Q1) Ten students got the following percentage of marks in Economics and

StatisticsCalculate the of correlation.

Roll No.
Marks in Economics
Marks in

Let the marks of two subjects be denoted by and respectively.Then the mean
for marks and the mean ofy marks

and are deviations ofx’s and ’s from their respective means, then the data may be
arranged in the following form:

x y X=x-65 Y=y-66 X2 Y2 XY

78 84 13 18 169 324 234

36 51 -29 -15 841 225 435

98 91 33 25 1089 625 825

25 60 -40 -6 1600 36 240

75 68 10 2 100 4 20

82 62 17 -4 289 16 -68

90 86 25 20 625 400 500

62 58 -3 -8 9 64 24

65 53 0 -13 0 169 0

39 47 -26 -19 676 361 494

Q 2) Compute coefficient of correlation by Karl Pearson Method for the following
data.

X: 1800 1900 2000 2100 2200 2300 2400 2500 2600

F: 5 5 6 9 7 8 6 8 9

A2)

Let the A.M.s and be 2200 and 6 for X and Y series respectively.

X Y dx (i=100) dy d d dxdy
dx
1800 5 -400 -4 -1 16 1 4

1900 5 -300 -3 -1 9 1 3

2000 6 -200 -2 0 4 0 0

2100 9 -100 -1 3 1 9 -3

2200 7 0 0 1 0 1 0

2300 8 100 1 2 1 4 2

2400 6 200 2 0 4 0 0

2500 8 300 3 2 9 4 6

2600 9 400 4 3 16 9 12
N=9

(Note: We can also proceed dividing X by 100)

Q 3) Let random variables. Define

X= ,,Y = where is real number in (-1, 1).

• Show that X and Y are bivariate normal.

• Find the joint PDF of X and Y.
• Find

(X, Y).
first, note that since are normal and independent, they are jointly normal, with
the joint PDF.

(a). we need to show that aX + bY is normal for all a,b R, we have

aX + bY =

Which is linear combination of and thus it is normal.

(b). we can use the method of transformations to find the joint PDF of X and Y.

The inverse transformation is given by

Where J = det = det

Thus, we conclude that

=
=

(c). To find (X, Y), first note that

Var(X) = Var(Z1) = 1,

Var(Y) =

Therefore, (X, Y) = Cov (X, Y)

= Cov(

= Cov( .

= .1 + .0

= .

Q 4) Let X and Y be jointly normal random variables with

parameters Find the conditional distribution of Y given X =x.

A4) one way to solve this problem is by using PDF formula. In particular, since X N ( x,
), we can use

Thus, given X = x we have,

and . .+ .

Since are independent, knowing does not provide any information on . We

have shown that given X =x, Y is a linear function of , thus it is normal. In particular

E[Y|X =x ]= + E

= ,
Var (Y|X = x) =

= (1 -

We conclude that given X = x, Y is normally distributed with mean and

variance (1 - .

Q 5: Let X and Y be jointly normal random variables with

parameters
• Find P(2X+Y

)
• Find Cov(X+Y,2X-Y)
• Find P(Y>1

A5) a. Since X and Y are jointly normal, the random variable V=2X+Y is normal. We have

Thus, V Therefore,

b. Note that Cov (X, Y) =

Cov (X+Y, 2X-Y) = 2Cov (X, X)-Cov (X, Y) +2Cov (Y, X)-Cov (Y, Y)

= 2-1+2-4 = -1.

d. Using Properties, we conclude that given X = 2, Y is normally distributed with

 Thus,

Q 6:Two Way Frequency Tables:

Student Grades in Science Projects.

Male Female
A 9 12
B 18 14
C 8 11
D 2 3
F 1 2

A6)

Male Female Total

A 9 12 21
B 18 14 32
C 8 11 19
D 2 3 5
F 1 2 3
Total 38 42 80

Q: How Many students earned a grade of A?

Ans: 21 Students

Q: How many males were surveyed?

Ans: 38 Male Students

Q: How many males earned a grade of A?

 Ans: 9 Male Students

Q: How many students earned a grade of B or C:

Ans: 51 Students

Q 7:

x 1 2 3 4 5 6 7
y 0.5 2.5 2.0 4.0 3.5 6.0 5.5

A7)

: y = 0.07143+0.8393x.
Q 8: Fit a least square line for the following data. Also find the trend values and show
that ∑(Y– )=0 ∑(Y– )=0.

X 1 2 3 4 5

Y 2 5 3 8 7

A8)

X Y XY X2 Y–
=1.1+1.3X
1 2 2 1 2.4 -0.4
2 5 10 4 3.7 +1.3
3 3 9 9 5.0 -2
4 8 32 16 6.3 1.7
5 7 35 25 7.6 -0.6
∑(Y- )=0
Trend
∑X=15 ∑Y=25 ∑XY=88 ∑X 2=55
Values

The equation of least square line Y=a +bX

Normal equation for ‘a’ ∑Y=na +b 25=5a+15b —- (1)

Normal equation for ‘b’ ∑XY = a∑X+b∑X2 88=15a+55b —-(2)

Eliminate a a from equation (1) and (2), multiply equation (2) by 3 and subtract from
equation (2).

Eliminate a from equation (1) and (2), multiply equation (2) by (3) and subtract from
equation (2). Thus, we get the values of a and b

Here a=1.1 and b=1.3, the equation of least square line becomes

Y=1.1+1.3X
Q 9: Using least square method to fit a straight line of the following data

X 8 2 11 6 5 4 12 9 6 1
Y 3 10 3 6 8 12 1 4 9 14

A9)

First, we calculate for the given data

Now we calculate

I
1 8 3 1.6 -4 -6.4 2.56
2 2 10 -4.4 3 -13.2 19.36
3 11 3 4.6 -4 -18.4 21.16
4 6 6 -0.4 -1 0.4 0.16
5 5 8 -1.4 1 -1.4 1.96
6 4 12 -2.4 5 -12 5.76
7 12 1 5.6 -6 -33.6 31.36
8 9 4 2.6 -3 -7.8 6.76
9 6 9 -0.4 2 -0.8 0.16
10 1 14 -5.4 7 -37.8 29.16

Calculate the slope

m= = -131/118.4

calculate the y-intercept

use the formula to calculate the y-intercept

b=
 = 7-(-1.1*6.4)

The required line equation is

Y= -1.1x+14.0

Q 10: Determine the constants a and b by the method of least square such that

X 2 4 6 8 10
Y 4.077 11.084 30.128 81.897 222.62

A10)

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx…. (1)

let,

log y = Y

x=X

log a = A

b=B

now we have,

…. (2)

…. (3)

Now we need to find

X=x Y =ln(y) xy
2 1.405 4 2.810
4 2.405 16 9.620
6 3.405 36 20.430
8 4.405 44 35.240
10 5.405 100 54.050

The normal equations to fit the straight line is

Y = logey

Y= ln(y)

17.025 = 5A +30B…. (4)

122.150 = 30A+220B…. (5)

By solving 4 and 5 we get

30A +180B = 102.15… (4)

30A+220B = 122.150… (5)

We get a = 0.405, b = 0.5

A =log a

a = 1.499

since we have X=x and Y=y

log y=Y,

And we know y= aebx

Y = (1.499) e0.5x is the required exponential curve.

Q 11) Fit the curve of the form y= aebx for the following data

X 0 2 4

Y 8.12 10 31.82

A11)

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx logee…. (1)

the required normal equations are,

…. (2)

…. (3), We have n=3

X y Y= logey xy X2
0 8.12 2.0943 0 0
2 10 2.3026 4.6052 4
4 31.82 3.4601 13.8404 16

The normal equations become

3A +6b = 7.8750

6A + 20 b = 18.4456

By solving the above two equations we get

A = 1.361 and b = 0.3415

Since A =logea a = e1.361 = 6.9317

The curve of the fit is

Thus, the required equation is,

Q 12:
X 1 2 3 4 5 6 7
Y 2 6 7 8 10 11 11
A12)
X
1 -4 2 16 -64 256 -8 32
2 -3 6 9 -27 81 -8 54
3 -2 7 4 -8 16 -14 28
4 -1 8 1 -1 1 -8 8
5 0 10 0 0 0 0 0
6 1 11 1 1 1 11 11
7 2 11 4 8 16 22 44
8 3 10 9 27 81 30 90
9 4 9 16 64 256 36 144
N=0

∑Y i =Na + b ∑X i +c∑

∑X i Y i =a ∑X i +b∑ +c∑

∑ Y i =a∑ +b∑ +c∑

The required parabola is of the form y= ax2+bx+c

∴74=9a+b (0) +60c∴9a+60c=74…(i)

51=a (0) +60b+0c ∴60b=51 ∴b=5160 =0.85411=60a+0b+708 c∴60a+708c=411…(ii)

Solving (i) and (ii) simultaneously, we get

a=10.004, c=-0.267

The Equation of parabola is therefore,

y=10.004+0.85X−0.267X 2

=10.004+0.85(x−5) −0.267(x−5) 2

=10.004+0.85x−4.25−0.267(x 2 −10x+25)

=10.004+0.85x−4.25−0.267x 2 +2.67x−6.675∴ y = −0.921+3.52x−0.267x 2

Q13) Find the least square approximation of degree two to the data
X 0 1 2 3 4
Y -4 -1 4 11 20

A13)

X y xy

0 -4 0 0 0 0 0
1 -1 -1 1 -1 1 1

2 4 8 4 16 8 16

3 11 33 9 99 27 81
4 20 80 16 320 64 256

the normal equations are:

Here,

n = 5,

by substituting all the above values in normal equations, we get,

30 = 5a+10b+30c

120=10a+30b+100c

434=30a+100b+354c
By solving the above equations, we get

a = -4, b=2, c=1.

Therefore, the required polynomial is

Y= -4x+2x+x2 and errors =0.

UNIT 2

Q1) Let X be a random variable with PDF given by

a. Find the constant c

b. Find EX and Var(X)

c. Find P(X )

A1)

a. To find c, we can use

Thus, we must have c=

b. To find EX, we can write

In fact, we could have guessed Ex=0 because the PDF is symmetric around x=0. To find
Var(X), We have
c. To find P(X ), we can write

Q2) Let X be a continuous random variable with PDF given by

If Y =

A2)

First, we note that

Thus,
Q3) Let X be a continuous random variable with PDF

Find P(X ).

A3)

We have

P(X )=

Q4) Consider tossing a fair coin 3 times.

Define X= the number of heads obtained.

Value 0 1 2 3
Probability 1/8 3/8 3/8 1/8

A: what is the probability of tossing at least two heads?

B: what is the probability of tossing fewer than three heads?

A4)

X = 0, TTT

X = 1, HTT THT TTH

X = 2, HHT HTH THH

X = 3, HHH

A: Ans:

B: Ans:

(Or)

Q 5) Let the random variable X have probability density function

Find the cumulative distribution function of Y = g(x) =

A5)

Q 6) There are very few continuous distributions that we can do much with without a
lot of math, but the simple uniform distribution is one of them. A uniform distribution
is one that is just that-uniform. If we image any possible number between o and 2 as
equally likely, we have an infinite possible number of numbers, so the probability of
any particular number is close to zero. Instead of focusing on particular numbers,
we have to focus on intervals.
The probability that some number between zero and two will come up will be one. To
find the probability of any interval, we find the length of the interval and multiply by
5. Hence, the probability of getting a number between zero and one is 5.

For the following, draw the picture and compute the probabilities. What is:

a): P(0<x<.8)

b): P(x>.8)

c): P(.5<x<1.5)

d): P(.5<x<1.2) or P(.8<x<1.3)

e): compute the probability that we will not be in the interval of .4 to 1.5

f): compute the middle interval in which we will be 90% of the time.

g): compute the tails (outside intervals) that we will be in 5% of the time

A6)

a): .8*.5 = .40

b): 1-.4 = .6

c): 1*.5 = .5

d): P(.5<x<1.5) = .8*.5 = .4

e): 1-P(.4<x<1.5) = 1-1.1*5 = 1-.55 = .45

f): 90% of 2 is 1.8, which centered will be from .1 to 1.9

g): 95% of the time we will be in the interval .05 to 1.95. 5% of the time we will

between 0 and .05 or 1.95 and 2.

Q7) Following probability distribution

X 0 1 2 3 4 5 6 7
P(x) 0

Find:(i) k (ii)

(iii) Distribution function (iv) If find minimum value of C

(v) Find

If P(x) is p.m.f –

X 0 1 2 3 4 5 6 7
P(x) 0
(v)

Q 8: The picture below shows a continuous probability distribution with the formula,
y = .5x for the range, x =0 to x = 2. In this case, area is probability.

To compute the probability of being between 0 and 1, for example, you need to know
y when x equals 1. The formula says it is .5. Hence, to find the probability that x is
between 0 and 1 Prob(0<x<1), we compute the area of the triangle with a length of 1
and a height of .5. The answer is 2.5 because .5HW = .5*.5*1.

a). Show that the area under the line is equal to 1.

b). Find the following probabilities:

Prob(0<x<.5)

Prob(0<x<1.5)

Prob(1.5<x<2)

Prob(.5<x<1.5)

Prob (x<.5 or x>1.5)

c). Suppose that we draw twice from this distribution and that probability of the
second draw is independent of the first. What is the probability of getting greater than
1.5 on both draws? What is the probability of getting less than 1.5 on both draws?
What is the probability of getting a result greater than 1.5 on one draw and less than
1.5 on the other draw?

A8)

a): Length *height*.5 is the area of a triangle. This triangle has length of 2 and height of 1,
so 2*1*.5= 1. The area is one, and for a continuous distribution, area is probability.

b): Prob(0<x<.5)

Prob(0<x<.5) = .5*.5*.25 = .0625

Prob(0<x<1.5) = .5*1.5*.75 = .5625

Prob(1.5<x<2) = 1-Prob(0<x<1.5) = 1-.5625 = .4375.

Prob(.5<x<1.5) = .5625 - .0625 = .5

Prob (x<.5 or x>1.5) = .0625+.4375 = .5

c): Both less than 1.5: .5625*.5625 = .3164: both greater than 1.5: .4375*.4375 = .1914; one
greater and one less: 1 - .3164 - .1914 = .4922.

Q9) The probability distribution of a discrete random variable X is shown below.

X 1 2 3 4 5 6
Pr(X=x) 0.02 0.13 0.38 0.17 0.05 0.25

Find: a) Pr(X ) b) Pr(X<5) c) Pr(X>0) d) Pr (2 )

e) Pr(X )

A9)

a): Pr(X ) = Pr (X=0)+Pr(X=1)+Pr(X=2)+Pr(X=3)

= 0.02+0.13+0.38+0.17

= 0.7. Working with Discrete Probability Distribution:

Q10)

X 0 1 2 3 4 or
more
P(X) 0.1 0.4 0.35 0.1 0.05

a): Verify that this is a legitimate probability distribution.

b): What is the probability that a customer does not rent any videos?

A10)

a): Confirm that the sum of the probability is 1.

0.1+0.4+0.35+0.1+0.05 = 1

b): This is P(X=0)=0.1 or 10% of customer do not rent any videos.

Please notice the notation.

Q11) Jean usually makes half of her free throws in basketball practice. Today, she
tries 3 free throws. What is the probability that Jean will make exactly 1 of her free
throws?

A11)

The Probability that Jean will make each free throw is or 0.5.

P(r) = Substitute 3 for n, 1 for r, 0.5 for p, and 0.5 for q

P (1) =

= 3(0.5)(0.25) = 0.375

The probability that Jean will make exactly one free throw is 37.5%.
 Q12) Suppose the probability of purchasing a defective computer is 0.02. What is the
probability of purchasing 2 defective computers in a group of 10?

A12)

X = 2, n = 10, and

P(X=2) =

= (45)(0.0004)(0.8508)

= 0.01531

Q13) A dozen eggs contain 3 defectives. If a sample of 5 is taken with replacement,

find the probability that.

a) Exactly 2 of the eggs sampled are defective.

b) 2 or fewer of the eggs sampled are defective

A13) If n = 5, P = ¼

“Success = Defective”

x = 0,1,2,3,4,5

f(x) =

P(X=2) = f (2) =

P (X
Q14) Suppose that a rare disease has an incidence of 1 in 1000 person-years.
Assuming that members of the population are affected independently, find the
probability of k cases in a population of 10,000(followed over 1 year) for k=0,1,2.

A14) The expected value (mean)

10 new cases expected in this population per year

Q 15: Suppose that the probability of a defect in a foot of magnetic tape is 0.002. Use
the Poisson distribution to compute the probability that 1500 feet roll will have no
defects.

A15)

Q16) The weekly incomes of shit foremen in the glass industry follow the normal
probability distribution with a mean of $1,000 and a standard deviation of $ 100. What
is the z value for the income, let’s call it X, of a foreman who earns $1,100 per week?
For a foreman who earns $900 per week?

A16)

For X = $ 1,100:
Q17) From the Below Diagram find the area of the both sides of the distribution.

A17)

Area to the left of the positive z: 0.05000+0.3000=0.8000.

Using Table E, z 0.84. X = 120+0.84(8) =126.72

Area to the left of the negative z= 0.5000-0.3000=0.2000

Using Table E, z 0.84. X = 120-0.84(8) =113.28

The middle 60% of readings are between 113 and 127.

 Unit 3

Q1) Use Simpson’s Rule to approximate the integral of

f(x) = on the interval [1,2].

The actual value of the integral is 3.75. This is exact because the error for Simson’s rule
is O( ) and we are integrating a cubic function.

Q2) Use Simpson’s Rule to approximate the integral of

on the interval [2,5].

The actual value of the integral is 2.122000934.

Q3) Suppose we can only evaluate a function at the integers (for example, when we
are periodically sampling a signal). Use three applications of Simpson’s rule and two
applications of Simpson’s 3/8 Rule to approximate the integral of response with a
decaying transient f(x) = cos(x)+x on the interval [0,6].

A3)

Thus, we calculate:

and,

The actual integral is -7exp (-6) + sin (6) +1 = 0.7032332366, and therefore, three
applications of Simpson’s rule with intervals of width 2 appears to be more accurate than
two applications of Simpson’s 3/8th rule with intervals of width 3.
Q 4: Estimation error of Newton Cotes formula of degree 4.

A4)

From Note (1) we get,

Again, that doesn’t mean that the error is zero. The error term can be obtained from the
next term in the Newton Polynomial. i.e.

Q 5: State the trapezoidal rule for finding an approximate area under the given curve.
A curve is given by the points (x, y) given below:

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

A5)

We construct the data table:

X 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Y 23 19 14 11 12.5 16 19 20 20

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method
Area of curve bounded on x axis =

Q6) Compute the value of ?

A6)

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

Here

For h=0.5, we construct the data table:

X 0 0.5 1
Y 1 0.8 0.5

By Trapezoidal rule

For h=0.25, we construct the data table:

X 0 0.25 0.5 0.75 1

Y 1 0.94117 0.8 0.64 0.5

By Trapezoidal rule
For h = 0.125, we construct the data table:

X 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1

Y 1 0.98461 0.94117 0.87671 0.8 0.71910 0.64 0.56637 0.5

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Q7) Evaluate, using trapezoidal rule with five ordinates

A7)

Here

We construct the data table:

X 0

Y 0 0.3693161 1.195328 1.7926992 1.477265 0

Q8) Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively.

For n=4, we have

Construct the data table:

X 1.0 1.5 2.0 2.5 3.0

Y=1/x 1 0.66666 0.5 0.4 0.33333

By Simpson’s Rule

For n = 8, we have

X 1 1.25 1.50 1.75 2.0 2.25 2.50 2.75 3.0

Y=1/x 1 0.8 0.66666 0.571428 0.5 0.444444 0.4 0.3636363 0.333333

By Simpson’s Rule

Q 9: Evaluate

Using Simpson’s 1/3 rule with .

For , we construct the data table:

X 0

0 0.50874 0.707106 0.840896 0.930604 0.98281 1

By Simpson’s Rule

Q10) Using Simpson’s 1/3 rule with h = 1, evaluate

A10)

For h = 1, we construct the data table:

X 3 4 5 6 7
9.88751 22.108709 40.23594 64.503340 95.34959

By Simpson’s Rule

= 177.3853
Q11) Evaluate

By Simpson’s 3/8 rule.

A11)

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

X 4.0 4.2 4.4 4.6 4.8 5.0 5.2

1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6487

By Simpson’s 3/8 rule

= 1.8278475

Q 12: Evaluate

A12)

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X 0 1 2 3 4 5 6
1 0.5 0.2 0.1 0.0588 0.0385 0.027

By Simpson’s 3/8 rule

+3(0.0385)+0.027]

=1.3571

Q13) Evaluate:

A13)

Calculate the values of f(x) =

x 0.2 0.25 0.3 0.35 0.4 0.45 0.5

f(x) 0.475 0.165 0.716 0.914 1.070 1.228 1.387
Q14)

x 0.25 0.5 0.75 1.0 1.25 1.5 1.75

Y= 0.8 0.6666 0.5714 0.5 0.4444 0.4 0.3636

A14)

Weddle’s Rule: It’s a 7-point quadrature formula, i.e. n=6

Unit – 4

Q1) Crisp logic (crisp) is the same as Boolean logic (either 0 or 1). Either a
statement is true (1) or it is not (0), meanwhile fuzzy logic captures the degree to
which something is true.

A1)

Consider the statement: “The agreed to met at 12’o clock but ben was not punctual.”

• Crisp logic: If Ben showed up precisely at 12, he is punctual, otherwise he is too early
or too late.
• Fuzzy logic: The degree, to which Ben was punctual, can be identified by on how much
earlier or later he showed up (e.g. 0, if he showed up 11:45 or 12:15, 1 at 12:00 and a
linear increase / decrease in between).

Q2) Crisp sets versus Fuzzy sets

• Heights of three adults: A is 178cm, B is 166cm, and C is 181 cm

Crisp set

Short

average under
170cm

tall 170 to
180cm

over
180cm

Characteristic Function

Short ave tall

B A C

short ave tall

A 0 1 0

B 1 0 0
C 0 0 1

Q3)
A3)

Q4) A fuzzy set A on R is convex iff A( )

For all and all where min denotes the minimum operator.

A4)

(i) Assume that A is convex and let Then and,

moreover, for any by the convexity o A. Consequently,

A( )

(ii) Assume that a satisfies. We need to prove that for any is convex. Now for
any (i.e., A ( ) and for any

A( )

i.e. .Therefore, is convex for any Hence, A is convex.

Q5) Consider three fuzzy sets that represent the concepts of a young middle-aged,
and old person. The membership functions are defined on the interval [0,80] as
follows:

A5)

Q6)

t 20 30 40
0.1 0.5 0.9

h 20 50 70 90
0.2 0.6 0.7 1

20 50 70 90
20 0.1 0.1 0.1 0.1
30 0.2 0.5 0.5 0.5
40 0.2 0.6 0.7 0.9

Calculated R (t,h) through min operator

A6)

Now suppose, we want to get information about the humidity when there is the following
premise about the temperature:
Temperature is fairly high.

This fact is rewritten as R(t): t is where

Where the fuzzy term is defined as below membership function of in T

(temperature)

t 20 30 40
0.01 0.25 0.81

R(h)= R(t) o R(t,h)

20 50 70 90
20 0.1 0.1 0.1 0.1
30 0.2 0.5 0.5 0.5
40 0.2 0.6 0.7 0.9
h 20 50 70 90
0.2 0.6 0.7 0.81

t 20 30 40
0.01 0.25 0.81

Q 7: Let P = {paris, berlin, Amsterdam} and Q = {Rome, Madrid, Lisbon} be two set
of cities, and E = {far, very far, near, very near}. Let R be the fuzzy soft relation over
the sets P and Q given by (R,E) = {R(for) = {(paris, Rome)/0.60, (paris, madrid/ 0.45,
(Paris, Lisbon)/0.40, (Berlin, Rome)/ 0.50, (Berlin, Madrid)/0.65), (Berlin,
Lisbon)/0.70,(Amsterdam, Rome)/0.75, (Amsterdam, Madrid)/0.50,(Amsterdam,
Lisbon)/0.80}}.This information can be represented in the form of two-dimensional
array(matrix) as shown in the Table 4. It is obvious from the matrix given in table 4
that a fuzzy soft relation may be considered as a parameterized fuzzy relation.

A7)

(1) Union:
(2) Intersection:

(3) Containment:
R(far) Rome Madrid Lisbon
Paris 0.60 0.45 0.40
Berlin 0.55 0.65 0.70
Amsterdam 0.75 0.50 0.80

Q8) Let U = { } be the set of watches and A = {cheap, costly}, B =

{Beautiful, in a golden locket}. Let (F, A) and (G,B) be two soft sets given by F (cheap)

= { .The
relation is given by the following membership matrices in table 5 and
6. The membership function of R can also be defined using other appropriate
techniques.

A8)

R(Costly
beautiful)
0.65 1 0.80 0.70 0.80 0.75
0.49 0.75 0.60 0.53 0.60 0.56
0.52 0.80 0.64 0.56 0.64 0.60
0.35 0.55 0.44 0.39 0.55 0.42
0.59 0.90 0.72 0.63 0.72 0.68
0.53 0.85 0.68 0.60 0.68 0.64

Unit – 5

Q1) It is necessary to compare 2 sensors based upon their detection levels & gain
settings the tails of gain settings & sensor detection level with a obtained item
being monitored providing typical membership values to reprocess the detection
levels for each sensor is given in tales.

Find following membership function:

a) b) c)
A1)

Gain setting Detection level of Detection level

sensor 1(D1) sensor 2(D2)
0 0 0
10 0.2 0.35
20 0.35 0.25
30 0.65 0.8
40 0.85 0.95
50 1 1

= max { }

b) = min { }

c) = 1-

Q2) Find the fuzzy cardinality of A defined as follows:

A2)

Here X= {0,1,2,3,4,5} s and its scalar cardinality are

 {0,2,4,5}

{0,1,2,3,4,5}

{2,5}

{0,2,3,4,5}

{2,4,5}

{5}

Thus,

Note: Fuzzy cardinality of Fuzzy set is a fuzzy s

Q3) Find fuzzy cardinality of fuzzy set A and B whose membership function are

given as follows , B(x)= where x

A3)

X 0 1 2 3 4

0 0.5 0.67 0.75 0.8

,

B(x)= 1 0.9 0.8 0.7 0.6

 {0,1,2,3,4}

{1,2,3,4}

{2,3,4}

{3,4}

{4}

{0}

{0,1}

{0,1,2}

{0,1,2,3}

{0,1,2,3,4}

Thus,

Thus,
Q4)

Their

Step 1: Find and

Step 2: Use decomposition theorem which is given as and arithmetic

operation on closed interval

Step 3: Find membership function.

Q5) The membership function of two fuzzy numbers A and B are given as follows

-1 < x
= =

= =
A(x) B(x)
=0 =0 otherwise

Find A+B and A-B

A5)

Step 1: Find and

For -1 < x
 By def. of ,

A(x)

For

= [2 -1, 3-2 for all

For

By def. of cut,

B(x)

For 3 < x

B(x)

= [2 +1, 5-2 for all

Step 2: By decomposition theorem & arithmetic operation on closed interval we
find as follows.

Step 3: Now find membership function

Let 4 Let 8-4

Thus, (A+B) (x) = ; 0<x

= 4

= 0; Otherwise

To find A-B
Step 4: By decomposition theorem and arithmetic operation on closed interval we
find follows

=[

= [4 for all

Step 5: Now find membership function to define fuzzy numbers A-B

Let 4 Let 2-4

Thus,

Q6) The membership function of two fuzzy numbers A and B are

 = x-1 1<x
=

=
=
A(x) B(x)
=0 =0 otherwise

Solve A+X =B

A6)

Let

A+X = B

Taking alpha cut on both side

Check

0.1 12.8 56.9

0.6 26.8 46.4

0.8 32.4 42.2

Thus, hence solution exists.

Let Let

Q7) The membership function of two fuzzy numbers A and B are

= x-3 3<x
=

=
=
A(x) B(x)
=0 =0 otherwise

Solve A.X =B

A7)
Let

A.X = B

Taking alpha cut on both side

Check

0.3 4.3636 6.0426

0.7 4.7568 5.4884

0.9 4.9231 5.1707

Thus, hence solution exists.

Let Let
Unit – 5

Q1) It is necessary to compare 2 sensors based upon their detection levels & gain
settings the tails of gain settings & sensor detection level with a obtained item
being monitored providing typical membership values to reprocess the detection
levels for each sensor is given in tales.

Find following membership function:

a) b) c)

A1)

Gain setting Detection level of Detection level

sensor 1(D1) sensor 2(D2)
0 0 0
10 0.2 0.35
20 0.35 0.25
30 0.65 0.8
40 0.85 0.95
50 1 1

= max { }

b) = min { }

c) = 1-
Q2) Find the fuzzy cardinality of A defined as follows:

A2)

Here X= {0,1,2,3,4,5} s and its scalar cardinality are

{0,2,4,5}

{0,1,2,3,4,5}

{2,5}

{0,2,3,4,5}

{2,4,5}

{5}

Thus,

Note: Fuzzy cardinality of Fuzzy set is a fuzzy s

Q3) Find fuzzy cardinality of fuzzy set A and B whose membership function are

given as follows , B(x)= where x

A3)
X 0 1 2 3 4

0 0.5 0.67 0.75 0.8

,

B(x)= 1 0.9 0.8 0.7 0.6

{0,1,2,3,4}

{1,2,3,4}

{2,3,4}

{3,4}

{4}

{0}

{0,1}

{0,1,2}

{0,1,2,3}

{0,1,2,3,4}
Thus,

Thus,

Q4)

Their

Step 1: Find and

Step 2: Use decomposition theorem which is given as and arithmetic

operation on closed interval

Step 3: Find membership function.

Q5) The membership function of two fuzzy numbers A and B are given as follows

-1 < x
= =

= =
A(x) B(x)
 =0 =0 otherwise

Find A+B and A-B

A5)

Step 1: Find and

For -1 < x

By def. of ,

A(x)

For

= [2 -1, 3-2 for all

For

By def. of cut,

B(x)

For 3 < x

B(x)
 = [2 +1, 5-2 for all

Step 2: By decomposition theorem & arithmetic operation on closed interval we

find as follows.

Step 3: Now find membership function

Let 4 Let 8-4

Thus, (A+B) (x) = ; 0<x

 = 4

= 0; Otherwise

To find A-B

Step 4: By decomposition theorem and arithmetic operation on closed interval we

find follows

=[

= [4 for all

Step 5: Now find membership function to define fuzzy numbers A-B

Let 4 Let 2-4

Thus,
Q6) The membership function of two fuzzy numbers A and B are

= x-1 1<x
=

=
=
A(x) B(x)
=0 =0 otherwise

Solve A+X =B

A6)

Let

A+X = B

Taking alpha cut on both side

Check
0.1 12.8 56.9

0.6 26.8 46.4

0.8 32.4 42.2

Thus, hence solution exists.

Let Let

Q7) The membership function of two fuzzy numbers A and B are

= x-3 3<x
=

=
=
A(x) B(x)
 =0 =0 otherwise

Solve A.X =B

A7)

Let

A.X = B

Taking alpha cut on both side

Check

0.3 4.3636 6.0426

0.7 4.7568 5.4884

0.9 4.9231 5.1707

Thus, hence solution exists.

Let Let
 Unit –6

Q1) 1 2 3 4 (Tasks)

Column Reduction

A 19 27 18 12
B 14 29 30 27
C 39 20 19 16
D 20 27 25 11

Workers

A1)

Row reduction:

19- 27- 18- 12-

12 12 12 12
14- 29- 30- 27-
14 14 14 14
39- 20- 19- 16-
16 16 16 16
20- 27- 25- 11-
11 11 11 11

column reduction:

7 15 6 0

0 15 16 13

23 4 3 0
9 16 14 0

7 11-3 3-3 0
0 11-3 13-3 13

23+3 0 0 0+3
9 12-3 11-3 0
7-0 15-4 6-3 0-0
0-0 15-4 16- 13-
3 0
23- 4-4 3-3 0-0
0
9-0 16-4 14- 0-0
3
7 11 3 0

0 11 13 13

23 0 0 0
9 12 11 0

7 8 0 0
0 8 10 13
0 0 3
26
9 9 8 0

workers tasks hours

A 3 18
B 2 14
C 1 20
D 4 11
Total 63
Q2) Three men are to be given 3 jobs and it is assumed that a person is fully capable
of doing job independently. The following table gives an idea of that cost incurred
to complete each job by each person

men jobs supply

20 28 21 1

15 35 17 1

8 32 20 1

Demand 1 1 1

Formulate as linear programming problem

A2)

The given problem can easily be formulated as a linear programming (transportation)

model as under.

Z = (20 +28 + 21 ) + (15 +35 + 17 )+ (18 +32 + 20 )

+ + =1

+ + =1 , where i =1, 2, 3, ……

+ + =1

Since each job can be assigned to only one person, therefore three equations for three
different persons three different jobs.

The given problem is just a special case of the transportation problem.

Q 3: you work as a sales manager for a toy manufacturer, and you currently have
three salespeople on the road meeting buyers. Your salespeople are in Austin, TX
Boston, MA; and Chicago, IL. You want them to fly to three other cities: Denver,
CO; Edmonton, Alberta; and Fargo, ND. The table below shows the cost of airplane
tickets in dollars between these cities.

From Denver Edmonton Fargo

To
Austin 250 400 350

Boston 400 600 350

Chicago 200 400 250

Where should you send each of your salespeople in order to minimize airfare?

A3)

Step 1: Subtract 250 from row 1, 350 from row2, and 200 from row 3.

Step 2: Subtract 0 from column 1, 150 from column 2, and 0 from column 3.

Step 3: Cover all the zeros of the matrix with the minimum
number of horizontal or vertical lines.

Step 4: Since the minimal number of lines is 3, an optimal assignment of zeros is possible
and we are finished. Since the total cost for this assignment is 0, it must be an optimal
assignment.
Here is the same assignment made to the original cost matrix.

Q4) A construction company has four large bulldozers located at four different
garages. The bulldozers are to be moved to four different construction sites are
given below.

Bulldozer A B C D

site
1 90 75 75 80
2 35 85 55 65
3 125 95 90 105
4 45 110 95 115

How should bulldozers be moved to the construction sites in order to minimize the
total distance travelled?

A4)

Step 1: Subtract 75 from row 1, 35 from row 2, 90 from row 3, and 45 from row 4.

Step 2: Subtract 0 from colum 1, 0 from column 2, 0 from column 3, and 5 from column 4.

Step 3: Cover all the zeros of the matrix with the minimum number of
horizontal or vertical lines.
Step 4: Since the minimal number of lines is less than 4, we have to proceed to step 5.

Step 5: Note that 5 is the smallest entry not covered by any line. Subtract from each
uncovered row.

Now add 5 to each covered column.

Now return to step 3.

Step 3: Cover all the zeros of the matrix with the minimum number of horizontal or
vertical lines.

step 4: Since the minimal number of lines is less than 4, we

have to return to step 5.

Step 5: Note that 20 is the smallest entry not covered by a line. Subtract 20 from each
uncovered row.

Then add 20 to each covered column.

 Now return to step 3

Step 3: Cover all the zeros of the

matrix with the minimum number of horizontal or vertical lines.

Step 4: Since the minimal number of lines is 4, an optimal assignment of zeros is possible
and we are finished.

Since the total cost for assignment is 0, it must be an optimal assignment.

Here is the same assignment made to the original cost matrix.

So, we should send Bulldozer 1 to site D, Bulldozer 2 to site C, Bulldozer 3 to site B, and
Bulldozer 4 to site A.

Q5) Using the following cost matrix, determine

) Optimal job assignment.

i) The cost of assignment.

A B C D

1 2 10 9 7
 2 15 4 14 8

3 13 14 16 11

4 4 15 13 9

A5)

Here, no.of rows = no.of columns, In other words, no.of workers = no.of jobs. So, it is a
balanced assignment problem.

Step 1: Subtract the smallest element of each row from all the elements of that row.

2 10 9 7
15 4 14 8
13 14 16 11
4 15 13 9

0 8 7 5
11 0 10 4
2 3 5 0
0 11 9 5

step 2: Subtract the smallest element of each column from all the elements of that column.

0 8 7 5
11 0 10 4
2 3 5 0
0 11 9 5
0 8 2 5
11 0 5 4
2 3 0 0
0 11 4 5
Step 3: cover all zeros

(a) Cover all zeros of the matrix with the minimum number of horizontal or vertical lines.
If the number of lines drawn is equal to the order of the matrix then the solution is
optimal.

8 2 5

0
11 0 5 4
2 3 0 0
11 4 5
0

The minimum number of lines that cover all the zeros = 3. And is not equal to the number
of rows/columns. Proceed to next step.

(b) Select all smallest elements from the matrix which is not covered by lines (here 2 is
the smallest and uncovered elements
Subtract the elements from all the uncovered elements and add this smallest element
to the element lying at the intersection of lines repeat step 3(a) and 3(b)

2 5

0 8
11 0 5 4
2 3 0 0
0 11 4 5

As the minimum number of lines that cover all zeros =4. And equal to the number of
rows/columns.

Therefore, the solution is optimal.

 3

0 8 0
11 0 3 2
4 5 0 0
0 11 2 3

Step 4: Assign the zeros.

(a) Examine the rows which contain only one ‘zero ‘element. Make an assignment to this
single zero by encircling it. Cross all other zeros, as these will not be considered for
any future assignment. Continue till all the rows have been examined.

(b) Now, examine the columns containing only one ‘zero’. Encircle this zero and make an
assignment there. Then cross any other zero and continue till all the columns are
eliminated

(c) Repeat the step 4(A) and 4(b) until all the assignments have been made.

(d) While assigning, if there is no single zero exists in the row or column, then select any
zero arbitrary and encircle it and cross all other zeros in the column and row in
respect of which the assignment is made.

Therefore, the optimal job assignment is (1-c), (2-B), (3-D), (4-A)

Total cost = 9+4+11+4 = 28.

Q6) solve the balanced assignment problem.

jobs 1 2 3 4
persons

A 10 12 19 11
B 5 10 7 8
C 12 14 13 11
D 8 12 11 9

A6)

Step 1: row reduction step 2: column reduction

0 2 9 1
0 5 2 3
1 3 2 0
0 4 3 1
0 0 7 1
0 3 0 3
1 1 0 0
0 2 1 1

person job total

A 2 12
B 3 7
C 4 11
D 1 8

Therefore, the optimum cost is Rs.38 (12+7+11+8 =38).

Q7) Recently, Yadaiah and haragopal published in the American journal of
operations Research a new approach to solving the unbalanced assignment
problem. They also provide a numerical example which they solve with their
approach and get a cost of 1550 which they claim is optimum. This approach might
be of interest; however, their approach does not guarantee the optimal solution. In
this short paper, we will show that solving this sample textbook formulation to
balance the problem and then solve it with the classic Hungarian method Of Kuhn
yields the true optimal solution with a cost of 1520.

jobs

machines

300 290 280 290 210

250 310 290 300 200
180 190 300 190 180
320 180 190 240 170
270 210 190 250 160
190 200 220 190 140
220 300 230 180 160
260 190 260 210 180

A7)

first sub-problem:

jobs

machines

180 190 300 190 180

320 180 190 240 170
270 210 190 250 160
190 200 220 190 140
 220 300 230 180 160

870 + 680 = 1550. It can easily be checked using the Hungarian method, that these sub-
problems were solved optimally. In their paper,

Yadaiah and Haragopal have a minor typo.

We will now solve the original problem of assigning these eight jobs to five machines
such that machine is used at least once, but not more than twice.

Second sub-problem

jobs

290 290 210

310 300 200

190 210 180

Final Hungarian method cost matrix

40 30 20 30 0 40 30 20 30 0
0 60 40 50 0 0 60 40 50 0
0 10 120 10 50 0 10 120 10 50
140 0 10 60 40 140 0 10 60 40
80 20 0 60 20 80 20 0 60 20
0 10 30 0 0 0 10 30 0 0
40 120 50 0 30 40 120 50 0 30
70 0 70 20 40 70 0 70 20 40
DUM 1 0 0 0 0 50 0 0 0 0 50
DUM 2 0 0 0 0 50 0 0 0 0 50
For a total minimum cost of 1520 (not 1550).

Q8) Let us consider a problem with 8worker and 4 tasks. Its worker-task
assignment cost matrix is shown in table

i/j 1 2 3 4
A 53 62 42 89
B 18 35 9 55
C 93 80 91 83
D 79 23 96 56
E 43 16 12 23
F 43 16 12 20
G 35 79 25 59
H 27 16 12 20

A8)

Step 1: Create an unbalanced matrix problem for worker task assignment cost matrix.

Step 2: Obtain table 2 by adding duplicate or one dummy task with all 1 assignment cost
to given table

i/j 1 2 3 4 5
A 53 62 42 89 1
B 18 35 9 55 1
C 93 80 91 83 1
D 79 23 96 56 1
E 43 16 12 23 1
F 43 16 12 20 1
G 35 79 25 59 1
H 27 16 12 20 1

Step 3: Calculate a reduced cost matrix by dividing each row by minimum cost of its
column. Reduced cost matrix obtained from table 2.

i/j 1 2 3 4 5
A 2.94 3.87 3.5 4.45 1
B 1 2.18 3.25 2.75 1
C 5.16 5 7.58 4.15 1
D 4.38 1.43 8 2.8 1
E 2.38 1 1 1 1
F 4.83 4037 7.25 1.55 1
G 1.94 4.93 2.083 2.95 1
H 1.5 1 1 1 1

Step 4: Look for the row with one reduced cost excluding 1 from dummy column. Assign
worker to task according to position of the choosen 1 and mark entire row to avoid later
redundant assignment as shown in above table.

i/j 1 2 3 4 5
A 2.94 3.87 3.5 4.45 1
B 2.18 3.25 2.75 1
1
C 5.16 5 7.58 4.15 1
D 4.38 1.43 8 2.8 1
E 2.38 1 1 1 1
F 4.83 4037 7.25 1.55 1
G 1.94 4.93 2.083 2.95 1
H 1.5 1 1 1 1

Step 5: Look for the column with one reduced cost excluding 1 from dummy row. Assign
worker to task according to position of the chosen 1 and mark entire column to avoid
later redundant assignment as shown in above table.

Step 6: Choose one of remaining 1 in the reduced cost matrix as a position to assign
worker to task. Mark row and colum of chosen 1 to avoid later redundant assignment.
After applying step 6 to the matrix in table 4, we get the new matrix as shown in table 5.
i/j 1 2 3 4 5
A 2.94 3.5 4.45 1

3.87
B 2.18 3.25 2.75 1
1
C 5.16 5 7.58 4.15 1
D 4.38 1.43 8 2.8 1
E 1 1 1 1
2.38
F 4.83 4037 7.25 1.55 1
G 1.94 4.93 2.083 2.95 1
H 1 1 1 1
1.5

Step 7: Assign dummy task to the first n-m agent from table 5, we assign dummy task to 8-
4 workers as shown in table 6

i/j 1 2 3 4 5
A 2.94 3.5 4.45 1

3.87
B 2.18 3.25 2.75 1
1
C 5.16 5 7.58 4.15 1
D 4.38 1.43 8 2.8 1
E 1 1 1 1
2.38
F 4.83 4037 7.25 1.55 1
G 1.94 4.93 2.083 2.95 1
H 1 1 1 1
1.5
Step 8: Count the number of assigned text excluding dummy task. If it is equal to the
number of tasks then go to step 11(5)

Step 9: Mark row assignment excluding dummy task assignment and mark dummy
column. The marked rows/column from our examples are shown in table 7.

i/j 1 2 3 4 5
A 2.94 3.5 4.45

3.87 1
B 2.18 3.25 2.75 1
1
C 5.16 5 7.58 4.15 1
D 4.38 1.43 8 2.8 1
E 1 1 1 1
2.38
F 4.83 4037 7.25 1.55 1
G 1.94 4.93 2.083 2.95 1
H 1 1 1 1
1.5

Step 10: Look for the minimum cost among remaining (unmarked) costs. Recalculate the
reduced cost table by dividing each remaining cost by the minimum cost and replace 0 at
intersection with the minimum cost. And go to step 4. Table 8 shows the modified
reduced matrix.

i/j 1 2 3 4 5
A 1.51 2.44 2.07 3.02 1
B 1 2.18 3.25 2.75 1.43
C 3.73 3.57 6.15 2.72 1
D 2.95 0 6.57 1.37 1
E 2.38 1 1 1 1.43
F 3.4 2.94 5.82 0.12 1
G 0.51 3.5 0.65 1.52 1
H 1.5 1 1 1 1.43
Step 11: Calculate total assignment cost.

Suppose B - >1, D - >2, E – >3 and H - >4. The total cost for this assignment is calculated
as follows.

Min(T) =

Q9) A travelling salesman has to visit five cities. He wishes to start from a particular
city, visit each city and then return to his starting point. The travelling cost (in Rs.)
of each city from a particular city is given below.

Travelling salesman problem:

From city A B C D E

To city
A a 2 5 7 1
B 6 a 3 8 2
C 8 7 a 4 7
D 12 4 6 a 5
E 1 3 2 8 a

What should be the sequence of the salesman’s visit, so that the cost is minimum?

A9)

The optimal solution is reached using Hungarian method.

From city A B C D E

To city
A a 1 3 6 0
B 4 a 0 6 0
C 4 3 a 0 3
D 8 0 1 a 1
E 0 2 0 7 a

The assignment shown in the above table gives the route sequence

A B, B C, C D, D E, E A

The travelling cost to this solution is 2000 + 3000 + 4000 + 5000 + 1000 = 15,000.

Therefore, the sequence feasible for this assignment is

A with the travelling cost of Rs. 15,000.

Q10) A sales man has to visit five cities A, B, C, D AND E.the intercity distances are
tabulated below

To/from A B C D E
A - 12 25 26 16
B 7 - 17 19 8
C 10 11 - 19 12
D 15 18 23 - 17
E 12 14 24 26 -

The above distances in km between two cities need not be same both ways. Which
route would you advise the salesman to take so that the total distance travelled by
him is minimum, if the salesman starts from city A and has to come back to city ‘A’.

A10)

step 1: Select smallest element in each row and subtract it from every element of that
row. Row reduction results in the following table.

To/from A B C D E
A - 0 13 14 4
B 0 - 10 12 1
C 0 1 - 9 2
D 0 3 8 - 2
E 0 2 12 14 -

Step 2: Select the minimum element in each column and subtract this element from every
element in that column, we get

To/from A B C D E
A - 0 5 5 3
B 0 - 2 3 0
C 0 1 - 0 1
D 0 3 0 - 1
E 0 2 4 5 -

Step 3: Cover all zeros and get the final minimum solution

To/from A B C D E
A - 0 5 5 3
B 0 - 2 3 0
C 0 1 - 0 1
D 0 3 0 - 1
E 0 2 4 5 -

Optimal schedule is B C, C D, D E, E A

Involving distance 12 + 17 + 19 + 12 = 77 km.

This route satisfies the travelling salesman problem.