# MGT613 - Production / Operations Management - Lecture Handout 44

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# PROJECT MANAGEMENT-I

## Learning Objectives

After learning about the network diagrams, the project life cycle and the responsibilities of project manager. We will now learn the important concept of time estimates (which is based on computing algorithms of Early Start, Early Finish, Late Start and Late Finish) and variances which are used to control the project activities. We will consider important aspects like the forward and backward path time estimates, Project Crashing, Time Cost Trade Offs, Project Management Software, Risk Management and develop a project management based Operations Strategy.

## Time Estimates

There are two common types of time estimates namely

1. Deterministic: Time estimates that are fairly certain
2. Probabilistic: Estimates of times that allow for variation

## Example: Hospital

We take the same hospital example and now place the time dimension to it .

The activities from locating the facility to making the hospital fully are represented in the form of a network diagram. The student should try to write down the activities along with the activity description then try to draw the network diagram using both the activity on node and activity on arrow as practice.

## Computing Algorithm

Network activities

1. ES: early start
2. EF: early finish
3. LS: late start
4. LF: late finish

Used to determine

1. Expected project duration
2. Slack time
3. Critical path

## Probabilistic Time Estimates

1. Optimistic time : Time required under optimal conditions
2. Pessimistic time: Time required under worst conditions
3. Most likely time: Most probable length of time that will be required

where

te = expected time
to = optimistic time
tm = most likely time
tp = pessimistic time

## Variance

The word variance reflects the square of standard deviation of activities on a path and represented by σ2. The size of variance reflects the degree of uncertainty associated with activity’s time, the larger the variance the larger the uncertainty.

## Example

Z indicates how many standard deviations of the path distribution the specified tine is beyond the expected path duration. If the value of “z” is +2.50 or more, treat the probability of path completion by the specified time as 100 percent.

Crash is the shortening activity duration
Procedure for crashing

• Crash the project one period at a time
• Only an activity on the critical path
• Crash the least expensive activity

Multiple critical paths: find the sum of crashing the least expensive activity on each critical path

## Project Crashing

Crashing a project involves paying more money to complete a project more quickly.
Since the critical path determines the length of a project, it makes sense to reduce the length of activities on the critical path.
Critical Path activities should be reduced until the project is reduced to the desired length or you are paying more per day than you save.
If you have multiple Critical Paths, they should be shortened simultaneously.

Example
The manager of a PHA is about to undertake a reforestation project throughout Pakistan. He is first asked to carryout a pilot project. The project will involve the following six activities:

 SR. # ACTIVITY PRECEEDS TIME ESITIMATES ( DAYS) OPTIMISTIC “ a” MOST LIKELY“ m” PESSIMISTIC “ b” START U,V U W 35 50 65 V W,X 28 40 52 W Z 26 35 44 X Y 28 40 52 Y Z 26 29 38 Z END 36 60 84

Solution: First of all, we construct network diagram based on Activity on Node followed by calculating the probabilistic time “t” and standard deviation “σ” using the formulas given below and then the ES,EF and LS, LF using the forward pass (progression) and backward pass (progression) respectively.
t = (a+4m+b)/6 and σ = (b-a)/6

The denominator of “6” reflects the concept of area under the curve that the range of data lies to + 3 Standard Deviations from mean also it shows the weighted average.

I. Time “t” = (a+4m+b)/6
Activity U = (35+4(50) +65)/6= (100+200)/6= 300/6= 50 days
Activity V = (28+4(40) +52)/6= (80+160)/6= 240/6= 40 days
Activity W = (26+4(35) +44)/6= (70+140)/6= 210/6= 35 days
Activity X = (28+4(40) +52)/6= (80+160)/6= 240/6= 40 days
Activity Y = (26+4(29) +38)/6= (64+116)/6= 180/6= 30 days
Activity Z = (36+4(60) +84)/6= (120+240)/6= 360/6= 60 days

### Standard Deviation “σ” = (b-a)/6

Activity U = (65-35)/6= (30)/6= 5 days
Activity V = (52-28)/6= (24)/6= 4 days
Activity W = (44-26)/6= (18)/6= 3 days
Activity X = (52-28)/6= (24)/6= 4 days
Activity Y = (38-26)/6= (12)/6= 2 days
Activity Z = (84-36)/6= (48)/6= 8 days

### Critical Path

The critical path is the longest path taken for the project to complete.
From Start to End there are three possible paths as from the Network Diagram
Start –U-W-Z-End = 50 + 35+60 = 145 days (logically incorrect)
Start-V-X-Y-Z-End= 40+40+30+60 = 170 days
Start-V-W-Z-End= 40+35+60 = 135 days (logically incorrect)
For the Critical Path, we also calculate the standard deviation of Project portfolio
Start-V-X-Y-Z-End=σ2= [(4)2+(4)2+(2)2+(8)2]= (16+16+4+64)= (100 )
Then σ = Square Root ( 100) = 10 days
Also individual sum of standard deviations = 4+4+2+8 = 18 days

Since portfolio project σ = 10 days is less than individual sum of 18 days, it shows our value of portfolio σ is correct
Normal Distribution
“z” = (X-μ)/σ , now since X= 200 days μ = 170 days,
σ= 10 days

Also Using the Normal Probability Curve

Z= (X – μ)/σ
= (200-170)/10= 30/10= 3.0
According to the standard normal table, the area at z = 3 is 0.4987. Adding 0.5 for left hand side of the standard normal curve, we get 0.9987.

Q.3: What is the estimated expected (mean) time for Project Completion?
135 days
145 days
170 days
180 days
255 days

### The answer is 170 Days (Choice C)

Q.4: What is the estimated slack time for activity W?
0 days
25 days
35 days
45 days
85 days
The answer is 25 Days (Choice B)

Q.5: What is the probability that the critical path for this project will be completed with in 200 days?
0.8413
0.9544
0.9772
0.9974
0.9987
Based on the calculations of critical path “σ” above, the answer comes out to be 0.9987 (Choice E).

Given the portion of the network shown above, what is the earliest finish time for activity 10-11, if the earliest start time of 8-10 is “12” and the earliest start time of 9-10 is “13”?
22
23
24
25
26

Q.1: What is the estimated expected (mean) time for activity Y?
30 days
29 days
38 days
26 days
35 days
The answer is 30 Days (Choice A)

 ACTIVITY FORWARD BACKWARD SLACK t ES EF LS LF n START 8 to 10 4 12 16 12 16 0 9 to 10 2 13 15 14 16 1 10 to 11 7 16 23 16 23 0

Q.2: What is the estimated standard deviation in the time for activity Z?
3 days
2 days
4 days
8 days
5 days
The answer is 8 Days (Choice D)

## Solved Examples

You have been hired as the Chief Project Manager, by your city’s Kabbadi Association for construction, renovation and repairs of the city Kabbadi Stadium. The Kabbadi Associations President had in the past hired an Indian Consultant to help him carry out the task of expanding and improving the hockey stadium. The Indian Consultant left the work after collecting the time (in days) associated with the activities and developing the forward path network diagrams.

### ACTIVITY OPTIMISTIC MOST LIKELY PESSIMISTIC IMMEDIATE PREDECESSOR

The Association President has asked you to calculate the following:

• Calculate the expected time and variance for each activity.
• Calculate the activity slacks and determine critical path using expected activity times?
• What is the probability of completing the project with in 550 days?

Solution
We first of all calculate the Expected times and variances for each activity using the formulae respectively

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

σ2= ((b-a)/6)2

The results are presented in the form of the table

### ACTIVITY EXPECTED TIME VARIANCE

We need to calculate the Earliest Start, Latest Start, Earliest Finish, Latest Finish represented by the symbols ES, LS, EF and LF respectively. We use the forward path network diagram as provided by the hockey association’s president.

### ACTIVITY ES EF t

As we can see from the table above the earliest time by which Activity G would finish is 20 days and requires 4.5 days of time to complete. We need to know calculate values of Latest Start and Latest Finish using the backward path. Please refer to the backward path diagram below, the direction of arrows have been reversed indicating that we are actually back tracing the activities with the same times as calculated above using forward path.

The critical path is B-C-E-G with total expected time of 20 days.

c. We first calculate the z value

Z = (t-te)/√σ2
= (23-20)/√3.89
= 3/1.972
= 1.5210

Using the Normal Distribution table, we calculate the probability of completing the project in 23 days to be 0.9357

## Project Management Software Tools

2. Groupware (Lotus Notes)
3. Project management software
• CA Super Project
• Harvard Total Manager
• MS Project
• Sure Track Project Manager
• Time Line

1. Imposes a methodology
2. Provides logical planning structure
3. Enhances team communication
4. Flag constraint violations
5. Automatic report formats
6. Multiple levels of reports
7. Enables what-if scenarios
8. Generates various chart types

## Project Risk Management

Risk: occurrence of events that have undesirable consequences

1. Delays
2. Increased costs
3. Inability to meet specifications
4. Project termination

## Risk Management

1. Identify potential risks
2. Analyze and assess risks
3. Work to minimize occurrence of risk
4. Establish contingency plans

## Operations Strategy

1. Many Organizations have setup a separate Project Management department or cell to administer unique and non repetitive activities.
2. The scope of the project decides whether to use a project management software tool or not.
3. Project teams normally operate as a matrix team with employees from different functional departments working with the project team. In such situations the organizations device a strategy that project manger should lead the team as he or she is more aware of the situation being faced by the whole organization as well as the constituent functional departments.

## Summary

1. Projects are unique set of activities established to given set of objectives in a limited time span.
2. PERT and CPM two commonly used techniques for developing and monitoring projects.
3. Two slightly different conventions can be used for constructing a network diagram.
4. The task of developing and updating project networks quickly becomes projects of even moderate size or PC applications.
5. A deterministic approach is useful for estimating the duration of the project, when activity times can be fairly well established.
6. In some instances, it may be possible to shorten or crash the length of a project by shortening one or more of the project activities.
7. Often Projects are shortened to the point where the cost of additional reduction would exceed the benefit of additional reduction to a specified time.