Spm Journal

MCT’s

Rajiv Gandhi Institute of Institute of Technology Versova, Andheri(W), Mumbai

Assignment 01

1. Develop a Network diagram using AON technique and calculate Critical path using following table 2. Enter the information in the table into MS project. What is critical path. Display the network diagram and

Critical path

Activity Estimated Duration

(days) Predecessor

A 1 None

B 3 A

C 4 B

D 2 B

E 1 C

F 3 C,D

G 3 E

H 1 F

I 2 G,H

J 5 I

Theory :

The expected project duration, 푇 , is determined by finding the longest path through the network. The longest path from the project start node to the end node is called the critical path and the activities that comprise it are called critical activities.

To reduce the project completion time, any reduction effort must be directed at the critical path.

A project can also have multiple critical path, to reduce project completion time in such situation one have to reduce time on critical activities in each of the critical path.

MCT’s


Solution 1:

Figure: AON diagram

Paths 푻풆 A-B-C-E-G-I-J 19 A-B-C-F-H-I-J 19 A-B-D-F-H-I-J 17

MCT’s


Solution 2:

Figure: AON diagram

Conclusion : Thus, we have found the critical path for the given predecessor table.

MCT’s


Assignment 02

1.Using the information below, complete the function point analysis in order to use the basic COCOMO model to estimate the duration of the number of people needed to develop application using C++. Assume that the project is relatively simple and straight forward and that the team is familiar with both the problem and technology. You can use perform the calculation by hands

Complexity

Files Low Average High Total

ILF __*7= __ __ *10= __ __ *15= __

EIF __*5=__ __*7=__ __*10=__

EI __*3=__ __*4=__ __*6=__

EO __*4=__ __*5=__ __*7=__

EQ __*3=__ __*4=__ __*6=__

Complexity

Low average High

ILF 4 2 0

EIF 0 1 0

EI 3 2 0

EO 5 7 3

EQ 2 5 2

Language Average source LOC per function point

Basic 107

C 128

C++ 53

COBOL 107

Delphi 29

Java 53

VB5 29

MCT’s


General System Chracteristics Degree of Influence

Data Communications 2

Distributed data Processing 3

Performence 3

Heavily used Configuration 4

Transaction rate 4

Online data entry 2

End user efficiency 2

Online update 2

Complex processing 2

Reusabilty 3

Installation ease 2

Operational ease 2

Multiple site 1

Faclitate change 1

Solution :

Since is is a simple and straight forward project, it is an ORGANIC project.

Complexity

Files Low Average High Total

ILF 4*7= 28 2*10= 20 0 *15= 0 48

EIF 0*5=0 1*7=7 0*10=0 7

EI 3*3=9 2*4=8 0*6=0 17

EO 5*4=20 7*5=35 3*7=21 76

EQ 2*3=6__ _5*4=20 __2*6=12_ 38 ∑ = 186

MCT’s


UAF =186

TDI =∑ 퐷푒푔푟푒푒 표푓 퐼푛푓푙푢푒푛푐푒 =33

VAF = (TDI X 0.01) +0.65

=(33 X 0.01) + 0.65

= 0.33 + 0.65

= 0.98

FP = UAF X VAF

= 186 X 0.98

= 182.28

Average source LOC per FP = 182.28 x 53

= 9660.84 ≈ 9661.

KDSI = Thousand of delivered source instructions

= 9.661

Person-Months = 2.4 x 9.661 .

= 25.9707

Duration = 2.5 x 25.9707 .

= 7.8165

People Required = 푬푭푭푶푹푻

푫푼푹푨푻푰푶푵 =

..

= 3.322 3 or 4 people.

MCT’s


Assignment 03

Using the following information, conduct an earned value analysis to determine whether you believe the project will remain within its original planned budget of $12,000. Use a spreadsheet software package to determine the Estimate at Completion (EAC) for both typical and atypical variances.

Planned

Value

Actual

Cost

Percent

Complete

Earned

Value

Phase 1: Test Planning

Develop Unit Test Plan $ 500 $ 500 100% $ 500

Develop integration test plan $ 600 $ 600 100% $ 600

Develop Acceptance test plan $ 550 $ 550 100% $ 550

Payment 1: $ 1,650 $ 1,650 $ 1,650

Phase 2: Unit Testing

Code walkthrough with team $ 600 $ 700 100% $ 600

Test software units $ 800 $ 1,200 100% $ 800

Identify programs that do not meet specifications $ 500 $ 500 100% $ 500

Modify code $ 1,200 $ 1,300 100% $ 1,200

Re-test units $ 300 $ 400 100% $ 300

Verify code meets standards $ 200 $ 200 100% $ 200

Payment 2: $ 3,600 $ 4,300 $ 3,600

Phase 3: Integration Testing

Test integration of all modules $ 600 $ 600 100% $ 600

Identify components that do not meet specifications $ 300 $ 500 75% $ 225

Modify code $ 1,400 $ 1,400 50% $ 700

Re-test integration of modules $ 400 $ 400 75% $ 300

Verify Components meet standards $ 200 $ 200 80% $ 160

Payment 3: $ 2,900 $ 3,100 $ 1,985

MCT’s


Phase 4: Acceptance Testing

Business Review with Client $ 400 0%

Client tests system $ 800 0%

Identify units and components that do not meet specifications

$ 550 0%

Modify code $ 800 0%

Retest units and components $ 700 0%

Verify that system meets standards $ 600 0%

Payment 4: $ 3,850

ANS:

The first phase of testing went as planned. The second phase has encountered some problems and the second instalment is more than planned. Therefore, analysis has to be done for the next phase.

Cumulative CPI = EV/AC = 7235/9050 = 0.799

Atypical variance:

Expected Time Complete (ETC) = (BAC – Cumulative EV to date)

= $12000-$7235

= $4765

Estimate at completion (EAC) = Cumulative AC + (BAC – Cumulative EV to date)

= $9050+$12000-$7235

= $13815

Typical variance:

Expected Time Complete (ETC) = (BAC – Cumulative EV to date) / Cumulative CPI

= ($12000-$7235)/0.799

= $5963.7

Estimate at completion (EAC)

= Cumulative AC+(BAC–Cumulative EV to date)/Cumulative CPI

= $9050 + $5936.7

= $15013.5

Final cost of the project will be somewhere between $13,815 and $15,013.5

MCT’s


Assignment 5

Case Study: Critical Path Analysis

This Activity is based around a decision by a food manufacturing business to install a new piece of machinery to cope with anticipated demand leading up to a busy time of the year. The machinery allows pre-packed food to be manufactured with 90% automation compared to the existing machinery which relied far more on human input, making the whole process only 40% automated.

In previous years, the business has always suffered shortages as a result of the inadequacy of the equipment they have had and this has caused some dissatisfaction amongst its customers. While most have remained loyal, there have been one or two who have taken their business elsewhere and this has prompted the business to act The Activity requires two groups to analyse competing pieces of machinery. Both machines will do the job that they are required to do, both have the same output ratios and initial cost but the crucial thing for the business is how quickly the machines can be installed and begin production.

You must use Critical Path Analysis to advise the business on the most appropriate piece of machinery out of the two.

You must consider the correct order for each task, draw out the network for the machine in question, and indicate the earliest start time, latest finishing time and the critical path. All the tasks below are those that need to be completed by the firm themselves - the new machine construction, for example, refers to the work that needs to be done following delivery of the equipment.

Machine A Task Activity Estimated Time

E Re-decoration of area 5 days B Lead time for new machine delivery 14 days C Re-routing of electrical supply 2 days H Delivery and inspection of new machine 1 day F Training of staff 2 days Program software 4 days

G Lead time for delivery of raw materials 5 days K Testing of new machine 2 days D Install and test network cabling for

computerized machine 3 days

L Complete hygiene treatment 1 day A Remove existing machinery 4 days J New machine construction 2 days

MCT’s


MCT’s


Machine B Task Activity Estimated Time

E Re-decoration of area 5 days B Lead time for new machine delivery 16 days C Re-routing of electrical supply 3 days H Delivery and inspection of new machine 1 day F Training of staff 3 days Program software 5 days

G Lead time for delivery of raw materials 3 days K Testing of new machine 2 days D Install and test network cabling for

computerized machine 2 days

L Complete hygiene treatment 1 day A Remove existing machinery 4 days J New machine construction 1 days

Machine A : critical Path A-C-D-E-L-F-I-J-K =25

Machine B : critical Path A-C-D-E-L-F-I-J-K =26

Conclusion : The manufacturing business should consider machine A.

MCT’s


MCT’s


Assignment 6

The activity information of a project is given in the table below:

Activity Predecessor Normal Time

(DAYS)

Normal Cost ($)

Crash Time (Days)

Crash Cost ($)

A - 7 1600 5 2600 B - 2 850 1 1000 C - 9 2500 7 3000 D B 10 2000 10 2000 E C 3 2000 2 2450 F A,B 9 950 6 2050 G D,E 13 1300 10 3000 H F,D 8 1000 6 1600

Identify the critical path in the project.

Determine the normal project completion time and cost. Calculate the cost for achieving a project completion time of 22 days. Determine which activities to crash and by how much. Determine what the minimum project completion time that can be achieved.

ANS:

F , 9

H , 8 END

G , 13 E , 3 C , 9

D, 10 B , 2

A , 7

START

MCT’s


Paths:

A-F-H 24

B-F-H 19

C-E-G 25*

B-D-H 20

B-D-G 25*

*Critical Path

Project completion time = 25 weeks Project completion cost = 12200 Cost Slope = Cc – Cn / Tc – Tn Crash activity G by 3 weeks. Therefore, Cost of completion of G = 566.67 * 3 = 1700 Total Cost = 12200 + 1700 = 13900 For minimum project completion time:-

Crash activity A by 2 weeks, F by 2 weeks and G by 2 weeks.

Total cost = 17700- 2866.68 = 14833.32

MCT’s


Assignment 7

1. A project consisting of 12 distinct activities is to be analysed by using PERT. Draw the PERT network. Indicate the expected total slack for each activity and hence indicate the average critical path. Within how many days would you expect the project to be completed with 99% chances? The following information is given ( the time estimates are in days)

ACTIVITY PREDECESSOR

ACTIVITY OPTIMISTIC TIME(days)

NORMAL TIME (days)

PESSIMISTIC TIME (days)

A - 2 2 2 B - 1 3 7 C A 4 7 8 D A 3 5 7 E B 2 6 9 F B 5 9 11 G C,D 3 6 8 H E 2 6 9 I C,D 3 5 8 J G,H 1 3 4 K F 4 8 11 L J,K 2 5 7

te = (a + 4m + b) / 6

V = [(b-a)/6]2

Paths te V A-C-G-J-L 22.16 2.074

A-C-I 13.83 1.13 A-D-G-J-L 20.49 2.07

A-D-I 12.16 1.13 B-E-H-J-L 39.32 4.66 B-F-K-L Te = 49.33 VP = 4.054

Z= (TS - Te) / √ VP = 2.3 = TS - 41.33 / √4.054 = 45.9 = 46

Hence, the project will be completed in 46 days with 99% chances.

MCT’s


2. Given the following information, is this project in trouble? Explain.

Task BCWS BCWP ACWP A $ 384.62 $ 384.62 $ 384.62 B $ 576.92 $ 576.92 $ 576.92 C $1,461.54 $1,461.54 $1,096.15

Total $2,423.08 $2,423.08 $2,057.69

Solution :

Cost Variance (CV) = EV – AC = 1461.54 – 1096.15 = 365.39 A positive CV indicates that the activity is under budget. Schedule variance(SV) = EV – PV = 1461.54 – 1461.54 = 0 Activity is right on schedule. WHOLE PROJECT CV = 2423.08 – 2057.69 = 365.48 A positive CV indicated that the project is under budget SV = 2423.08 – 2423.08 = 0 The project is right on schedule. Conclusion : The project is on schedule and under budget, therefore it is not in trouble.

MCT’s


3. Three proposals (W,X and Y) have been rated on six criteria as follows : 1-poor 2- average 3- good Choose between the three proposals

A. By simple rating method B. By weighted rating method

Criteria weight W X Y Attention to

quality .25 2 1 3

Cost .2 3 3 1 Project Plan .2 2 2 1

Project Organization

.15 3 2 3

Likelihood of success

.1 2 3 3

Contractor Credentials

.1 2 2 3

Having received proposals from multiple contractors, the customers must now review and evaluate them. Choosing the best, reaching an agreement with the contractor, and committing funds are all part of the “project selection” process.

Among the many approaches for selecting the best proposals, one is a prescreening method that rejects proposals that fail to meet minimal requirements such as a too-high price tag, a too-low rate of return, or a contractor with insufficient experience.

Proposals that survive pre screening are subjected to closer scrutiny; a common method employs a checklist to rate proposals according to several evaluation criteria.

MCT’s


Simple Rating Method :

Each proposal is reviewed and given a score 푠 for each criterion j. the overall score for the proposal is the sum of

the scores for all criteria,

S = ∑푆 , where j = 1,2,…,n

The proposal that receives the highest overall score wins. This method is called simple rating.

Criteria W X Y Attention to

quality 2 1 3

Cost 3 3 1 Project Plan 2 2 1


3 2 3


2 3 3


2 2 3

Total 14 13 14

According to the sum of simple ratings, W and Y were rated the best.

Weighted rating :

The limitation of simple rating is that all the evaluation criteria are treated equally as equally important. When some criteria are clearly more important than others, a method called weighted rating is used instead wherein the relative importance of each criterion j is indicated with an assigned weight 푤 . After a given criterion has

been scored, the score is multiplied by the weight of the criterion,

푠 . 푤 , The overall score for the proposal is the sum of the 푠 . 푤 for all criteria,

S = ∑ 푠 .푤 , where j = 1,2,…,n

푤 = 1 푎푛푑 0 ≤ 푤 ≤ 1.0

All the three proposals were considered too close to make an objective decision. Therefore all three are considered for weighted ratings.

MCT’s


Criteria weight Attention to

quality .25

Cost .2 Project Plan .2


.15


.1


.1

Criteria Weight W X Y (w) (s) (w)(s) (s) (w)(s) (s) (w)(s)

Attention to quality

.25 2 0.5 1 0.25 3 0.75

Cost .2 3 0.6 3 0.6 1 0.2 Project Plan .2 2 0.4 2 0.4 1 0.2


.15 3 0.45 2 0.3 3 0.45


.1 2 0.2 3 0.3 3 0.3


.1 2 0.2 2 0.2 3 0.3

Total 1 2.35 2.05 2.2

Using the sum of the weighted ratings, the proposal W is superior to other proposals.

Spm Journal

Documents

Transcript of Spm Journal