Priya Bothra Roll No. 264

8/9/2019 Priya Bothra Roll No. 264

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An

Assignment

On

QUANTITATIVE TECHNIQUES AND

BUSINESS STATASTICS

Submitted towards partial fulfillment of the

II semester of MBA/MBL degree course

for the subject

Submitted by : Submitted to:

Priya Bothra Dr. R.N Aggarwal

MBA- MBL(IISem) Faculty In Charge

FACULTY OF MANAGEMENT STUDIES

NATIONAL LAW UNIVERSITY, JODHPUR

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INDEX

S.N

O

PARTICULARS PAGE-NO.

1. Acknowledgment 3

2. Mean deviation 4

3. Assignment 5

4. Transportation 8

5. Bayes theorm 9

6. Co-relation 10

7. Linear programming 12

8. Replacement 15

9. Binomial distribution 17

10. Sampling 18

11. Queuing Theory 19

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ACKNOWLEDGEMENT

This project owes its accomplishments to the inputs, help, and assistance received fromvarious website and various books. A great sense of gratitude is also expressed towards our

guiding faculty, Dr. R.N. Agarwal who very ably guided me in the completion of this

project. The project could not have been completed without his persistent feedback and

knowledge imparted about the subject.

Thanks to the library staff of the National Law University.

In the end I also extend thanks to our classmates who have always helped me whenever Ineeded their help.

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MEAN DEVIATION

Q.1 In a Company named ARIL Ltd., at the end of the first month; the salaries are

distributed to the managing officials. Given below are the different ranges of salary in

thousands and the respective number of officials entitled or lying in a particular salary

bracket. Calculate the mean deviation by taking assumed mean.

Class-Interval

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

Frequency 15 18 20 23 27 21 25 16 24 30

SOLUTION:

Class-Interval(C.I.)

Frequency(F)

Mid- Value(X)

D=X-A/C.I. FD Dx =X-X FDx

0-10 15 5 -4 -60 48.5 727.5

10-20 18 15 -3 -54 38.5 693

20-30 20 25 -2 -40 28.5 570

30-40 23 35 -1 -23 18.5 425.5

40-50 27 45 0 0 8.5 229.5

50-60 21 55 1 21 1.5 31.5

60-70 25 65 2 50 11.5 287.5

70-80 16 75 3 48 21.5 344

80-90 24 85 4 96 31.5 756

90-100 30 95 5 150 41.5 1245

From the above distribution;

F= Number of employees

N= summation of frequency = 219

Class-Interval= Salary Bracket

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D= Deviations From X

A= Assumed mean

FD=188

Dx=250

FDx=5309.5

Mean

X = A +fd/*i

=45 + 188/219 *10

= 45 +8.58=53.3

Mean deviation=

=5309.5/219

=24.24

Coefficient of mean deviation=

=24.24/53.5

=0.45

Conclusion:

Hence, the mean deviation of the salaries of managing officials at ARIL LTD Is Rs.24,240.

ASSIGNMENT

Q.2 In the Rajasthan handlooms Corporation, There are four processes to be carries outnamely- Ginning, Spinning, Weaving, Packaging. For each work, there are five differenttypes of machines employed MI, M2, M3, M4, M5 which render different volumes ofproduction units. Make the assignment of each work in such a way that there is maximumoutput.

M1 M2 M3 M4 M5GINNING 40 38 45 50 46

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SPINNING 35 42 37 56 47WEAVING 49 60 44 58 54PACKAGING 53 51 52 48 55

SOLUTION:

The given question is one of the types of maximization assignment problem.

It requires the addition of a dummy row as the number of jobs to be assigned is not

equal to the number of machines.

Let ginning, spinning, weaving and packaging be denoted by G, S, W, P

respectively.

STEP 1- Addition of a dummy row

M1 M2 M3 M4 M5G 40 38 45 50 46S 35 42 37 56 47W 49 60 44 58 54P 53 51 52 48 55D.R. 0 0 0 0 0

STEP 2- Selecting the maximum value from the grid and subtracting the rest of thevalues of the grid from it..the maximum value-60

M1 M2 M3 M4 M5G 20 22 15 10 14S 25 18 23 4 13W 11 0 16 2 6P 7 9 8 12 5

D.R. 60 60 60 60 60

STEP 3- Selecting the minimum element from each row and subtracting it from the rest ofthe elements in each row. The minimum elements have been highlighted in the above grid.

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M1 M2 M3 M4 M5G 10 12 5 0 4S 21 14 19 0 9W 11 0 16 2 6

P 2 4 3 7 0D.R. 0 0 0 0 0

STEP 4- The same operation is repeated selecting minimum from each column. 0 is theminimum element in each column. Thereafter, cancelling all the zeroes with minimumlines.

M1 M2 M3 M4 M5

G 10 12 5 0 4S 21 14 19 0 9W 11 0 16 2 6P 2 4 3 7 0D.R. 0 0 0 0 0

STEP 5- Selecting the minimum element from the uncovered elements and subtracting itfrom the rest of the uncovered elements and adding that minimum element at junctionpoints. The rest elements remain same. The minimum element is-4

M1 M2 M3 M4 M5G 6 8 1 0 0S 17 10 15 0 5W 11 0 16 6 6P 2 4 3 11 0D.R. 0 0 0 4 0

STEP 6- Repeating the same operation

M1 M2 M3 M4 M5G 5 7 0 0 0S 16 9 14 0 5W 11 0 16 7 7P 1 3 2 11 0

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D.R. 0 0 0 5 1

STEP 7-Assigning the jobs by selecting zeroes starting from the column with minimumnumber of zeroes and cancelling all the zeroes parallel to it horizontally.

M1 M2 M3 M4 M5G 5 7 0 0 0S 16 9 14 0 0W 11 0 16 7 7P 1 3 2 11 0D.r. 0 0 0 5 1

FINAL SOLUTION:

M1- No job as dummy row wont be consideredM2- Weaving

M3-Ginning

M4- Spinning

M5- Packaging

TOTAL PRODUCTION

M1- 0M2- 60 UNITS

M3- 45 UNITS

M4- 56 UNITS

M5- 55 UNITS

TOTAL= 216 UNITS

Hence, total production by use of all four processes and five machines are 216 units.

TRANSPORTATION PROBLEM

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Q3 Mafatlal textiles have three manufacturing units. The monthly demand of warehouses is

being given in the table below. The transportation cost per tonne from different

manufacturing units to meet the demand is given below. Calculate and suggest the optimum

transportation schedule and the minimum transportation costs.

W1 W2 W3 W4 SUPPLYD1 7 3 8 6 60D2 4 2 5 10 100D3 2 6 5 1 40DEMAND 20 50 50 80 200

SOLUTION

1 2 3 4 SUPPLY P1 P2 P3A --- 20 --- 40 60/20/0 3 3 4

B 20 30 50 --- 100/70/20/0 2 2 2

C --- --- --- 40 40/0 1 -- --

DEMAND 20/0 50/30/0 50/0 80/40/0 200

P1 2 1 0 5

P2 3 1 3 4

P3 3 1 3 4

TOTAL COST = 3 x 20 + 6 x 40 + 4 x 20 + 2 x 30 + 5 x 50 + 1 x 40

= Rs 730

Hence, the above given solution is optimal transportation schedule for the company and it

will satisfy demand of each and every distribution point minimum and the optimum cost ofRs 730.

BAYES THEOREM

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Q4. SHIV industries is a manufacturer of bicycles and tricycles for childrens and for this

they have two different suppliers for wheels of cycle and they are named chawla & co. and

kamal industries .and these wheels are placed in a heap and from the past record it has been

found out that 5% of parts supplied by chawla & co. are defective and 9% from kamal

industries are defective. Wheel is selected from the heap at random and is found to be

defective. What is the probability that it was supplied by chawla & co?

Solution:-

Using the Bayes Theorem

H1 = Steering supplied by chawala & co. = P (H1) = 1/2

H2 = Steering supplied by kamal industries = P (H2) = 1/2

E = Defective part is found

P( E/ H1) = 5/100

P( E/ H2) = 9/100

P(H1/E) = (P( H1 ) (P E/H1 ) /[ P( H1 ) (P E/H1 ) ] + [ P( H2 ) (P E/H2 ) ]

Hence Substituting bayes theorm

* 5/100 / * 5/100 + *9/100

P (H1/E) = 0.3571

CONCLUSION:

The probability that the steering wheel would be supplied by chawla and co. is 0.3571.

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CO-RELATION

Q5. Food and Agricultural Association has ordered a probe into the deaths of general public

due to consumption of toxic crops. The following data has been obtained from the Araria

district of Bihar which produces wheat. Calculate Karl Pearsons coefficient of correlation

between amount of pesticide & wheat crop in a district Araria in year 08-09.

Pesticide(in 000

Kg.)10 12 15 18 25 35 45 50 55 65

ProductionOf Wheat(in metric

ton)

5 7 13 15 20 21 29 30 36 44

Solution:-

Size ofitem

(pesticide)X

Deviationfrom mean

X= 33Dx

Squaring ofdeviation

dx

Size ofitem

(productionin metricton) Y

Deviationfrom mean

Y=22

Square ofdeviation

d ydxdy

10 -23 529 5 -17 289 391

12 -21 441 7 -15 225 31515 -18 324 13 -9 81 162

18 -15 225 15 -7 49 105

25 -8 64 20 -2 4 16

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35 2 4 21 -1 1 -2

45 12 144 29 7 49 84

50 17 289 30 8 64 136

55 22 484 36 14 196 308

65 32 1024 44 22 484 704

X= 330 0 dx= 3528 Y= 220 0 dy= 1442dxdy=

2219

X-SERIES Y-SERIES

X = X/ N=330/10 =33 Y= Y/N= 220/10 = 22

x = dx/N = 3528/10 = 18.78 y= dy/N =1442/10

=12.01

r= dxdy/Nx.y = 2219/1018.7812.01= 0.98

Hence, there is a positive correlation between the pesticides used on wheat crop and deathdue to use of pesticides is 0.98.

LINEAR PROGRAMMING PROBLEM

Q6.A Company named Jyoti Bhog Ltd owns two flour mills, A and B, which have different

productions capacities for high, medium, and low grade flour. This company has entered to

supply flour to a firm every week with 12, 8, & 24 quintals of high, medium and low grade

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respectively. It costs the company Rs 1000 and Rs 800 per day to run a mill A & B

respectively. On a day, Mill A produces 6, 2 and 4 quintals of high, medium and low grade

flour respectively; Mill B produces 2, 2 and 12 quintals of high, medium and low flour

respectively. How many days per week should each mill operated in order to meet the

contract order most economically.

SOLUTION:

Let number of days per week Mill A be = x

Let number of days per week Mill B operates be = y

The data can be represented in the following tabular manner

Produce/Mills

Mill A

x

Mill B

y

Requirements

High Grade 6 2 12

Medium Grade 2 2 8

Low Grade 4 12 24

Cost (Rs)/ day 1000.00 800.00

Thus the given data can be given in the table can be expressed in Linear Programming form

in the following way

ZMIN = 1000x + 800y

Subject to

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6x +2y 12

2x + 2y 8

4x + 12y 24

Condition,

x 0, y 0

The inequalities are converted to equalities

6x +2y = 12

X 0 2

Y 6 0

The data can represented in the form of equations of a line in the following manner

2x + 2y = 8

X 0 4

Y 4 0

4x + 12y = 24

X 0 6

Y 2 0

Now, substituting the values of all the above given 3 equations in to graph to find outfeasible area to minimize the c ost of operating both mills owned by company.

Conclusion

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points Co-ordinates Zmin = 1000x +800y

A (6,0) 6000

B (3,1) 3800

C (1,3) 3400

D (0,6) 4800

Thus minimum cost of operating 2 mills is 3400 at point c hence, the company will operate

first mill for 1 day and second mill for 3 days in a week to attain the minimum cost of

operation.

REPLACEMENT:

Q7. In order to project the financial position how much expenses are being incurred by the

company in maintaining its machine and also to find out best period to replace its existing

machine being used in the bottling plant of bislery (mineral water) unit. The followinggiven below information is ascertained by the accountant of the company.

Year

Maintenance Cost (in

Rs.)

Resale Price (in

Rs.)

1 8000 145000

2 9000 132000

3 10500 122000

4 13000 114000

5 15000 109000

6 20000 90000

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7 25000 70000

8 30000 50000

In which year should the processing machine be replaced?

Solution:

Costs of owning and operating the machine

Year (n) Mt Mt S C-S

T(n)=C-S+

Mt A(n)=T(n)/ n

1 8000 8000

14500

0 15000 23000 23000

2 9000 17000

13200

0 28000 45000 22500

3

1050

0 27500

12200

0 38000 65500 21833.3

4

1030

0 40500

11400

0 46000 86500 21625

5

1500

0 55500

10900

0 51000 106500 21300

620000 75500 90000 70000 145500 24250

7 2500 10050 70000 90000 190500 27214.28

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0 0

8

3000

0

13050

0 50000

11000

0 240500 30062.5

* Mt =Maintenance Cost

Mt= Cumulative maintenance cost

S= Resale Price

C=Purchase Price

T (n) = Total Cost for n years

A (n) = Average Cost for n years

From the above table we can conclude that the machine should be replaced at the end of 5th

year. If it is not replaced at this period of time the average cost per year will start increasing.

BINOMIAL DISTRIBUTION:

Q8. Apple is supplying its macbook to its distributors in India who is located in Bombay

and wants to conduct an enquiry into the inventory of technical support mechanism in all its

branches located throughout India. Each laptop has an 8% probability of not working. 20

laptops were randomly selected for the enquiry:

(a) What is the likelihood that 5 will be broken? (b) What is the likelihood that they

will all work?

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Solution:-

An item can either be defective or not defective. Thus, there are only two outcomes. The

probability (p) of a defective item being produced is 0.08.

p=0.08

q=1-p =1- 0.08= 0.92

n=20

a) Probability of 5 broken laptops [ p(5)] n r

205 (.08)5 (.92)15

= .0145

b) Probability of all laptops working [p(0)]=20

5 (.08)0 (.92)20

= .1887

Hence, the probability of five laptop being defective out of randomly selected 20 laptop is

0.0145 and the all being in working condition is 0.1887.

SAMPLING:

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Q9. In order to admit students to its Help Underprivileged Children education program,shiksha high school has conducted a survey of household income of its 3000 students. Howlarge a sample size should be taken in order to estimate the mean annual earning within plusand minus Rs. 1000 and at 95% confidence level? The standard deviation of annualearnings is known to be Rs. 3000.

Solution:

As desired upper and lower limit is Rs. 1000 i.e. we want to estimate the annual earningswithin plus and minus Rs.1000.

As the confidence level is 95% value of is 1.96.

Where is Standard Error

is standard deviation i.e. 3000

Therefore, the desired sample size is about 35.

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QUEUING THEORY:

Q10. In order to decide the number of breaks to be given to employees according toTaylors fatigue study model, Perfect Footwear Companys Jaipur outlet has found thefollowing data: That on an average 20 customer arrive in an hour and their service rate is 23

customers per hour. Find out the average time the employees are free on a 10 hour workingday. And also find the expected number of customers at the outlet.

Solution:

Service rate () = 23/ hr

Arrival Rate () = 20/ hr

Where is utilization parameter

The average time the employees are free on a 10 hr day

1.30 hrs

Expected number of customers at the outlet

= 6.692

Hence, from the solution we can conclude that average time for which employee remainsfree in an hour is 1.30 hr and expected no. of customer arrives at outlet is 7.

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Documents

Priya Bothra Roll No. 264