qt assignment - nitin gupta (260)

8/7/2019 QT Assignment - Nitin Gupta (260)

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AnAssignment

Submitted towards the partial fulfilment of the II Semester of theMBA-LLM Degree Course, for the subject

QUANTITAVE TECHNIQUES AND BUSINESS STATASTICS

Submitted By: SubmittedTo:Nitin Gupta Dr. R. N. AgarwalMBA-LLM Assistant ProfessorII Semester Faculty of ManagementStudies

R. No. 260 National Law University, Jodhpur

NATIONAL LAW UNIVERSITY


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JODHPUR

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Table of Content s

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S. No. Name of the Problem Page Nos.

1 Standard Deviation 3-6

2 Co-Relation 7-9

3 Transportation 10-11

4 Assignment 12-155 Replacement 16-17

6 Linear Programming Problem (Solution by GraphicalMethod) 18-21

7 Conditional Probability (Bayes Theorem) 22-23

8 Poisson Distribution 24-25

9 Normal Distribution 26-27

10 Sampling 28-30


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Feb.02, 2010 QT Problem No. 1: Standard Deviation Nitin Gupta

Roll No.260

Sem: MBA-LLM II

The company has the policy to pays to its workers, working in its production unit, as per their duration of association with

the company. The company has two production units. The given data provides us with the various time bands of association,

number of workers in that time band of association and their wages.

Calculate the mean wages and standard

deviation of the wages of all the workers of the

company together.

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Time Bands

Unit I Unit II

Wages(Rs.oo)

No of Workers Wages

(Rs.oo)

No of workers

0-4 30 15 32 18

4-8 34 18 35 20

8-12 36 25 38 34

12-16 40 24 42 36

16-20 42 20 46 27

20-24 46 12 50 25

24-28 50 6 55 20


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UNIT I

f 1x1= 4614 f 1= 120

f 1x12= 180780

x 1= f 1x1/ f 1 = 4614/120 = 38.451 = [ f 1x12/ f 1 (f 1x1/ f 1)2]

= [ 180780/ 120 (38.45) 2]

= (1506.5 1478.5) = 28

12 = 28

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Time Bands (x 1) (x 12) (f 1) (f 1 x1) (f 1 x12)

0- 4 30 900 15 450 13500

4-8 34 1156 18 612 20808

8-12 36 1296 25 900 32400

12-16 40 1600 24 960 38400

16-20 42 1764 20 840 3528020-24 46 2116 12 552 25392

24-28 50 2500 6 300 15000

TOTAL 120 4614 180780


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UNIT II

f 2x2= 7672 f 2= 180

x 2= f 2x2/ f 1 = 7672/180 = 42.622 = [ f 2x22/ f 2 (f 2x2/ f 2)2]

= [ 335664/180 (42.62) 2]

= (1864.8 1816.4) = 48.4

22 = 48.4

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Time Bands (x 2) (x 22) (f 2) (f 2x) (f 2x22)

0- 4 32 1024 18 576 18432

4-8 35 1225 20 700 24500

8-12 38 1444 34 1292 49096

12-16 42 1764 36 1512 63504

16-20 46 2116 27 1242 5713220-24 50 2500 25 1250 62500

24-28 55 3025 20 1100 60500

TOTAL 180 7672 335664


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Combined Mean

x 12= (n 1x 1 + n 2x 2)/( n 1 + n 2)= 120*38.45 + 180*42.62/300

= 4614 + 7672/300 = 12286/300 = 40.953

d 1 = x 1 - x 12 = 38.45- 40.45 = -2.5 d 2 = x 2 - x 12 =42.62 40.95 = 1.67

d 12 = 6.25 d 22 = 2.79

Combined Standard Deviation

12 = [(n 112 + n 222 + n 1d 12 + n 2d 22)/(n 1 + n 2)]

= [(120*28 + 180*48.4 + 120*6.25 + 180*2.79)/(300)]

= [(3360+8712+750+502.2)/(300)]

= [13324.2/300]

= 44.41

= 6.66

Therefore, the mean wages of all the workers of the company is Rs. 4095.3 and standard deviation is of Rs. 666.43

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Checked on: February 02, 2010

Feb.09, 2010 QT Problem No. 2: Co-Relation NitinGupta

Roll No.260

Sem: MBA-LLM II

The efficiency of 51 salesmen were measured by collecting data relating to the time spent by them in the sales promotionand units of diaries sold by them in a month. The data so collected was then tabulated on the daily average basis. Data isgiven below.

S.No. Ave.Time(Hrs)

Ave. Sales S.No. Ave.Time(Hrs)

Ave. Sales S.No. Ave.Time(Hrs)

Ave. Sales

1 2.5 42 18 5 49 35 8 462 7 35 19 8.4 54 36 9 543 5 32 20 1.2 41 37 6 654 12 84 21 2.8 49 38 10 595 1.5 27 22 8.5 56 39 8 426 10.9 71 23 4.9 36 40 4 357 4.2 49 24 8 62 41 4.7 428 5.5 60 25 7.5 62 42 10.5 609 6.7 52 26 3 3.6 43 7 75

10 2.6 24 27 11 63 44 9.5 4811 5.2 26 28 1 16 45 3.6 3512 7.7 53 29 9.5 61 46 9 6613 0.5 19 30 6.1 54 47 7 3614 9.5 70 31 6 55 48 10 8515 7.2 57 32 7.7 47 49 1.7 33

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16 2.75 31 33 9 63 50 4.8 4417 11.5 70 34 2 23 51 3.5 35

Find the coefficient of correlation between the factors i.e. average time spent and average unit sale by using groupingmethod.

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Average TimeSpent(Hrs.)

Average Unit Sale

15-25

25-35

35-45

45-55

55-65

65-75

75-85

y 20 30 40 50 60 70 80

xd y -3 -2 -1 0 1 2 3

f f.d x f.d x2 f.d x.d yd x

0-2 1 -4 3 2 1 -- -- -- -- 6 -24 96 56

2-4 3 -2 1 4 2 1 -- -- -- 8 -16 32 264-6 5 0 -- 2 3 3 2 -- -- 10 0 0 0

6-8 7 2 -- 1 2 5 3 1 -- 12 24 48 2

8-10 9 4 -- -- -- 3 4 2 1 10 40 160 44

10-1211 6 -- -- -- -- 2 2 1 5 30 180 54

f 4 9 8 12 11 5 2 f =51 f.d x=54 f .d x2=516 f.d x.d y=182

f.d y -12 -18 -8 0 11 10 6 f.d y= -11

f.d y2 36 36 8 0 11 20 18 f.d y2=129

f.d x.d y 42 28 4 0 34 44 30 f.d x.d y=182


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dx = x-A = x-5

dy = y-B/10 = (y-50)/10

r = f.d x.d y (f.d x*f.d y)/[nf .d x2-(f.d x)2*nf .d y2-(f.d y)2]

= 182 (54*-11) /51*516-(54)2*51*129-(-11)2]

= 776/[26316-2916*6579-121]

= 776/ [152.97*80.36]

= 776/12292

= 0.063

This means that there is negligible positive co-relation between time spent by the salesmen and unit sold by them and it, toa great extent, depends upon the skill of the salesman, used by him in the market, to serve his purpose.

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Checked on: February 09, 2010

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Mar.09, 2010 QT Problem No. 3: Transportation Nitin GuptaRoll No.

260

Sem: MBA-LLM II

A construction company provided its 2500 labours with housing facilities at 4 different places. Currently the company isengaged in construction projects at 5 different places, where these labours have to reach daily. The distance matrix (in km)between the place of residence and the place of project is given below.

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Project Place

Residence P1 P2 P3 P4 P5 Total

Residents

R1 15 12 3 5 2 800

R2 11 10 19 2 6 300

R3 4 9 7 11 16 500

R4 2 8 5 1 15 900

Total LaboursRequired 600 400 100 1200 200 2500


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Advice the company, as to how many labours, from which residential colony to allocate to each project, so that the labourshave to spent the minimum time in travelling. Also calculate the total minimum distance which the labours have to traveldaily.

This problem can be solved by using five methods. Although, all the methods were tried but the minimum result came outby using the Row Minimum Method. Hence, below is the solution of the problem by the use of Row Minimum Method.

3*100 + 5*500 + 2*200 + 2*300 + 4*500 + 2*100 + 8*400 + 1*400 = 9600

Therefore, the total minimum distance, which the labours have to travel daily, is 19200 km.

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Project PlaceResidence P1 P2 P3 P4 P5

TotalResidents

R1 100 500 200800 600

500 0

R2 300 300 0

R3 500 500 0

R4 100 400 400900 500

400 0

Total LaboursRequired 600 100 0 400 0 100 0

1200 700400 0 200 0 2500

155

12

3

1115

4

2

2

2

5

5

155

10

19

1

11

16

9

7

8

6


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The number of labours, which are required to be deployed from different residential colonies, should be equal to the numbercalculated and mentioned in the table above.

Checked on: March 09, 2010

Mar.13, 2010 QT Problem No. 4: Assignment Nitin Gupta

Roll No.260

Sem: MBA-LLM II

Aakash Deep Pvt. Ltd. ordered Rajdeep Ltd. for two consignments to be delivered at its warehouses, situated one in Delhiand another in Bhopal via e-mail. Unfortunately, the order, which was to be delivered at Delhi, was delivered at Bhopal andvice versa. Now Akash Deep has to hire four trucks, in order to get the consignment placed at the right place. It took 2 hoursto unload the truck and 3 hours to load a truck. So there is a limitation that next truck cannot move before 3 hours and the

truck cannot return back before 5 hours. There is a set schedule of the movement of the trucks between the two places, due to the traffic rules. The trucks cannotmove out of the time table. Below is the time table.

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Bhopal to Delhi

Fleet No. Departure Arrival

101 7:00 15:00

102 10:00 18:00

103 16:00 00:00

104 21:00 5:00

Truck owners charge the extra amount for waiting time. Advice Aakash Deep, whether to hire trucks from Delhi or Bhopal orfrom Delhi and Bhopal both; so that the waiting can be reduced.

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Delhi to Bhopal

Fleet No. Departure Arrival

1 6:00 14:00

2 12:00 20:003 15:00 23:00

4 22:00 6:00


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When Row Minimum action performed

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When Trucks are hired from Bhopal

101 102 103 104

1 15 12 6 25

2 21 18 12 7

3 24 21 15 10

4 7 28 12 17

When Trucks are hired from Delhi

101 102 103 104

1 17 20 26 7

2 11 14 20 25

3 8 11 17 22

4 25 28 10 15

When Trucks are hired from Delhi andBhopal

101 102 103 104

1 *15 *12 *6 7

2 11 14 *12 *7

3 8 11 *15 *10

4 *7 *28 10 15


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When Column Minimum action performed

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101 102 103 104

1 *9 *6 *0 1

2 4 7 *5 *0

3 0 3 *7 *24 *0 *21 3 8

101 102 103 104

1 *9 *3 *0 1

2 4 4 *5 *0

3 0 0 *7 *2

4 *0 *18 3 8

101 102 103 104

1 *9 *3 1

2 4 4 *5

3 0 *7 *2

4 *18 3 8

*0

*0

*0

0


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Therefore, by interpreting this matrix, it is advisable that Aakash Deep should hire three trucks from Bhopal and one fromDelhi.

The truck from Delhi should leave Delhi at 15:00 hrs and the same should return back from Bhopal at 10:00 hrs nextday. ( 3-102 )

The first truck from Bhopal should leave Bhopal at 7:00 hrs and the same should return back from Delhi same day at22:00 hrs. ( 101-4 )

The second truck from Bhopal should leave Bhopal at 16:00 hrs and the same should return back from Delhi next dayat 6:00 hrs. ( 103-1 )

The last truck from Bhopal should leave Bhopal at 21:00 hrs and the same should return back from Delhi next day at12:00 hrs. ( 104-2 )

The total waiting time would be 11 + 7 + 6 +7 = 31 hrs.


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Mar.18, 2010 QT Problem No. 5: Replacement Nitin Gupta

Roll No.260

Sem: MBA-LLM II

Bharat Chemicals Limited has been using the furnaces which can be used to carry out reactions at 1000 0C, since long time

and has the experience, which can be used to determine the cost of maintenance, which the furnace demands in every year

after its purchase. Now the company is planning to purchase a new furnace, costing around Rs.9,50,000/- and installation

charges would cost the company around Rs.50,000/-. The maintenance cost and the salvage value of the furnace after each

year is tabulated below. Advice the company that when should the company get the furnace replaced.

Year 1 2 3 4 5 6 7

Maintenance Cost 2000 4,000 8,000 16,000 32,000 64,000 128,000

Salvage Value 8,50,000 7,50,000 6,80,000 6,00,000 5,60,000 5,50,000 5,50,000

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Cost of furnace = 9,50,000 + 50,000 = Rs. 10,00,000/-

Year (n)

Maintenance

Cost (M t)

CumulativeMaintenance

Cost (M t)

Surplus Value

(S) (C-S)

T(n)

(C-S+M t)

A(n)

[(C-S+M t)/n]1 2,000 2,000 8,50,000 1,50,000 1,52,000 1,52,000

2 4,000 6,000 7,50,000 2,50,000 2,56,000 1,28,000

3 8,000 14,000 6,80,000 3,20,000 3,34,000 1,11,333.3

4 16,000 30,000 6,00,000 4,00,000 4,30,000 1,07,500

5 32,000 62,000 5,60,000 4,40,000 5,02,000 1,00,400

6 64,000 1,26,000 5,50,000 4,50,000 5,76,000 96,000*7 128,000 2,54,000 5,50,000 4,50,000 7,04,000 1,00,571.4

After making all this calculations, I would like to advice Bharat Chemicals Ltd., that they should replace the furnace after theend of six years.


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Mar.23, 2010 QT Problem No. 6: Linear Programming Problem Nitin Gupta(Solution by Graphical Method) Roll No.

260

Sem: MBA-LLM II

Titan India Limited designed a new kind of wrist watch, in which one can see the time as well as the date and month of the

day as well. Actually, there is pair of watch included in one. One of the watch functions, as any other watch performs, while

in another watch, the hour hand tells about the month, while the minute hand about the date of the day. Company also

came up with similar type of table calendar, but that does not consisted of time showing watch in it. Whereas, for the wrist

watch, the company requires two pairs of hands, the table calendar requires single pair of hands only. The watch requires

single button cell port installed in it, but as the calendar has installed in itself an extra feature that it glows in the dark,demands two ports of cell. However the company has only 1600 pairs of hands and 1200 button cell ports available with

them. The glasses, which are required for the wrist watch, are only 800, while the LED bulb required one for each calendar,

are only 1000 available with the company. If the company earns Rs.2000 profit on the wrist watch whereas the table

calendar earns a profit of Rs. 1500 for the company, advice the company about the number of wrist watches and calendars,

that it should manufacture, in order to earn maximum profit.

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Zmax = 2000X + 1500Y

S.T.:

2X + Y < 1600

X + 2Y < 1200

X < 800

Y < 1000

2X + Y = 1600 X + 2Y = 1200 X = 800 Y = 1000

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Wrist Watch (X) Table Calendar (Y)Profit 2000 1500

Pair of Hands 2 1 1600Port of Cells 1 2 1200

Glass 1 - 800LED Bulb - 1 1000

X 0 800

Y 1600 0

X 800 800

Y 0 800

X 0 1000

Y 1000 1000

X 0 1200

Y 600 0


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1600 (0,1600) Scale: 1cm = 100 units

1500

1400

1300

1200 (2X + Y = 1600)

1100

1000 (0,1000) (Y = 1000)

900

800 (X= 800)

700

600 (0,600)

500 (X + 2Y = 1200)

400

300 (667,266)

200

100 (800,0) (1200,0)

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600

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2X + Y = 1600

(X + 2Y = 1200).2

2X + 4Y = 2400

-3Y = -800

Y= 800/3

= 266.66 = 266

X = 1200-533

= 667

Zmax = 2000.667 + 1500.266 = 17,33,000

Therefore, if Titan manufactures 667 Watches and 266 Calendars, it will earn the maximum profit of Rs. 17,33,000


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Mar 30, 2010 QT Problem No. 7: Conditional Probability Nitin Gupta(Bayes Theorem) Roll

No. 260

Sem: MBA-LLM II

With the increase in temperature in the city, frequency of people, visiting the electronics shop for the purchase of Air-

Conditioners has increased multi-folds. Kunal is also one such visitor, who visited an electronic showroom for the purchase

of an Air-Conditioner for himself, where various types of Air-Conditioners viz. Split ACs, Window ACs, Slim (Floor standing)

ACs and Cassette ACs of various companies as per the following matrix were displayed:

At last, Kunal decided to purchase a Slim AC; but he is confused as to which company he should go for. Calculate the

probability of his, selecting the Slim AC of O-General for himself.

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Carrier (C) O-General (O-G) Hitachi (H) Blue Star (BS) Voltas (V)

Split AC (S) 6 4 3 3 6Window AC (W) 8 7 6 5 9

Slim AC (Sl) 3 2 3 2 2Cassette AC (C) 3 1 2 2 3


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P (C) = 1/5 P (O-G) = 1/5 P (H) = 1/5 P (BS) = 1/5 P (V) = 1/5

P (Sl/C) = 3/20 P (Sl/O-G) = 2/14 P (Sl/H) = 3/14 P (Sl/BS) = 2/12 P(Sl/V) = 2/20

P (O-G/Sl) = [P (O-G). P (Sl/O-G)]/[P (C). P (Sl/C) + P (O-G). P (Sl/O-G) + P (H). P (Sl/H) + P (BS). P (Sl/BS) + P (V). P(Sl/V)]

= [1/5*3/20]/[1/5*3/20 + 1/5*2/14+ 1/5*3/14 + 1/5*2/12 + 1/5*2/20]

= [1/5*3/20]/1/5[3/20 + 2/14 + 3/14 + 2/12 + 2/20]

= [3/20]/ [3/20 + 2/14 + 3/14 + 2/12 + 2/20]

= [3/20]/ [(63 + 60 + 90 + 70 + 42)/420]

= [3/20]/ [325/420] = [3]/ [325/21] = 63/325 = .0193846 = 19.38%

Therefore, there is the probability of 19.38% that Kunal will opt for the Slim AC of O-General.


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April 03, 2010 QT Problem No. 8: Poisson Distribution Nitin Gupta

Roll No.260

Sem: MBA-LLM II

Prashant, who was in hurry, purchased 10 packets of food stuffs for the fishes, lying in the aquarium at his place. He missed

to see that the expiry date stamp was not there on the packets. He served the fishes with the food and after two days found

few of the fishes dead. He approached to the shopkeeper, but the shop was closed and there he came to know that a raid

was conducted at his shop and 60% of the items at the shop were found expired.

Keeping the facts into consideration, find out the probability, by using poisson approximation to binomial distribution, that

out of the packets purchased by Prashant, 4 packets were expired.

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n = 10

p = 60%

= np = 0.6 * 10 = 6

r = 4

P(r) = ( re - )/r

= (6 4*0.0025)/4

= 3.24/24

= 0.135

= 13.5%

Therefore, it can be said that the probability of having 4 packets expired out of 10 packets, so purchased by Prashant, is

that of 13.5%

Checked on: April 03, 2010

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April 08, 2010 QT Problem No. 9: Normal Distribution Nitin Gupta

Roll No.260

Sem: MBA-LLM II

KRBL is in the market with its product Rice under various brand names. The product is available in the market in the

packing of 1kg, 5kg, 10 kg, 25 kg, and 50kg. When the company received the complaints from the market about the fault in

weights of the packets, weighing 50kgs, the company picked up the samples of 100 bags of rice at random. It was found

that however, the total weight of all the rice bags in sample is correct, as it should be and hence the mean is as per the

desire; but there is the standard deviation of 5kgs. Now, the company wants to determine the proportion of bags that are

weighing between 49kg to 52 kg assuming that the bags are normally distributed.

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n = 100

x = 50

= 5

x1 = 49

x2 = 52

z1 = (x 1 - x)/ = (49-50)/5 = -0.2

z2 = (x 2 - x)/ = (52-50)/5 = 0.4 0.2 0 0.4

Area ( z1 to z 2) = Area (-0.2 to 0) + Area (0 to 0.4)

= 0.0793 + 0.1554

= 0.2347

This implies that out of the sample of 100 bags picked by the company, minimum 23 and maximum 24 bags are

weighing in between 49 to 52 kg.


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April 13, 2010 QT Problem No. 10: Sampling Nitin Gupta

Roll No.260

Sem: MBA-LLM II

The Ministry of Human Resource Development of India conducted a survey upon the students of management at graduation

as well as that of at the post graduation level across the territory of India. The reason for conducting the survey was to

determine the level of difficulty as well as the interests of the students with respect to the Quantitative Techniques and

Business Statistics as a compulsory subject of management studies. A random sample of 8000 students was taken, out of

which 5000 students found the subject easy as well as interesting. If there are in total 45000 students of management in

India; calculate how many students find the subject easy and interesting.

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N = 45,000

n = 8000

n = 5000

p = n/n = 5000/8000 = 0.625

n/N = 8000/45000 = 0.177778 i.e. >0.05

S.E. (p) = [p(1-p)/n]

= [0.625(1-0.625)/8000]

= [0.625*0.375/8000]

= 0.234375/8000

= 0.005413

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C.I. = P + Z *S.E.(p)

= 0.625 + 3*0.005413

= 0.625 + 0.016239

= 0.608761 0.641239

By multiplying the above interval by 45000

= 27394.25 28855.76

This implies that out of 4500 students of management across the territory of India, 27394 to 28856 students find

the subject easy and interesting.


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qt assignment - nitin gupta (260)

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