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11th Business Maths - Operations Research Model Question Paper

Published on: 13/12/2019

Operations Research

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Questions:

Generic 3

5 x 3 = 15

Develop a network based on the following information.

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

Solve the following LPP graphically, Minimize \(Z=3{ x }_{ 1 }+5{ x }_{ 2 }\)
Subject to the constraints \({ x }_{ 1 }+3{ x }_{ 2 }\ge 3,\quad { x }_{ 1 }+{ x }_{ 2 }\ge 2\quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0.\)

Solve the following LPP graphically. Minimize\(Z=-3{ x }_{ 1 }+4{ x }_{ 2 }\)
Subject to the constraints \({ x }_{ 1 }+2{ x }_{ 2 }\le 8\quad ,{ 3x }_{ 1 }+{ 2x }_{ 2 }\le 12\quad and\quad \quad { x }_{ 1 }\ge 0,{ x }_{ 2 }\ge 2.\)

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

A soft drink company has two bottling plants C₁ and C₂. Each plant produces three different soft drinks S₁, S₂ and S₃. The production of the two plants in number of bottles per day are:

Product	Plant
Product	C₁	C₂
S₁	3000	1000
S₂	1000	1000
S₃	2000	6000

A market survey indicates that during the month of April there will be a demand for 24000 bottles of S₁, 16000 bottles of S₂ and 48000 bottles of S₃. The operating costs, per day, of running plants C₁ and C₂ are respectively Rs.600 and Rs.400. How many days should the firm run each plant in April so that the production cost is minimized while still meeting the market demand? Formulate the above as a linear programming model.

Generic 5

5 x 5 = 25

The following table use the activities in a building project.

Activity	1-2	1-3	2-3	2-4	3-4	4-5
Duration (days)	21	26	11	13	5	11

Draw the network for the project, calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and find the critical path. Compute the project duration.

Every gram of wheat provides 0.1 g of proteins and 0.25 g of carbohydrates. The corresponding values of rice are 0.05 g and 0.5 g respectively. Wheat cost Rs.4 per kg and rice cost Rs.6 per kg. The minimum daily requirements of proteins and carbohydrate for an average child are 50 g and 200 g respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrate at minimum cost. Frame an LPP and solve it graphically.

A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (min) required for each toy on the machine is given below:

Type	Machine I	Machine II	Machine III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of 6 hours/day. If the profit on each toy of type A is Rs.7.50 and for B is Rs.5. Show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

One kind of the cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of other ingredients used in making the cakes?

10.

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

Generic 2

4 x 2 = 8

11.

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.

12.

A toy company manufactures two types of dolls A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is atmost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of n2 and n6 per doll, how many of each should be produced weekly in order to maximize the profit. Formulate the above as mathematical LPP.

13.

Draw the event oriented network for the following data:

Events	1	2	3	4	5	6	7
Immediate Predecessors	-	1	1	2,3	3	4,5	5,6

14.

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.

Multiple Choice Question

5 x 1 = 5

15.

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

16.

In critical path analysis, the word CPM mean ______.

(a)

Critical path method

(b)

Crash project management

(c)

Critical project management

(d)

Critical path management

17.

The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤ 20, x + 2y ≤ 20, x > 0 and y > 0 is ______.

(a)

(b)

(c)

(d)

18.

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

19.

In a network while numbering the events which one of the following statement is false?

(a)

Event numbers should be unique

(b)

Event numbering should be carried out on a sequential basis from left to right

(c)

The initial event is numbered 0 or 1

(d)

The head of an arrow should always bear a number lesser than the one assigned at the tail of the arrow

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Answers Key:

Generic 3

5 x 3 = 15

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

Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane.Consider the equations
\({ x }_{ 1 }+3{ x }_{ 2 }=3\)

\({ x }_{ 1 }\)	0	3
\({ x }_{ 2 }\)	1	0

\({ x }_{ 1 }+{ x }_{ 2 }=2\)

\({ x }_{ 1 }\)	0	2
\({ x }_{ 2 }\)	2	0

The feasible region is ABC and its co-ordinates are A(3, 0) C(0,2) and B its the point of intersection of the lines
\({ x }_{ 1 }+3{ x }_{ 2 }=3\) ...(1) and \({ x }_{ 1 }+{ x }_{ 2 }=2\) ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+3{ x }_{ 2 }=3\)
\( \quad \quad (-)\quad (-)\quad \quad (-)\)
\((2)\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=2\)
\( \quad -----------\)
\(2{ x }_{ 2 }=1\Rightarrow { x }_{ 2 }=\frac { 1 }{ 2 } \)
\( From(2),\ { x }_{ 1 }+\frac { 1 }{ 2 } =2\ \Rightarrow { x }_{ 1 }=2-\frac { 1 }{ 2 } \Rightarrow \frac { 3 }{ 2 } \therefore B\quad is\quad \left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 } \right) \)

Corner Points	\(Z=3{ x }_{ 1 }+5{ x }_{ 2 }\)
A(3,0)	9
B\(\left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 } \right) \)	\(\frac { 9 }{ 2 } +\frac { 5 }{ 2 } =7\)
C (0, 2)	10

Minimum of Z occurs at B(3/2, 1/2)
Hence, the solution is x₁= 3/2, x₂ = 1/2 and Z_min = 7

Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane. Consider the equations
\({ x }_{ 1 }+2{ x }_{ 2 }= 8\)

\({ x }_{ 1 }\)	0	8
\({ x }_{ 2 }\)	4	0

\({ 3x }_{ 1 }+{ 2x }_{ 2 }=12\)

\({ x }_{ 1 }\)	0	4
\({ x }_{ 2 }\)	6	0

The feasible region is OABC and its co-ordinates are 0(0, 0) A( 4, 0) qo, 4) and B is the point of intersection of the lines
\({ x }_{ 1 }+2{ x }_{ 2 }=8\) ... (1) \(and\quad { 3x }_{ 1 }+{ 2x }_{ 2 }=12\) ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+2{ x }_{ 2 }=8\\ \quad \quad (-)\quad (-)\quad \quad (-)\\ (2)\Rightarrow 3{ x }_{ 1 }+2{ x }_{ 2 }=12\\ -----------\\ -2{ x }_{ 1 }=-4 \Rightarrow { x }_{ 1 }=2\)
\(From(1), 2+2{ x }_{ 2 }=8\Rightarrow 2{ x }_{ 2 }=6\Rightarrow { x }_{ 2 }=3\)

∴ B is (2,3)

Corner Points	\(Z=-3{ x }_{ 1 }+4{ x }_{ 2 }\)
O(0,0)	0
A(4, 0)	-12
B (2, 3)	6
C(0,4)	16

Minimum of Z occurs at A(4, 0).
Hence, the solution is x₁= 4, x₂ = 0 and Z_min = - 12.

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

(i) Variables: Let x₁ be the number of days required to run plant C₁and x₂ be the number of days required to run plant C₂
Objective function: Minimize Z = 600 x₁ + 400 x₂
(ii) Constraints: 3000 x₁ + 1000 x₂ ≥ 24000 (since there is a demand of 24000 bottles of drink A, production should not be less than 24000)
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁+ 6000 x₂ ≥ 48000
(iii) Non-negative restrictions: Since be the number of days required of a firm are non-negative, we have x₁, x₂ ≥ 0
Thus we have the following LP model.
Minimize Z = 600 x₁ + 400 x₂
subject to 3000 x₁ + 1000 x₂ ≥ 24000
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000 and x₁, x₂ ≥ 0

Generic 5

5 x 5 = 25

E₁= 0	L₅= 48
E₂= 0+21=21	L₄= 48 -11 = 37
E₃ =(21 + 11) or (0 + 26) Whichever is maximum =32	L₃= 37 - 5 = 32
E₄= (32 + 5) or (21 + 13) =Whichever is maximum = 37	L₂= (37 - 13) or (32 - 11) Whichever is minimum =21
E₅=37 + 11 = 48	L₁=(21 - 21) or (32 - 26) Whichever is minimum = 0

Activity	Duration	EST	EFT = EST + t_ij	EFT = EST - t_ij	LFT
1-2	21	0	21	21-21=0	21
1-3	26	0	26	32-26=6	32
2-3	11	21	32	32-11=21	32
2-4	13	21	34	37-13=24	37
3-4	5	32	37	37-5=32	37
4-5	11	37	48	48-11=37	48

EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .

Let x₁g of wheat and x₂ g of rice be mixed in the daily diet. Let Z be the minimum cost of diet.

	Proteins	Carbohydrates	Cost
1 g of Wheat	0.1 g	0.25 g	Rs.4/kg
1 g of Rice	0.05 g	0.5 g	Rs.6/kg
Minimum Requirement	50 g	200 g

Thus, the mathematical formulation of the LPP
Minimize \(Z=\frac { 4{ x }_{ 1 } }{ 1000 } +\frac { 6{ x }_{ 2 } }{ 1000 } \quad \Rightarrow \quad Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
Subject to the constraints
\(0.1{ x }_{ 1 }+0.05{ x }_{ 2}\ge 50 \Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }\ge 1000\)
\( 0.25{ x }_{ 1 }+0.5{ x }_{ 2\quad }\ge 200 \Rightarrow { x }_{ 1 }+2{ x }_{ 2 }\ge 800\)
\(and\ { x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equation
\(2{ x }_{ 1 }+{ x }_{ 2 }= 1000\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

\({ x }_{ 1 }+2{ x }_{ 2 }= 800\)

\({ x }_{ 1 }\)	0	500
\({ x }_{ 2 }\)	1000	0

The feasible region is ABC an its co-ordinates are A(800, 0), C(0, 1000) and B is the point of intersection of the lines 2x₁+ x₂ = 1000 ... (1) x₁+ 2x₂ = 800 ... (2)
Verification of B:
\(\Rightarrow (1)\times 2 4{ x }_{ 1 }+2{ x }_{ 2 }=2000\)
\(\quad (-)\quad (-)\quad \quad (-)\)
\(\Rightarrow (2) 15{ x }_{ 1 }+6{ x }_{ 2 }=30 \Rightarrow { x }_{ 1 }=400\)
\(--------------\)
Substracting, \(3{ x }_{ 1 }=2000\)
From (2), \(\quad 400+2{ x }_{ 2 }=800\)
\(2{ x }_{ 2 }=400\quad \Rightarrow \quad { x }_{ 2 }=200\)

Corner Points	\( Z=\frac { { x }_{ 1 } }{ 250 } +\frac { 3{ x }_{ 2 } }{ 500 } \)
A(800,0)	\(\frac { 800 }{ 250 } =3.2\)
B(400, 200)	\(\frac { 400 }{ 250 } +\frac { 600 }{ 500 } =2.8\)
C(0,1000)	\(\frac { 3000 }{ 500 } =6\)

Minimum of Z occurs at B(400, 200). Hence, the solution is x₁= 400, x₂ = 200 and Z_rnin = 2.8

Let x₁ toys of type A and x₂ toys of type B are produced. Let Z be the maximum profit on two types of toys A and B.

Type	Machine I	Machine II	Machine III	Profit
A	12	18	6	Rs. 7.50
B	6	0	9	Rs. 5
Time available	6h = 360 min	6h = 360 min	6h = 360 min

Thus, the mathematical formulation of the LPP is maximize Z = 7.50x₁ + 5x₂
Subject to the constraints
12x₁ + 6x₂ ≤ 360,18x₁ ≤ 360,6x₁ + 9x₂ ≤ 360,x₁,x₂ ≥ 0
Consider the equations
12x₁ + 6x₂ ≤ 360

\({ x }_{ 1 }\)	0	30
\({ x }_{ 2 }\)	60	0

\(18{ x }_{ 1 }=360\)

x₁

\({ 6x }_{ 1 }+9{ x }_{ 2 }=360\)

\({ x }_{ 1 }\)	0	60
\({ x }_{ 2 }\)	40	0

X₁ = 20 is a line parallel to x₂-axis at a distance of 20 units from it

The feasible region is OABCD and its co-ordinates are O(0, 0) A(20, 0) D(0, 40), B is the point of intersection ofthe lines x₁ = 20 and 12x₁+ 6x₂ = 360
\(\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=60\)
\(\Rightarrow 40+{ x }_{ 2 }=60\)
\( \Rightarrow { x }_{ 2 }=20\)
Verification of Band C:
\(\therefore \quad B\quad (20,20)\)
And C is the point of intersection of the lines
\(2{ x }_{ 1 }+{ x }_{ 2 }=60...(1)\)
\( (-)\quad \quad (-)\quad \quad (-)\)
\(2{ x }_{ 1 }+3{ x }_{ 2 }=120 \left[ \because \quad 6{ x }_{ 1 }+{ 9x }_{ 2 }=360 \right] ...(2)\)
\(-------------\)
\( -2x_{ 2 }=-60 \Rightarrow { x }_{ 2 }=30\)
\(2{ x }_{ 1 }+30=60 [\because \quad From\quad (1)] \Rightarrow { x }_{ 1 }=\frac { 30 }{ 2 } =15\)
\( \therefore \quad C\quad is\quad (15,30)\)

Corner Points	Z = 7.5x₁ + 5x₂
O(0,0)	0
A(20,0)	150
B(20, 20)	250
C(15,30)	262.5
D(0, 40)	200

Maximum of Z occurs at (15,30)
Hence, the solution is x₁ = 15, x₂ = 30 and Z_max= 262.5

Let x₁cakes of the one kind and x₂ cakes of another kind are made. Let Z be the maximum number of cakes

Ingredients	x₁(g)	x₂(g)	Total (kg)
Flour	200	100	5
Fat	25	50	1

Thus, the mathematical formulation of the LPP is Maximize Z = x₁+ x₂
Subject to the constraints
\(200{ x }_{ 1 }+100{ x }_{ 2 }\le 5000\)
\(25{ x }_{ 1 }+50{ x }_{ 2 } \le 1000\)
\({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equations
\(200{ x }_{ 1 }+100{ x }_{ 2 }= 5000\)

\({ x }_{ 1 }\)	0	25
\({ x }_{ 2 }\)	50	0

\(25{ x }_{ 1 }+50{ x }_{ 2 }=1000\)

\({ x }_{ 1 }\)	0	25
\({ x }_{ 2 }\)	50	0

The feasible region is OABC and its co-ordinates are O(0, 0) A(25, 0) C(O, 20) and B is the point of intersection of the lines
200x₁ + 100x₂ = 1000 .... (1)
and 25x₁ + 50x₂ = 1000 ... (2)
Verification of B:
\((1) \Rightarrow 200{ x }_{ 1 }+100{ x }_{ 2 }=5000\)
\( (-)\quad \quad \quad (-)\quad \quad (-)\)
\( (2)\times 5\Rightarrow 50{ x }_{ 1 }+100{ x }_{ 2 }=2000\)
\(------------------\)
\(150x_{ 1 }=3000 \Rightarrow { x }_{ 1 }=20\)
\(From(2), 25(20)+50{ x }_{ 2 }=1000\Rightarrow 500+50{ x }_{ 2 }=1000 \Rightarrow 50{ x }_{ 2 }=500\)
\(\Rightarrow { x }_{ 2 }=10\)
\(\therefore B\ is\ (20,10)\)

Corner Points	Z=x₁ +x₂
O(0,0)	0
A(25, 0)	25
B(20, 10)	30
C(0,20)	20

Maximum of Z occurs at B(20, 10)
Hence, the solution is x₁ = 20, x₂ = 10 and Z_max = 30.

10.

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.

Generic 2

4 x 2 = 8

11.

(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.

12.

(i) Variables:
Let x₁, x₂represent the dolls of A and B produced in a week.
(ii) Objective function:
Let Z be the total profit in a week.
\(\therefore Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Since we have to maximize the profit, we have maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
(iv) Non-negative restictions:
Since the number of dolls on type A and B cannot be negative, we have \({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Hence, the mathematical formation of LPP is
Maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
and x₁, x₂≥ 0.