Assignment 1 Masih

MULTIMEDIA UNIVERSITY

GIM

FACULTY OF MANAGEMENT

BQA 7074 – QUANTITATIVE ANALYSIS

ASSIGNMENT 1

STUDENT IDENTIFICATION NO

QUESTION (1)

Thompson Manufacturing make three components for sale to refrigeration companies. The components are processed on two machines: a shaper and a grinder. The times (in minutes) required on each machine are as follows:

Machine

Component Shaper Grinder

1 6 4

2 4 5

3 4 2

The shaper is available for 120 hours, and the grinder is available for 110 hours. No more than 200 units of component 3 can be sold, but up to 1000 units of each of the other components can be sold. The company already has orders for 600 units of component1 that must be satisfied. The profit contributions for component 1, 2, and 3 are RM 8, RM 6, and RM 9 respectively.

a) Formulate a linear programming model and solve for recommended production quantities.

Max 8X1 + 6X2 + 9X36X1 + 4X2 + 4X3 <= 7200 4X1 + 5X2 + 2X3 <= 6600X3 <= 200X2 <= 1000X1 <= 1000X1 >= 600X1,X2,X3 >= 0

Solution X1=600, X2=700, X=200Optimal sale 10800

b) What are the ranges of optimality for the profit contribution of three components? Interpret these ranges for company management

Components Lower bound Upper boundX1 -Infinity 9X2 5.3333 9X3 6 Infinity

c) What are the ranges of feasibility for the right-hand sides? Interpret these ranges for company management.

Constraint Upper bound Lower bound1 4400 74402 6300 Infinity3 100 9004 700 Infinity5 600 Infinity6 514.2857 1000

d) If more time could be made available on the grinder, how much would it be worth?

It cant make change because its dual is 0, so even if we spend 200000 time on it , it cant make any change on our sale benefit in optimal point

e) If more units of component 3 can be sold by reducing the sales price by RM 4, should the company reduce the price?

We are going to change Max X3 from 9 to 4 which makes change on final answer from 10800 to 10114.29.

QUESTION 2 A

The following table shows all necessary information on the availability of supply at each warehouse, the requirement of each market and the unit transportation cost in RM from each warehouse to each market.

Market

Warehouse

P Q R S Supply

A 6 3 5 4 22

B 5 9 2 7 15

C 5 7 8 6 8

Requirement 7 12 17 9 45

The shipping officer has worked out the following schedule from experience: 12 units from A to Q, 1 unit from A to R, 9 units from A to S, 15 units from B to R, 7 units from C to P and 1 unit from C to R.

a) What is the total cost of transportation for the shipping officer’s schedule?

The current shipment schedule would be:

Shipment list

From To Shipment Cost per unit Shipment cost

A Q 12 3 36

A R 1 5 5

A S 9 4 36

B R 15 2 30

C P 7 5 35

C R 1 8 8

Total Costs 150

So the total cost for the shipping officer’s schedule would be 150 RM.

b) Find the optimal schedule and minimum total cost of transportation

The optimal schedule and minimum cost of transportation, using minimum cost method:

Shipment list

From To Shipment Cost per unit Shipment cost

A Q 12 3 36

A R 2 5 10

A S 8 4 32

B R 15 2 30

C P 7 5 35

C S 1 6 6

Total Costs 149

Using this method we find out that the current shipping schedule could still be optimized by 1RM less.

c) If the officer is approached by a courier to route C to Q, who offers to reduce his rate in the hope of getting some business, by how much the rate should be reduced such that the officer will offer him some business?

If they want to approach some business from the shipping officer to their rout they need to reduce their price for sure, but if they could minus 3RM from their current price they could only get 1 shipment, means if they could reduce their price to 4RM they will get a deal of one shipment from their rout, but if they could manage to reduce their price to 2RM they could attract 7 more shipments to their rout which would be total of 8 shipments.

QUESTION 2 B

XYZ electronics supplies television components to various TV manufacturers. The past records indicate that a certain number of components assembled by the workers were returned because of product defects. The average number of defects produced by each worker per month is 82. Various component production jobs along with their defects produced by workers A,B,C,D,E, and F are presented in the table below:

Jobs

Worker1 2 3 4 5 6

A 23 29 15 31 21 14

B 31 25 17 27 31 23

C 15 23 19 24 22 31

D 19 17 26 15 13 23

E 26 19 15 17 17 29

F 24 25 11 15 29 21

Determine the optimal assignment of the workers to various jobs to minimize the total defects.

Create data for assignment →Number of jobs (6). Number of machines(6)

After that we enter data

Press solve→ Assignment list

A/6→14B/3→17C/1→15D/5→13E/2→19F/4→15Total→93

QUESTION 3 [ 12 Marks ]

Ariff Corporation has undertaken a project that involves twelve main activities. For easy reference, the activities are named A to L. The management team has identified the immediate predecessor activity/activities and the most likely time of completion for each activity. Besides the most likely time, the optimistic and pessimistic times for completion are also obtained for each activity. The information summary is given in the table below.

Activity ImmediateActivity

Time (weeks)Optimistic Most

LikelyPessimistic

A - 13 18 20B - 8 9 10C A 4 6 7D B 10 14 18E C 10 13 14F E 6 8 9G C 12 14 16H D, F, G 6 8 10I E 9 10 11J H 4 5 7K I, J 4 6 8L K 3 8 13

a) Draw a network diagram to represent the project.

Activity ImmediateActivity

Time (weeks) (b) 1MO M P t S2

A - 13 18 20 5 0.444B - 8 9 10 9 0.111C A 4 6 7 5.8 0.250D B 10 14 18 14 1.778E C 10 13 14 12.7 0.444F E 6 8 9 7.8 0.250G C 12 14 16 14 0.444H D, F, G 6 8 10 8 0.444I E 9 10 11 10 0.111J H 4 5 7 5.2 0.250K I, J 4 6 8 8 0.444L K 3 8 13 10 0.444

62.5 2.972 (c) 1M (e) 1M

b) Compute the expected time and the variance for each activity time.

Draw a network diagram to represent the project.

0,0

5,5

9,17.3

10.8,10.8

31.3,31.3

23.5,23.5

39.3,39.3

44.5,44.5 52.5,52.5

A(5)

B(9)

C(5.8)

D(14)

G(14) F(7.8)

H(8)

I(10)

K(8)

62.5,62.5

L(10)

E(12.7)

J(5.2)

c) Obtain the expected completion time for the project.

62.5 Daysd) Determine the critical path.

The critical path: A-C-E-F-H-J-K-Le) Calculate the variance in the expected completion time for the project.

f) Find the probability that the project will be completed within 60 weeks.

P[T<60] = P[Z<-1.4501] = 0.0735

(Total: 12 marks)

QUESTION 4

Small business owners often look to payroll service companies to handle their employee payroll. Reasons are that small business owners face complicated tax regulations and penalties for employment tax errors are costly. According to Internal Revenue Service, 26% of all small business employment tax returns contained errors that resulted in a tax penalty to the owner. The tax penalty for a sample of 30 small business owners are as follows:

820 270 450 1010 890 700 1350 350 300 1200

390 730 2040 230 640 350 420 270 370 620

740 350 560 790 820 940 360 290 1200 680

a) At 0.05 level of significance, is there evidence that the population mean tax penalty is greater than RM 670?

One-Sample Test

Test Value = 670

t df Sig. (2-tailed) Mean Difference

95% Confidence Interval of the

Difference

Lower Upper

TAx .014 29 .989 1.000 -149.32 151.32

No the population mean is not greater than 670

b) Interpret the meaning of p value in (a)

The p value is greater than 0.05, therefore, we accept the null hypothesis that the population mean is equal to 670

c) Estimate the 95% confidence interval of the mean difference and interpret its meaning

The 95% confidence interval of the difference is – 149.32 and 151.32. The lower bound is 520.68 and the upper bound is 821.32. Zero falls between -149.32 and 151.32 indicating that we can’t reject the null hypothesis

d) Is the highest penalty , RM 2040 , an outlier?

The Box and Whisker plot shows that the highest value (13) 2040 is an outlier

e) Compare the results obtained in (b) and (c)

The results obtained in (b) and (c) are the same

f) Test the assumption required to perform the test

Tests of Normality

Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

TAx .142 30 .127 .862 30 .001

a. Lilliefors Significance Correction

Shapiro-Wilk test results show that data is not normally distributed

QUESTION 5

The college Board provided comparisons of Scholastic Aptitude Test (SAT) scores based on the highest level of education attained by the test taker’s parents. A research hypothesis was that students whose parents had attained a higher level of education would on average score higher on the SAT. During 2008, the overall mean SAT verbal score was 507. SAT verbal score for independent samples of students are as given below. The first sample shows the SAT verbal test scores for students whose parents are college graduates with a bachelor’s degree. The second sample shows the SAT verbal test scores for students whose parents are high school graduates but do not have a college degree.

Student’s Parents

College Grads High School Grads

485

534

650

554

550

572

497

592

487

533

526

410

515

578

448

469

442

580

479

486

528

524

492

478

425

485

390

535

a) Is there sufficient evidence at 5% significance level that students show higher population mean verbal score on the SAT if their parents attained a higher level of education.

M

Equal variances

assumed

Equal variances

not assumed

Levene's Test for Equality of

Variances

F .365

Sig. .551

t-test for Equality of Means t 1.767 1.804

df 26 25.340

Sig. (2-tailed) .089 .083

Mean Difference 38.000 38.000

Std. Error Difference 21.501 21.067

95% Confidence Interval of

the Difference

Lower -6.195 -5.359

Upper 82.195 81.359

There is no evidence that parents education level influenced students’ SAT scores

b) Estimate with 95% confidence the difference in mean marks obtained by two categories of students and interpret its meaning

95% confidence interval of the difference is -6.195 and 82.195. Since the range contains 0 we accept the null hypothesis that there is no difference in SAT scores between two group of students

c) Test the assumptions required to perform the test

a) Levene’s test for equality of variance shows ( p = 0.551) the variance of the two groups are the same. The data is normally distributed

Tests of Normality



M .085 28 .200* .986 28 .964


*. This is a lower bound of the true significance.

d) Compare the results obtained in (a) and (b)

Both the tests show that there is no difference in SAT marks between two groups.

QUESTION 6 [ 14 Marks ]

An amusement park studied methods for decreasing the waiting time ( minutes) for rides by loading and unloading riders more efficiently. Two alternative loading/unloading methods have been proposed. To account for potential differences due to the type of ride and the possible interaction between the method of loading and unloading and the type ride an experiment was designed. The results of the experiment are presented in the table below.

Type of ride

MethodRoller Coaster Screaming

DemonLog Flume

1 31 40 38

33 36 37

229 30 30

33 32 28

352 48 48

44 44 49

a) Test if there is a difference in loading and unloading time due to the method at 5% significance level .

b) Test if there is a difference in loading and unloading time due to the type of ride at 5% significance level

Tests of Normality



Time .178 18 .135 .912 18 .093


Levene's Test of Equality of Error Variancesa

Dependent Variable:Time

F df1 df2 Sig.

. 8 9 .

Tests the null hypothesis that the error variance of

the dependent variable is equal across groups.

a. Design: Intercept + TR + Method + TR *

Method

Tests of Between-Subjects Effects

Dependent Variable:Time

Source

Type III Sum of

Squares df Mean Square F Sig.

Corrected Model 978.778a 8 122.347 17.478 .000

Intercept 25840.222 1 25840.222 3691.460 .000

TR 7.111 2 3.556 .508 .618

Method 922.111 2 461.056 65.865 .000

TR * Method 49.556 4 12.389 1.770 .219

Error 63.000 9 7.000

Total 26882.000 18

Corrected Total 1041.778 17

a. R Squared = .940 (Adjusted R Squared = .886)

c) Perform the post hoc test and explain the results

Multiple Comparisons

(I) TR (J) TR

Mean Difference

(I-J) Std. Error Sig.

95% Confidence Interval

Lower Bound Upper Bound

rc sd -1.33 1.528 .670 -5.60 2.93

LF -1.33 1.528 .670 -5.60 2.93

sd rc 1.33 1.528 .670 -2.93 5.60

LF .00 1.528 1.000 -4.26 4.26

LF rc 1.33 1.528 .670 -2.93 5.60

sd .00 1.528 1.000 -4.26 4.26

(I)

Method

(J)

Method

Mean Difference

(I-J) Std. Error Sig.

95% Confidence Interval

Lower Bound Upper Bound

M1 M2 5.50* 1.528 .014 1.24 9.76

M3 -11.67* 1.528 .000 -15.93 -7.40

M2 M1 -5.50* 1.528 .014 -9.76 -1.24

M3 -17.17* 1.528 .000 -21.43 -12.90

M3 M1 11.67* 1.528 .000 7.40 15.93

M2 17.17* 1.528 .000 12.90 21.43

d) Test the assumption required to perform the test

e) From the plot determine whether there is an interaction between type of ride and method of loading and unloading

QUESTION 7

The owner of Showtime Movie Theaters, Inc., would like to estimate weekly gross revenue as a function of advertising expenditures. Historical data for a sample of eight weeks is as follows:

Weekly Gross Revenue( RM 1000)

TV Advertising( RM 1000)

Newspaper Advertising( RM 1000)

96 5.0 1.590 2.0 2.092 4.0 1.595 2.5 2.594 3.0 3.394 3.5 2.395 2.5 4.294 3.0 2.5

a) Develop an estimated regression equation with amount of TV advertising as the independent variable

Regression

Descriptive Statistics

MeanStd.

Deviation N

Revenue 93.7500 1.90863 8

TV Advertising

3.1875 .96130 8

Correlations

RevenueTV

Advertising

Pearson Correlation Revenue 1.000 .419

TV Advertising

.419 1.000

Sig. (1-tailed) Revenue . .151

TV Advertising

.151 .

N Revenue 8 8

TV Advertising

8 8

Variables Entered/Removed

ModelVariables Entered

Variables Removed Method

1 TV Advertising

. Enter

a. All requested variables entered.

b. Dependent Variable: Revenue

Model Summary

Model R R SquareAdjusted R

SquareStd. Error of the Estimate

Durbin-Watson

1 .419a .175 .038 1.87233 1.806

a. Predictors: (Constant), TV Advertising


ANOVA

ModelSum of Squares df Mean Square F Sig.

1 Regression 4.466 1 4.466 1.274 .302a

Residual 21.034 6 3.506

Total 25.500 7

a. Predictors: (Constant), TV Advertising


Coefficients

Model

Unstandardized Coefficients

Standardized Coefficients

t Sig.B Std. Error Beta

1 (Constant) 91.101 2.438 37.366 .000

TV Advertising

.831 .736 .419 1.129 .302

a. Dependent Variable: Revenue

Coefficients

Model



t Sig.B Std. Error Beta

1 (Constant) 91.101 2.438 37.366 .000

TV Advertising

.831 .736 .419 1.129 .302

Residuals Statistics

Minimum Maximum MeanStd.

Deviation N

Predicted Value 92.7633 95.2560 93.7500 .79877 8

Residual -2.76328 1.82126 .00000 1.73344 8

Std. Predicted Value

-1.235 1.885 .000 1.000 8

Std. Residual -1.476 .973 .000 .926 8


Correlations

RevenueTV

Advertising

Revenue Pearson Correlation 1 .419

Sig. (2-tailed) .302

N 8 8

TV Advertising

Pearson Correlation .419 1


N 8 8

Revenue TV Advertising

Revenue Pearson Correlation 1 .419


N 8 8

TV Advertising

Pearson Correlation .419 1


N 8 8

Correlations

RevenueTV

Advertising

Spearman's rho Revenue Correlation Coefficient

1.000 .236

Sig. (2-tailed) . .574

N 8 8

TV Advertising

Correlation Coefficient

.236 1.000

Sig. (2-tailed) .574 .

N 8 8

b) Develop an estimated regression equation with amount of TV advertising and News paper advertising as the independent variables

Regression

Descriptive Statistics

Mean Std. Deviation N

Revenue 93.7500 1.90863 8

TV Advertising 3.1875 .96130 8

Newspaper Advertising 2.4750 .91144 8

Correlations

Revenue TV AdvertisingNewspaper Advertising

Pearson Correlation Revenue 1.000 .419 .300

TV Advertising .419 1.000 -.556

Newspaper Advertising .300 -.556 1.000

Sig. (1-tailed) Revenue . .151 .235

TV Advertising .151 . .076

Newspaper Advertising .235 .076 .

N Revenue 8 8 8

TV Advertising 8 8 8

Newspaper Advertising 8 8 8

Variables Entered/Removed

Model Variables Entered Variables Removed Method

1 Newspaper Advertising, TV Advertising

. Enter

a. All requested variables entered.

Model Summary

Model RR

SquareAdjusted R

SquareStd. Error of the

Estimate

Change Statistics

R Square Change

F Change df1 df2

Sig. F Change

1 .766a .586 .420 1.45307 .586 3.539 2 5 .110

a. Predictors: (Constant), Newspaper Advertising, TV Advertising


ANOVA

Model Sum of Squares df Mean Square F Sig.

1 Regression 14.943 2 7.471 3.539 .110a

Residual 10.557 5 2.111

Total 25.500 7

a. Predictors: (Constant), Newspaper Advertising, TV Advertising


Coefficients

Model



t Sig.

Collinearity Statistics

B Std. Error Beta Tolerance VIF

1 (Constant) 84.387 3.559 23.711 .000

TV Advertising 1.683 .688 .848 2.448 .058 .690 1.448

Newspaper Advertising

1.615 .725 .771 2.228 .076 .690 1.448


Co linearity Diagnostics

ModelDimension Eigenvalue Condition Index

Variance Proportions

(Constant) TV AdvertisingNewspaper Advertising

1 1 2.846 1.000 .00 .01 .01

2 .140 4.510 .00 .16 .29

3 .014 14.494 1.00 .83 .71


Residuals Statistics

Minimum Maximum Mean Std. Deviation N

Predicted Value 90.9840 95.3794 93.7500 1.46106 8

Residual -1.54250 2.36675 .00000 1.22807 8

Std. Predicted Value -1.893 1.115 .000 1.000 8

Std. Residual -1.062 1.629 .000 .845 8


Correlations

RevenueTV

AdvertisingNewspaper Advertising

Revenue Pearson Correlation 1 .419 .300

Sig. (2-tailed) .302 .471

N 8 8 8

TV Advertising Pearson Correlation .419 1 -.556

Sig. (2-tailed) .302 .152

N 8 8 8

Newspaper Advertising

Pearson Correlation .300 -.556 1

Sig. (2-tailed) .471 .152

N 8 8 8

c) Is the estimated regression equation coefficient for TV advertising expenditures the same in part (a) and in part (b) ? Interpret the coefficient in each case.

The estimated regression for TV advertising is the same in both part (a) and (b) r=0.419

pvalue=0.302In part (a)

The Pearson’s Correlation Coefficient, r = 0.419, with p-value = 0.302 > 0.05.

The Spearman’s Correlation Coefficient, r = 0.236, with p-value = 0.574 > 0.05.

Based on Spearman’s and Pearson’s there exist a positive relationship between revenue and TV advertising in part (a)

That is, as TV advertising increases the revenue also increases.

In part (b)

Based on Pearson’s there exist a positive relationship between revenue, Newspaper advertising and TV advertising in part (b)

That is, as TV advertising and newspaper advertising increase the revenue also increases.

d) Interpret the value of R2 in each case.

As it can be seen from Model Summary table in part (a) the R-Square is 0.175 and in part (b) is 0.586, it means that in part (a) 17.5 percent of revenue is from TV advertisement and in part (b) 58.6 percent of revenues is from Newspaper advertisement.

e) What is the estimate of the weekly gross revenue for a week when RM 3500 is spent on TV advertising and RM 1800 spend on newspaper advertising?

Assignment 1 Masih

Documents

shaper

sold

grinder

components