i)

(1 mark)

True.

2.

(3.5 marks) Corner points (0 , 0) (150 , 0) (0 , 80) (125 , 75) (150 , 50) (100 , 80)

Z = 40X1 + 30X2 0 6000 2400 7250 7500 6400

Optimal solution: X1 = 150, X2 = 50. Maximum total profit = RM7500

(1.5 marks)

DEC 2018 (PB) a) X1 = number of tables to be produced X2 = number of chairs to be produced Maximize Z = 7X1 Subject to: 4X1 + 3X2 2X1 + X2 X1 X1 , X2

+ 5X2 ≤ ≤ ≤ ≥

240 100 4X2 0

(5 marks)

b) i) Cj 5 0 7 Zj C j - Zj

7 0

5 0

1.6875 -1.6875

0 0

0.0625 -0.0625

405

(3 marks) ii) X 1 = 15 S1 = 0 X 2 = 60 S 2 = 10 Z = 405 S3 = 0 (3 marks) iii) S 2 = 10 (1 mark) iv) U1 = 1.6875 U2 = 0 U3 = 0.0625 (1 mark) v) There is no alternative optimal solution because none of the non-basic variables has zero value in Cj - Zj row. (2 marks)

JUN 2018 (PA) a) i) b)

True.

(1 mark)

(4 marks) Corner points (20 , 10) (32 , 10) (8 , 22) (8 , 40)

C = 24x + 28y 760 1048 808 1312

The optimal solution C = 760, x = 20 and y = 10.

(1 mark)

JUN 2018 (PB) a) x1 = number of strawberry flavoured cake. x2 = number of chocolate flavoured cake Maximize Z = 1.1x1 + 1.2x2 Subject to: 0.3x1 + 0.3x2 ≤ 45

(1/2 mark) (1 mark) (1 mark)

0.1x1

≤ 12 0.1x2 ≤ 10 x1 , x2 ≥ 0

(1 mark) (1 mark) (1/2 mark)

b) i) Cj

Solution Mix

1.1 0 1.2

x1 S2 x2 Zj C j - Zj

x1

x2

S1

S2

S3

1.1 1 0 0 1.1 0

1.2 0 0 1 1.2 0

0 3.33 -0.033 0 3.67 -3.67

0 0 1 0 0 0

0 -10 1 10 1 -1

Quantity 50 7 100 175

ii) The maximum total profit = RM175 iii) No alternative optimal solution since non-basic variables in C j - Z j row are not equal to zero. marks) iv) S 1 = 0, S 3 = 0 S 1 and S 3 are fully utilized? v) y 1 = 3.67, should pay at most RM3.67 for an additional unit of the first resource. y 2 = 1, should pay at most RM1 for an additional unit of the third resource.

(3 marks) (1 mark) (2

(2 marks) (2 marks)

JAN 2018 (PA) a)

b) The solution is Unbounded. JUL 2017 (PA) a)

(4 marks) (1 mark)

(4 marks) (1 mark)

b) Special case: Unbounded Solution. JUL 2017 (PB) a) X1 = number of one-bedroom units to build X2 = number of two-bedroom units to build X3 = number of three-bedroom units to build Maximize Z = 550X1 + 750X2 + 900X3 Subject to: 650X1 + 850X2 + 1000X3 ≤ 50,000 X1 ≤ 12 X2 ≤ 28 X3 ≤ 14 X1 + X2 + X3 ≤ 50 X1 , X2 , X3 ≥ 0

(1 mark) (1 mark) (1/2 mark) (1/2 mark) (1/2 mark) (1 mark) (1/2 mark)

b) i) Cj

Solution Mix

0 0 50

S1 S2 X3 Zj C j - Zj

30

24

50

0

0

0

X1 -16 -1 2 100 -70

X2 -1 11 1 50 -26

X3 0 0 1 50 0

S1 1 0 0 0 0

S2 0 1 0 0 0

S3 -13 -1 1 50 -50

Quantity 320 140 10 500 (3 marks)

ii) X 1 = 0 X2 = 0 X 3 = 10 Total profit = RM500 iii) Resources 1 and Resources 2 are not fully utilized. iv) Maximum amount to pay for Resource 3 is RM50. v) No alternate optimal solution. There is no zero value for non-basic variables in C j – Zj row. marks) DEC 2016 (PA) a)

(2 marks) (2 marks) (1 mark) (2

Corner points (0 , 15) (4 , 10) (24 , 0) (0 , 30)

Z = 8X + 3Y 45 62 192 90 (3 marks) (2 marks)

b) The optimal solution Z = 45, X = 0 and Y = 15.

DEC 2016 (PB) a) x1 = number of air conditioners to be produce x2 = number of large fans to be produce Maximize Z = 25x1 + 15x2 Subject to: 3x1 + 2x2 ≤ 240 (wiring) 2x1 + x2 ≤ 140 (drilling) x1 ≥ 20 (minimum supply) x1 , x2 ≥ 0

(1/2 mark) (1 mark) (1 mark) (1 mark) (1 mark) (1/2 mark)

b) i) Cj 10 0 20 Zj C j - Zj

10 0

20 0

15 0

2.5 -2.5

0 0

2.5 -2.5

1525 (3 marks)

ii) X 1 = 37.5 S1 = 0 Total profit = RM1525 X 2 = 57.5 S2 = 5 X3 = 0 S3 = 0 iii) The ingredient that is not fully utilized is ingredient B. The amount unused is 5 units. iv) Y1 = RM2.5 Y2 = RM0 Y3 = RM2.5 Total cost = RM1525 marks) JUN 2016 (PA) a)

(2 marks) (1 mark)

(4

Graph Corner points C = 86X + 16Y (0.71 , 2.58) 102.34 (1.5 , 1) 145 (3.33 , 1) 302.38 b) The minimum C = 102.34, with the value of X = 0.71 and Y = 2.58.

(3 marks)

(2 marks)

JUN 2016 (PB) a) i) Zj C j - Zj

90 -85

20 0

25 0

20 -20

25 -25

0 0

1550 (3 marks)

ii) X 1 = 0 X 2 = 40 X 3 = 30 S1 = 0 S2 = 0 S 3 = 30 Maximum profit = 1550 iii) Resources 1 and 2. iv) A new profit function = 1550 + (20)(10) = RM1750 DEC 2015 (PA)

(3 marks) (2 marks) (2 marks)

Graph

(2½ marks)

Corner points Z = 5x + 7y (0,0) 0 (0,6.3) 44.1 (5,3) 46 (6,2) 44 (6,0) 30 Maximum Z = RM46, when x = 5 and y = 3.

(2½ marks)

DEC 2015 (PB) a)

X1 = Quantity of component A X2 = Quantity of component B X3 = Quantity of component C Maximize Z = 8x1 + 6x2 + 9x3 Subject to: 16x1 + 5x2 + 8x3 ≤ 7200 12x1 + 6x2 + 8x3 ≤ 6600 x3 ≤ 200 x2 ≤ 1000 x1 = 500 x1 , x 2 , x3 ≥ 0

(1 mark) (1 mark) (1/2 mark) (1/2 mark) (1/2 mark) (1/2 mark) (1/2 mark) (1/2 mark)

b) i) Cj

Basic

4 0 2

X1 S2 X3 Zj C j - Zj

4

1

2

0

0

0

X1 1 0 0 4 0

X2 -1 -2 6 8 -7

X3 0 0 1 2 0

S1 0.25 0 -0.25 0.5 -0.5

S2 0 1 0 0 0

S3 -0.25 -1 1.25 1.5 -1.5

Quantity 2.5 20 7.5 25 (3 marks)

ii) X 1 = 2.5 Maximum profit = RM25 X2 = 0 X 3 = 7.5 iii) If one unit of resource 2 is increase, the profit remains the same or does not change. iv) Minimize C = 20Y1 + 30Y2 + 10Y3 Subject to: 5Y1 + Y2 + Y3 ≥ 4

(2 marks) (1 mark)

Y1 + 3Y2 + 5Y3 ≥ 1 Y1 + Y2 + Y3 ≥ 2 Y1 , Y2 , Y3 ≥ 0

(4 marks)

DEC 2014 (PB) a)

Graph Special case : The solution is unbounded (Unboundedness) or Redundancy.

(4 marks) (1 mark)

b) i) Cj

Basic

25 20 0

X3 X2 S2 Zj C j - Zj

10 X1 3 2 -1 115 -105

20 X2 0 1 0 20 0

25 X3 1 0 0 25 0

0 S1 0 1 0 20 -20

0 S2 0 0 1 0 0

0 S3 1 0 -1 25 -25

Quantity 10 40 10 1050 (4 marks)

ii) X 1 = 0 Maximum profit = RM1,050 X 2 = 40 X 3 = 10 iii) Resource 1 and Resource 2 are fully utilized? iv) The shadow price for Resource 2; Y2 = 0 That means an additional unit of Resource 2 will not affect the profit. The shadow price for Resource 3; Y3 = 25 That means an additional unit of Resource 3 will increase the profit by RM25.

(2 marks) (2 marks) (1 mark) (1 mark)

JUN 2014 (PA) 1. a) False

(1 mark)

2.

Graph

(2½ marks)

Corner points Z = 8x + 6y (0,5) 30 (0,9) 54 max (4,1) 38 (4,3) 50 The optimal solution is when x = 0 and y = 9 with maximum profit of RM54.

(2½ marks)

JUN 2014 (PB) a) x1 = number of Calculator A to be produced x2 = number of Calculator B to be produced Maximize Z = 4x1 + 6x2 Subject to: x1 + 3x2 ≤ 61,000 0.2x1 + 0.35x2 ≤ 8000 x1 ≥ 35,000 x1 , x 2 ≥ 0

(5 marks)

b) i) Cj

Solution mix

100 0 90

X1 S2 X2 Zj C j - Zj

100

90

0

0

0

X1 1 0 0 100 0

X2 0 0 1 90 0

S1 2/3 -1/3 -1/3 36.67 -36.67

S2 0 1 0 0 0

S3 -1/3 -1/3 2/3 26.67 -26.67

Quantity 6 1 7 1230 (3 marks)

ii) X 1 = 6 X2 = 7 Optimal profit = RM1230 iii) S 1 = S 3 = 0. The manufacturing time (S1) and steel (S3) are fully utilized. iv) If there were one additional hour of manufacturing time, the profit will increase by RM36.67 mark) v) Y1 = RM36.67 Y2 = RM0 Y3 = RM26.67

(2 marks) (2 marks) (1

(2 marks)

DEC 2013 (PA)

Graph

(2½ marks) Corner points (0,10)

Z = 8x + 2y 20

(0,6) 12 (4,5) 42 (4,4) 40 Optimal solution, x = 4, y = 5. Maximum Z = 42.

(2½ marks)

DEC 2013 (PB) a) x1 = number of laser printers assembled x2 = number of inkjet printers assembled Maximize Z = 300x1 + 250x2 Subject to: 2x1 + x2 ≤ 40 x1 + 3x2 ≤ 45 x1 ≤ 12 x1 , x 2 ≥ 0 b) i)

(6 marks)

Cj 55 30 0 Zj C j - Zj

55 0

35 -10

30 0

25 -25

5 -5

0 0

2500 (4 marks)

ii) X 1 = 40 S1 = 0 Maximum profit = RM2,500 X2 = 0 S2 = 0 X 3 = 10 S 3 = 30 iii) The resource that is not fully utilized is Resource C. iv) If one additional unit of Resource A is made available, the profit will increase by RM25.

(3 marks) (1 mark) (1 mark)

AUG 2013 (PA) 1. a) True.

(1 mark)

2.

Graph Corner points

(2 marks) Z = 3X1 + 4X 2

(0,0) (20,0) (0,24) (10,20)

3.

0 60 96 110

Maximum profit = RM110 when X1 = 10, X2 = 20

(2 marks) (1 mark)

X = number of item X produced Y = number of item Y produced Maximize Z = 20X + 30Y Subject to: 15X + 10Y ≤ 40 x 60 = 2400 20X + 25Y ≤ 35 x 60 = 2100 X ≥ 10 X,Y ≥ 0

(5 marks)

4.

Graph Corner points (0,0) (0,2.5) (2,2)

(2½ marks) Z = 5X + 6Y 0 15 22 (1½ marks) (1 mark)

Maximum profit = RM22 when x = 2, y = 2 AUG 2013 (PB) a)

b) c)

d)

e)

Optimal solution: X1 = 0 S 1 = 60 Total profit = RM10,250 X 2 = 105 S2 = 0 X 3 = 230 S3 = 0 The resources that are fully utilized are S2 and S3 There is no alternate optimal solution since none of the non-basic variables has zero value in C j – Zj row. marks) The dual model for the primal LP problem: Minimize C = 500Y1 + 460Y2 + 840Y3 Subject to: Y1 + 3Y2 + 2Y3 ≥ 30 2Y1 + 8Y3 ≥ 10 Y1 + 2Y2 ≥ 40 Y1 , Y2 , Y3 ≥ 0 Dual optimal solution: Y1 = 0, means 1 unit increase in Constraint 1 will not change the total profit. Y2 = 20, means 1 unit increase in Constraint 2 will increase the total profit by RM20. Y3 = 1.25, means 1 unit increase in Constraint 3 will increase the total profit by RM1.25.

AUG 2013 (PB) a) Solution

45

50

60

0

0

0

(4 marks) (1 mark) (2

(5 marks)

(3 marks)

Cj 50 60 0

Mix X2 X3 S3 Zj C j - Zj

X1 -0.25 1.5 4 77.5 -32.5

X2 1 0 0 50 0

X3 0 1 0 60 0

S1 0.5 0 -4 25 -25

S2 -0.25 0.5 2 17.5 -17.5

S3 0 0 1 0 0

Quantity 50 300 500 20,500 (4 marks)

b)

c) d)

e)

f)

X1 = 0 X 2 = 50 X 3 = 300 Profit = RM20,500 S 3 = 500 Minimize C = 400Y1 + 600Y2 + 900Y3 Subject to: Y1 + 3Y2 + 2Y3 ≥ 45 2Y1 + 8Y3 ≥ 50 Y1 + 2Y2 ≥ 60 Y1 , Y2 , Y3 ≥ 0 Y1 = 25, means 1 hour increase in labor hours will increase the total profit by RM25. Y2 = 17.5, means 1 hour in machine hours will increase the total profit by RM17.50. Y3 = 0, means 1 hour increase in processing time will not change the total profit. marks) New total profit = RM20,500 + (3)(25) = RM20,575

(2 marks) (1 mark)

(4 marks)

(2

(2 marks)

JUN 2013 (PA) a)

Graph Corner points Z = 300x + 250y (20,0) 6000 (15,10) 7000 (0,15) 3750 Maximum profit = RM7000 when x = 15, y = 10 b)

(3 marks)

(2 marks)

i) Cj 2 0 3

Solution mix X2 S2 X1 Zj C j - Zj

3 X1 0 0 1 3 0

2 X2 1 0 0 2 0

1 X3 3 6 -1 3 -2

0 S1 1.5 2.5 -0.5 1.5 -1.5

0 S2 0 1 0 0 0

0 S3 -0.5 -1.5 0.5 0.5 -0.5

Quantity 7.5 22.5 2.5 22.5 (4 marks)

ii)

iii)

X 1 = 2.5 X 2 = 7.5 X3 = 0 Z = 22.5 U 1 = 1.5

(2 marks)

U2 = 0 U 3 = 0.5 iv) Profit increase by = 3 x 1.5 = 4.5 New profit = 22.50 + 4.50 = 27.00

(2 marks) (2 marks)...