Operations Research

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Questions:

Generic 5

4 x 5 = 20

Draw the network and calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Jobs	1-2	1-3	2-4	3-4	3-5	4-5	4-6	5-6
Duration	6	5	10	3	4	6	2	9

Solve the following linear programming problems by graphical method.
(i) Maximize Z = 6x₁ + 8x₂ subject to constraints 30x₁+20x₂≤300; 5x₁+10x₂≤110; and x₁, x₂ ≥ 0 .
(ii) Maximize Z = 22x₁+ 18x₂ subject to constraints 960x₁+ 640x₂≤15360; x₁≥ x≤20 and x₁, x₂ ≥ 0.
(iii) Minimize Z = 3x₁+ 2x₁ subject to the constraints 5x₁+ x₂≥10; x₁+x₂≥6; x₁+ 4, x₂ ≥12 and x₁, x₂≥ 0.
(iv) Maximize Z = 40x₁+ 50x₂ = subject to constraints 30x₁+x₂≤9; x₁+2x₂≤8 and x₁, x₂≥ 0
(v) Maximize Z = 20x₁+ 30x₂ subject to constraints 3x₁+3x₂≤36; 5x₁+2x₂≤50; 2x₁+6x₂≤60 and x₁,x₂≥ 0
(vi) Minimize Z = 20x ₁+ 40x₂ subject to the constraints 36x₁+ 6x₂≥108, 3x₁+12x₂≥36, 20x₁+10x₂≥100 and x₁,x₂≥ 0.

Maximize Z = 3x₁ + 4x₂ subject to x₁ – x₂ ≤ –1; –x₁ + x₂ ≤ 0 and x₁, x₂ ≥ 0

Solve the following LPP. Maximize Z = 2 x₁ + 5x₂ subject to the conditions x₁+ 4x₂ ≤ 24. 3x₁+x₂ ≤ 21, x₁+x₂ ≤ 9 and x₁, x₂ ≥ 0.

Generic 2

8 x 2 = 16

Construct the network for the projects consisting of various activities and their precedence relationships are as given below: A, B can start simultaneously
A < D, E; B < F; E < G, D < C, F < H.

Draw a network diagram for the project whose activities and their predecessor relationships are given below

Activity	A	B	C	D	E	F
Predecessor activity	-	-	D	A	B	C

A dealer whises to purchase a number of fans and sewing machines. He has only Rs.5760 to invest and has a space for atmost 20 items. A fan costs him Rs.360 and a sewing machine Rs. 240. His expectation is he can sell a fan at a profit of Rs.22 and a sewing machine at a profit of ns. Formulate this as an LPP to maximize his profit?

A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods X and Y. To produce one unit of X, 2 unit of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of Y. If X and Yare priced at HOO and H20 per unit respectively, how should the producer use his resources to maximize the total revenue? Formulate the LPP for the above.

An aeroplane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. Formulate the mathematical LPP for the above.

10.

Construct the network for the projects consisting of various activities and their precedence relationships are as given below:
A, B, C can start simultaneously A < F, E; B < D, C; E, D < G

11.

Draw the network for the project whose activities with their relationships are given below:
Activities A, D, E can start simultaneously; B, C > A; G, F > D, C; H > E, F.

12.

Draw the logic network for the following:
Activities C and D both follow A, activity E follows C, activity F follows D, activity E and F precedes B.

Generic 3

4 x 3 = 12

13.

Solve the following LPP graphically. Maximize \(Z={ x }_{ 1 }+{ x }_{ 2 }\)
Subject to the constraints \({ x }_{ 1 }-{ x }_{ 2 }\le -1,{ -x }_{ 1 }+{ x }_{ 2 }\le 0\quad and\quad { x }_{ 1 }+{ x }_{ 2 }\ge 0\)

14.

Solve the following LPP graphically. ∴ Maximize Z = 3x₁ + 4x₂subject to the constraints x₁ + x₂ ≤ 4 and x₁,x₂ ≥ 0.

15.

Construct a network diagram for the following situation:
A < D, E; B, D < F; C < G and B < H.

16.

Draw a network diagram for the project whose activities and their predecessor relationships are given below:

Activity:	A	B	C	D	E	F	G	H	I	J	K
Predecessor activity:	-	-	-	A	B	B	C	D	F	H,I	F,G

Multiple Choice Question

6 x 1 = 6

17.

Given an L.P.P maximize Z = 2x₁ + 3x₂ subject to the constrains x₁ + x₂ ≤ 1, 5x₁ + 5x₂≥ 0 and x₁ ≥ 0, x₂ ≥ 0 using graphical method, we observe ______.

(a)

No feasible solution

(b)

unique optimum solution

(c)

multiple optimum solution

(d)

none of these

18.

Network problems have advantage in terms of project _______.

(a)

Scheduling

(b)

Planning

(c)

Controlling

(d)

All the above

19.

Which of the following is not correct?

(a)

Objective that we aim to maximize or minimize

(b)

Constraints that we need to specify

(c)

Decision variables that we need to determine

(d)

Decision variables are to be unrestricted

20.

The maximum value of the objective function Z = 3x + 5y subject to the constraints x > 0 , y > 0 and 2x + 5y ≤ 10 is ______.

(a)

(b)

(c)

(d)

21.

A solution which maximizes or minimizes the given LPP is called ______.

(a)

a solution

(b)

a feasible solution

(c)

an optimal solution

(d)

none of these

22.

The critical path of the following network is________.

(a)

1 – 2 – 4 – 5

(b)

1 – 3 – 5

(c)

1 – 2 – 3 – 5

(d)

1 – 2 – 3 – 4 – 5

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Answers Key:

Generic 5

4 x 5 = 20

E₁ = 0	L₆ = 31
E₂ = 0 + 6 = 6	L₅ = 31 - 9 = 22
E₃ = 0 + 5 = 5	L₄ = min of {22- 6, 31-2} = 16
E₄ = max of {6 + 10, 5 + 3}	L₃ = min of {16- 3, 22- 4} = 13
E₄ = 16	L₂ = 16 -10 = 6
E5 = max of { 5 + 4, 16 + 6 } = 22	L₁ = min of { 6 - 6, 13 - 5} = 0

Activity	Duration t_ji	EST	EFT = EST + t_ji	LST = LFT - t_ij	LFT
1-2	6	0	6	6-6 = 0	6
1-3	5	0	5	18-5 = 13	13
2-4	10	6	16	16-10 = 6	16
3-4	3	5	8	16-3 = 13	16
3-5	4	5	9	22-4 = 18	22
4-5	6	16	22	22-6 = 16	22
4-6	2	16	18	31-2 = 29	31
5-6	9	22	31	31-9 = 22	31

Since EFT and LFT is same on 1 - 2, 2 - 4, 4 - 5 and 5 - 6, the critical path is 1 - 2 - 4 - 5 - 6 and duration time taken is 31 days.

(i) First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
\(\therefore\) We have the lines 30x₁ + 20x₂≤ 300;
5x₁ + 10x₂ ≤ 110
30x₁ + 20x₂≤ 300; is a line passing through the points (0, 15) and (10, 0).
[(0, 15) is obtained by taking x₁ = 0 in 30x₁ + 20x₂ = 300 (10,0) is obtained by taking x₂ = 0 in 30x₁ + 20x₂ = 300]
Any point lying on or below the line 30x₁ + 20x₂= 300. Satisfies the constraint 30x₁ + 20x₂≤ 300
We follow the same steps for the following
30x₁ + 20x₂= 300

x₁	0	10
x₂	15	0

5x₁+10x₂=110

x₁	0	22
x₂	11	0

Now we draw the graph
30x₁+20x₂=300 is a line passing through the points (0,15) and (10,0).
Any point lying on or below the line 30x₁+20x₂=300 satisfies the constrint 30x₁+20x₂≤300.

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(10, 0), B(4,9), C(0,11).
The Corner points and corresponding Z value are

Corner points	Z = 6x₁+ 8x₂
O(0,0)	0
A(10,0)	60
B(4,9)	24 + 72 = 96
C(0,11)	88

The optimal solution is B(4,9)
\(x_1=4, x_2=9, Z_{\max }=96\)
Verification
\(3 x_1+2 x_2 =30 \)
\(x_1+2 x_2 =22 \)
(-) (-) (-)
__________
2x₁ = 8
x₁ = 4
4 + 2x₂ = 22
2x₂ = 18
x₂ = 9
(4, 9) is the point of inch
(ii) First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
\(\therefore\) We have the lines \(960 x_1+640 \dot{x}_2 \leq 15360; x_1+x_2 \leq 20\)
\(960 x_1+640 x_2=15360\) is a line passing through the points (0,24) and (16,0).
[(0,24) is obtained by taking x₁ = 0 in 960 x₁+ 640 x₂ = 15360, (16,0) is obtained by taking x₂= 0 in \(960 x_1+640 x_2=15360]\)
Any point lying on or below the line \(960 x_1+640 x_2=15360\). Satisfies the constraint \(960 x_1+640 x_2 \leq 15360\).
We follow the same steps for the following
960x₁ + 640x₂= 15360

x₁	0	16
x₂	24	0

x₁ + x₂ = 20

x₁	0	20
x₂	20	0

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(16, 0), B(8,12), C(0,20).

Corner Points	Z = 22x₁ + 18x₂
O(0,0)	0
A(16,0)	352
B(8,12)	176+216=392
C(0,20)	360

The optimal solution is at B(8,12)
\(x_1=8, x_2=12, Z_{\max }=392\)
Verification
\(960 x_1+640 x_2=15360 \) ...(1)
\(\div 320 \quad 3 x_1+2 x_2=48 \quad 8+x_2=20 \)
\(\text { (2) } \times 2 \frac{2 x_1+2 x_2=40}{x_1=8}\) \( x_2=12\)
\( \therefore B(8,12)\)
(iii) Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
consider the equations \(5 x_1+x_2=10; x_1+x_2=6 \ and \ x_1+4 x_2=12\).
5x₁ + x₂ = 10 is a line passing through the points (0,10) and (2,0). Any point lying on or above the line 5x₁ + x₂ = 10. Satisfies the constraint \(5 x_1+x_2 \geq 10.\)
We follow the same steps for the following
5x₁ + x₂ = 10

x₁	0	2
x₂	10	0

x₁+ x₂= 6

x₁	0	6
x₂	6	0

x₁ + 4x₂= 12

x₁	0	12
x₂	3	0

The feasible region is ABCD (since the problem is of minimization type we are 0 moving towards the origin.

Corner Points	Z = 3x₁ + 2x₂
A(0,10)	20
B(1,5)	3+10=13
C(4,2)	12+4=16
D(12,0)	36

The minimum value of z occurs at C(4, 2).
Hence the optimal solution is x₁ = 4, x₂ = 2, Z_min = 13
Verification

\(5 x_1+x_2=10 \)	\(x_1+4 x_2=12 \)
\(x_1+x_2=6 \) \(4 x_1=4 \Rightarrow x_1=1 \) \(x_2=6-1=5 \) \(B(1,5) \)	\(x_1+x_2=6 \) \(3 x_2=6 \) \(x_2=2 \) \(x_1=4 \) C(4,2)

(iv) First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
\(\therefore\) We have the lines \(3 x_1+x_2 \leq 9; x_1+2 x_2 \leq 8\)
3x₁ + x₂ = 9 is a line passing through the points (0,9) and (3,0).
[(0,9) is obtained by taking x₁ = 0 in \(3 x_1+x_2=9,(3,0)\) is obtained by taking \(x_2=0 \ in \ \left.3 x_1+x_2=9\right]\)
Any point lying on or below the line 3x₁ + x₂ = 9. Satisfies the constraint \(3 x_1+x_2 \leq 9\).
We follow the same steps for the following

3x₁ + x₂= 9

x₁	0	3
x₂	9	0

x₁ + 2x₂= 8

x₁	0	8
x₂	4	0

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(3, 0), B(2,3), C(0,4).

Corner Points	Z = 40x₁+ 50x₂
O(0,0)	0
A(3,0)	120
B(2,3)	80+150=230
C(0,4)	200

The optimal solution is at B(2,3)
Solution is \(x_1=2, x_2=3, Z_{\max }=230\)
Verification
\(3 x_1+x_2 =9 \)    (1)
\(x_1+2 x_2 =8 \)   (2)
(1)   \(3 x_1+x_2 =9 \)
(2) x (3) \(3 x_1+6 x_2 =24 \)
\(-5 x_2 =-15 \)
\(x_2 =3 \)
\(3 x_1+3 =9 \)
\(3 x_1 =6 \)
\(x_1 =2 \)
(2, 3)
(v) First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations:
\(\therefore\) We have the lines \( 3 x_1+3 x_2 \leq 36; 5 x_1+2 x_2 \leq 50 ; 2 x_1+6 x_2 \leq 60.\)
\(3 x_1+3 x_2=36 \Rightarrow x_1+x_2=12\) is a line passing through the points (0,12) and (12,0).
[(0,12) is obtained by taking \(x_1=0 \ in \ x_1+x_2=12,(12,0)\) is obtained by taking \(x_2=0 \ in \ \left.x_1+\dot{x}_2=12\right]\)
Any point lying on or below. the line 3x₁ + 3x₂ = 36. Satisfies the constraint \(3 x_1+3 x_2 \leq 36\)
We follow the same steps for the following
3x₁ + 3x₂ = 36
x₁ + x₂ = 12

x₁	0	12
x₂	12	0

5x₁ + 2x₂= 50

x₁	0	10
x₂	25	0

2x₁ + 6x₂= 60; x₁ + 3x₂= 30;

x₁	0	30
x₂	10	0

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(10, 0), B(8.7,3.3), C(3, 9), D(0,10)

Corner Points	Z = 20x₁+ 30x₂
O(0,0)	0
A(10,0)	200
B(8.7, 3.3)	273
C(3,9)	330
D(0,10)	300

The optimal solution is at C(3,9)
\(x_1=3, x_2=9, Z_{\max }=330\)
Verification

\(2 x_1+2 x_2 =24 \)	\(x_1+x_2=12 \)
\(5 x_1+2 x_2 =50 \) \(-3 x_1 =-26 \) \(x_1 =8.7 \) \(8.7+x_2 =12 \) \(x_2 =3.3 \) B(8.7,3: 3)	\(x_1+3 x_2=30 \) \(-2 x_2=-18 \) \(x_2=9 \) \(x_1=3 \) \(C(3,9) \)

(vi) Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
consider the equations \(36 x_1+6 x_2=108; 3 x_1+12 x_2=36 \ and \ 20 x_1+10 x_2=100. 36 x_1+6 x_2=108 \Rightarrow 6 x_1+x_2=18\) is a line passing through the points (0,18) and (3,0). Any point lying on or above the line \(36 x_1+6 x_2=108\) satisfies the constraint \(36 x_1+6 x_2 \geq 108\).
We follow the same steps for the following
36x₁+ 6x₂= 108
6x₁ + x₂ = 108

x₁	0	3
x₂	18	0

3x₁ + 12x₂ = 36
x₁ + 4x₂ = 12

x₁	0	12
x₂	3	0

20x₁ + 10x₂= 100;
2x₁ + x₂= 10;

x₁	0	5
x₂	10	0

Draw the graph using the constraints

The feasible region is ABCD (since the problem is of minimization type we are moving towards the origin.

Corner Points	Z = 20x₁ + 40x₂
A(12,0)	240
B(4,2)	160
C(2,6)	280
D(0,18)	720

The minimum value of Z occurs at B (4, 2).
The optimal solution is \(x_1=4, x_2=2, Z_{\min }=160\)
Verification

6x₁ + x₂ = 18	x₁ +4x₂ = 12
2x₁+x₂=10 4x₁= 8 x₁= 2 4 + x₂= 10 x₂= 6 C(2, 6)	2x₁+x₂=10 2x₁+8x₂=24 2x₁+x₂=10 x₂= 14 x₂= 2 x₁= 4 B(4 ,2)

Since both the decision variables x₁, x₂ are non-negative, the solution lies in the first quadrant of the plane.
Consider the equations x₁ – x₂ = –1 and – x₁ + x₂ = 0
x₁ – x₂ = –1 is a line passing through the points (0,1) and (–1,0)
–x₁ + x₂ = 0 is a line passing through the point (0,0)
Now we draw the graph satisfying the conditions x₁ – x₂ ≤ –1; –x₁ + x₂ ≤ 0 and x₁, x₂ ≥ 0

There is no common region(feasible region) satisfying all the given conditions. Hence the given LPP has no solution.

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines x₁ + 4x₂= 24 ; 3x₁ + x₂ = 21; x₁ + x₂= 9
x₁+ 4x₂= 24 is a line passing through the points (0, 6) and (24, 0)
[(0, 6) is obtained by taking x₁ = 0 in x₁ + 4x₂ = 24 , (24, 0) is obtained by taking x₂ = 0 in x₁+ 4x₂ = 24].
Any point lying on or below the line x₁ + 4x₂ = 24 satisfies the constraint x₁ + 4x₂ ≤ 24 .
3 x₁ + x₂= 21 is a line passing through the points (0, 21) and (7, 0). Any point lying on or below the line 3 x₁ + x₂ = 21 satisfies the constraint 3 x₁ + x₂ ≤ 21.
x₁ + x₂ = 9 is a line passing through the points (0, 9) and (9, 0). Any point lying on or below the line x₁ + x₂ = 9 satisfies the constraint x₁ + x₂ ≤ 9.
Now we draw the graph

The feasible region satisfying all the conditions is OABCD.The co-ordinates of the points are O(0,0) A(7,0);B(6,3) [the point B is the intersection of two lines x₁ + x₂ = 9 and 3 x₁ + x₂= 21]; C(4,5) [the point C is the intersection of two lines x₁ + x₂ = 9 and x₁+ 4x₂ = 24] and D(0,6).
Corner points Z = 2x₁ + 5x₂ O(0,0) 0 A(7,0) 14 B(6,3) 27 C(4,5) 33 D(0,6) 30 Maximum value of Z occurs at C. Therefore the solution is x₁ = 4, x₂ = 5, Z_max = 33

Generic 2

8 x 2 = 16

Using the precedence relationship and following the rules of network construction, the required network is shown in the following diagram

(i) Variables:
Let x₁and x₂ represent the number of fans and sewing machines.
(ii) Objective functions:
Let Z be the profit of the dealer.
∴ Maximize Z = 22x₁ + 18x₂ is the objective function.
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
(iv) Non-negative restrictions:
Since the number of fans and sewing machine cannot be negative, we have x₁, x₂≥ 0.
Hence, mathematical formulation of the LPP is
Maximize \(Z=22{ x }_{ 1 }+18{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
and x₁, x₂≥ 0.

(i) Variables:
Let x₁, x₂represent the number of units of X and Y.
(ii) Constraints:

	Labour	Capital
X	2	3
Y	3	1

∴ 2x₁ + 3x₂ ≤ 30 and 3x₁ + x₂ ≤ 17
(iii) Non-negative restrictions:
Since the number of units of X and Y cannot be negative,x₁, x₂≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=100{ x }_{ 1 }+120{ x }_{ 2 }\)
Subject to the constraints
\( { 2x }_{ 1 }+3{ x }_{ 2 }\le 30\)
\({ 3x }_{ 1 }+{ x }_{ 2 }\le 17\)
and x₁, x₂≥ 0.

(i) Variables: Let x₁be the executive class tickets and x₂ be the economy class ticket.
(ii) Objective function: Let Z be the maximum profit
Maximize Z=1000x₁+600x₂
(iii) Constraints:\({ x }_{ 1 }\ge 20\)
\(\Rightarrow { x }_{ 2 }\ge 4{ x }_{ 1 }\ and\ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
\( \Rightarrow 4{ x }_{ 1 }\le { x }_{ 2}\ and\ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
\(\Rightarrow 4{ x }_{ 1 }-{ x }_{ 2 }\le 0 \ and \ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
(iv) Non-negative restrictions:
Since the number of tickets cannot be negative, \({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Hence, the mathematical formulation of the LPP is Maximize Z=1000x₁+600x₂
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\( { x }_{ 1 } \ge 20\)
\(4{ x }_{ 1 }-{ x }_{ 2 } \le 0 \ and \ { x }_{ 1 },{ x }_{ 2 }\ge 0\)

10.

Using the precedence relationships and following the rules of network construction, the required network is shown in the following figure.

11.

The required network for the above information.

12.

The required network for the above information.

Generic 3

4 x 3 = 12

13.

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\({ x }_{ 1 }-{ x }_{ 2 }=-1\)