This document contains responses to questions about project management. It discusses key aspects of project organization including temporary job assignments, a command structure with flexibility, and good information flow. It also covers work breakdown structures, resource constraints, critical path analysis, float utilization, and updating project schedules as new information becomes available.

1. Chapters 34 & 35: Project Management – I & II
Responses to Questions:
1. In a project once a job is completed, the employees working on the job can
move over to another job or another project. A project appointment, therefore,
differs from the regular manufacturing appointment; it is a roll-over kind of an
assignment. Despite this limitation, the jobs are to be done within a given time
with good quality in line with the objectives of the project. In projects there has
to be a good command structure combined with local flexibility. Another
important aspect is that the organization should be conducive to excellent
coordination between different connected / related activities. The project
organization needs to design for the necessary flow of information.
Thus a project has jobs / people that are temporary yet highly committed; and
has an organization that has a command structure, yet it has much flexibilities
at all levels.
Today’s manufacturing is imbibing some of these good qualities of a project
organization.
2. Network is for control. Control cannot be successful unless there is clear-cut
demarcation of responsibilities and accountabilities. Hence, a proper network
has to base itself on the organizational considerations. Work Breakdown
Structure is an important starting point in drawing a network.
3. Project identification needs much introspection (What do we want to do?
What is our business?) by the organization. The ideas that are generated on
introspection have to be whetted by considering the external environment,
other limitations and priorities.
4. Corporate planning process and project identification and appraisal process
are very similar. Both processes start with introspection and extrospection.
They consider constraints, capacities already available etc. and arrive at a
feasible action plan. It is an integrated process in both cases.
5. Without a clear thinking and definition of objectives, a project would be a
largely wasteful exercise. The effect of this is strongly felt after the completion
of the project (and the commencement of the process). By then the die has
been cast.
For instance, unless Reliance conceptualized the various value creation
opportunities right at the inception, the Refinery project would not have been
the success that it has been.
Many irrigation projects in our country are lacking in success because the

2. 2
objectives have not been considered and incorporated into the project right at
the beginning. It then becomes just a civil engineering exercise done to
precision and no more.
6. All projects need a social cost benefit analysis (SCBA) done during their initial
formulation stage. The need is even more acute for projects of the
government/public system. Ultimately, all projects have to be for the good of
the society. This needs to be assessed right in the formulation stage.
Impact analysis is about (i) environmental / ecological impact, (ii) economic
impact and (iii) socio-cultural impact. ‘Impact’ is the long term effect. Along
with or as a part of SCBA, this impact analysis needs to be carried out.
7. Project Evaluation is the ‘hindsight’, a retrospective appraisal of: (i) What was
intended versus what has been achieved? (ii) What went right and what went
wrong? (iii) What lessons can be learnt for future projects / programmes?
Whereas, Project Appraisal is about ‘selection’ of projects through technical
and financial evaluation. Thus, Project Appraisal is at the beginning of a
project management process while Project Evaluation is after the project is
over.
8. An ‘activity’ is the smallest subdivision of the work of the project that makes
‘control’ sense.
Without management control, a breakdown of the packets of work would be
rather meaningless. The jobs would be rather meaningless. The jobs would
be unimplementable and uncontrollable. A network could turn out to be
farcical.
9. Without proper estimation of time for the activities, a realistic time schedule
cannot be drawn. The analysis of the slacks and the time scheduling of
resources will be exercises in futility.
10. The 3 time estimates are used in PERT, whereas CPM uses only 1 single
time estimate. The merit of PERT (and 3 time estimates) is that it comes up
with the ‘probabilities’ of completion of activities/milestones/project by a given
date. PERT (and 3 time estimates) provides this additional dimension. The
probabilities would also tell us as to which activities are more liable to become
’critical’. Some people see a merit in “3 estimates are better than just 1”; this
is, of course, a simplistic view.

3. 3
The de-merits are:
(i) In practice, it is difficult to get 3 time estimates - that too optimistic,
pessimistic and modal.
(ii) Probabilities and probabilistic critical paths make the issue
complicated and, hence, less practicable.
11. More than the ‘path’, we are concerned about critical activities which are the
bottleneck jobs. These cannot be delayed without delaying the entire project.
So, the activities on the critical path need to be watched carefully and
controlled.
Yes, there can be more than 1 critical path.
12.The ‘Total Float’, if used completely, would make the succeeding activities
critical. For this reason, it is not desirable to utilize this float completely,
although the information that so much float is there is helpful. Whereas, ‘Free
Float’ can be utilized completely without disturbing the succeeding activities.
Therefore, in case of slippages in time, Free Float is first made use of before
resorting to Total Float.
One may use the floats judiciously. The person responsible for only one
activity, such as a supervisor can be given only the Independent Float
information or at the most the Free Float information. The information
regarding Total Float is liable not to be understood properly and may
generate unnecessary complacency. At other levels of management also only
the Free Float may be used first, and the Total float may be used only
exceptionally and with the permission of the Project Manager.
13.The final schedule can be obtained only after the Resource Analysis is
performed. The initial Time Analysis is done with the assumption of the
availability of the resources without any constraints. The practical schedule
can only be obtained after the limits on resources are also taken into account
through the Resource Analysis.
14.CPM considers only the Finish to Start relationship between two activities.
This can be a significant handicap while considering civil works.
Secondly, a CPM analysis is only as good as its estimates of time and
resource requirements. Estimates, particularly the ‘time estimates’ are at the
hear of any CPM analysis.
15.CPM charts need to be updated frequently when the project is in force,
PERT/CPM charts can never be totally stable. The students’ visit to the
project cannot wait until the charts are stable, because they never will be.

4. 4
16.The CPM diagram is drawn and time analysis performed as given in the
network below.
9 A 11 D 5 H 14 J
O O O O
20 B 4 F 2 G
O O O
10 C 10 E 18 I
O O
24 K 6 L
O
The floats are as in the table below:
Activity TF FF IF
A 4 0 0
B 0 0 0
C 5 0 0
D 4 4 0
E 5 4 -1
F 0 0 0
G 3 0 0
H 0 0 0
I 1 1 0
J 0 0 0
K 1 0 0
L 13 13 12
The total float is zero for activities B, F, H and J. These are the critical activities.
O O 9 13
24 25
24 2510 15
20 20
29 29 43 4324 24
24 24 26 29

5. 5
17. As per the given network diagram activity E has immediate predecessors of C,
B and A. This is wrong. E should have had only B & C as its immediate
predecessors.
The correct network diagram is as follows:
B D
O O O
A C E F
O O O O O
Notice how the end event of B is split up to allow correct relationship to be
depicted.
18. We draw the network showing the requirements of both skilled and unskilled
workers for each of the activities. The requirements are at first worked out on
an early start of all activities. Whenever the daily requirements exceed the
restrictions on the availability, we try to reschedule the activities taking into
consideration their respective floats. In impossible situations, we may exceed
the duration of the project.
The initial (early start & no rescheduling) virgin network is drawn as below:
2 2 2 2 2
5 5 5 5 5
4 4 2 2 2 2 3 3 3 3
2 2 4 4 4 4 4 4 4 4
O O O O
4 4 4
3 3 3
6 6 8 8 8 2 3 3 3 3 skilled required
7 7 12 12 12 4 4 4 4 4 unskilled required

6. 6
For the first constraint of 9 unskilled and 6 skilled (on any day), the rescheduled
network is as follows (total duration = 12 days).
2 2 2 2 2
5 5 5 5 5
4 4 2 2 2 2 3 3 3 3
2 2 4 4 4 4 4 4 4 4
O O O O
1 2 3 4
4 4 4
3 3 3
6 6 4 4 4 6 4 4 3 3 3 3 skilled required
7 7 9 9 9 7 3 3 4 4 4 4 unskilled required
___________________________
For the second constraint of 7 unskilled and 6 skilled, the rescheduled network is as
follows (total duration = 13 days).
2 2 2 2 2
5 5 5 5 5
4 4 2 2 2 2 3 3 3 3
2 2 4 4 4 4 4 4 4 4
O O O O
4 4 4
3 3 3
6 6 2 2 2 6 6 6 2 3 3 3 3 skilled required
7 7 5 5 5 7 7 7 4 4 4 4 4 unskilled required
_________________________________
19. When job interruptions are not allowed, the Early Start (ES) and Early Finish
(EF) are calculated by taking a Forward Pass. The following relationships hold
good.
ES : For jobs with no predecessor, set ESj=0.
For all other jobs: for each predecessor arc ij, calculate ESj as follows:

7. 7
Predecessor type Calculation
FiSj ESj = EFi + FiSj
SiSj ESj = ESi + SiSj
FiFj ESj = EFi + FiFj – tj
Finally, set ESj equal to the maximum of all the ESj’s calculated above.
EF: For all jobs: EFj = ESj + tj
T : Project duration T = maximum of EFj
Accordingly, ES of jobs A, D, C and P are zero each.
For Job B, we have:
ESB = ESA + SA SB = 0 + 3 = 3
ESB = ESA + dA + FA FB – tB = 0 + 43 + 2 – 18 = 27
The maximum is 27.
Hence, ES of job B = 27.
For Job E, we have three predecessor relationships:
ESE = ESD + SD SE = 0 + 2 = 2
ESE = ESD + tD + FD FE – tE = 0 + 56 + 0 – 80 = - 24
ESE = ESC + tC + FC SD = 0 + 18 - 0 = 18
The maximum of the above is 18. This is the ES of job E.
For Job N: ESN = ESP + SP SN = 0 + 4 = 4
ESN = ESP + tP + FP FN – tN = 0 + 80 + 4 – 90 = - 6
The maximum being 4, it is the ES of Job N.
For Job F: ESF = ESB + SB SF = 27 + 3 = 30
ESF = ESB + tB + FB FF – tF = 27 + 18 - 3 – 90 = - 48
ESF = ESE + SE SF = 18 + 28 = 46
ESF = ESE + tE + FE FF – tF = 18 + 80 + 28 – 90 = 36
The maximum of the above is 46. This is the ES of job F
For Job G: ESG = ESF + tF + FF FG – tG = 46 + 90 + 2 – 30 = 108
ES of job G is 108.
For Job H: ESH = ESG + SG SH = 108 + 2 = 110
ESH = ESG + tG + FG FH – tH = 108 + 30 + 2 – 30 = 110
ES of job H is, therefore, 110.
For Job I: ESI = ESG + SG SI = 108 + 2 = 110
ESI = ESG + tG + FG FI – tI = 108 + 30 + 2 – 30 = 110
ES of job I is, therefore, 110.

8. 8
For Job J: ESJ = ESN + SN SJ = 4 + 28 = 32
ESJ = ESN + tN + FN FJ – tJ = 4 + 90 + 28 – 80 = 42
ESJ = ESH + SH SJ = 110 + 5 = 115
ESJ = ESH + tH + FH FJ – tJ = 110 + 30 + 1 – 80 = 61
ESJ = ESI + SI SJ = 110 + 2 = 112
ESJ = ESI + tI + FI FJ – tJ = 110 + 30 + 2 – 80 = 62
The maximum of the above is 115. This is the ES of job J.
For Job M: ESM = ESF + SF SM = 46 + 7 = 53
ESM = ESF + tF + FF FM – tM = 46 + 90 + 7 - 15 = 128
The maximum of the above being 128, it is the ES of job M.
For Job K: ESK = ESJ + tJ + FJ SK = 115 + 80 + 0 = 195
ESK = ESM + tM+ FM SK = 128 + 15 + 0 = 143
The maximum of the above is 195. This is the ES of job K.
For Job L: ESL = ESK + tK + FKSL = 195 + 20 + 0 = 215
The ES for Job L is 215.
For Job L: EFL = ESL + tL = 215 + 15 = 230
The project completion time is, therefore, 230
Late Start Schedule:
This is obtained by performing a Backward Pass.
LF: For jobs with no successor, LFj = Project Completion Time.
For all other jobs: for each successor arc jk, calculate LFj as follows:
Successor type Calculation
Fj Sk LFj = LSk – Fj Sk
Fj Fk LFj = LFk – Fj Fk
Sj Sk LFj = LSk - Sj SK + tj
LFj is equal to the minimum of all the LFj’s calculated above.
LS: For all jobs: LSj = LFj - tj
We perform the Backward Pass, starting with job L.
For job L: LSL = LFL – tL = 230 –15 = 215
For job K: LFK = LSL – FK SL = 215 – 0 = 215
LSK = LFK – tK = 215 – 20 = 195
The late start time for job K is 195.

9. 9
For job M: LFM = LSK – FM SK = 195 – 0 = 195
LSM = LFM – tM = 195 – 15 = 180
The late start time for job M is 180.
For job J: LFJ = LSK – FJ FK = 195 – 0 = 195
LSJ = LFJ – tJ = 195 – 80 = 115
The late start time for job J is 115.
For job I: LFI = LSJ – SI SJ + tI = 115 – 2 + 30 = 143
LFI = LFJ – FI FJ = 195 – 2 = 193
The minimum of the above is 143, which is the LF for job I.
LSI = LFI – tI = 143 – 30 = 113.
For job H: LFH = LSJ – SH SJ + tH = 115 – 5 + 30 = 140
LFH= LFJ – FH FJ = 195 – 1 = 194
The minimum of the above is 140, which is the LF for job H.
LSH = LFH – tH = 140 – 30 = 110.
For job N: LFN = LSJ – SN SJ + tN = 115 – 28 + 90 = 177
LFN = LFJ – FN FJ = 195 – 28 = 167
The minimum of the above is 167, which is the LF for job N.
LSN = LFN – tN = 167 – 90 = 77.
For job G: LFG = LSI – SG SI + tG = 113 – 2 + 30 = 141
LFG = LFI – FG FI = 143 - 2 = 141
LFG = LSH – SG SH + tG = 140 – 2 + 30 = 168
LFG = LFH – FG FH = 140 - 2 = 138
The minimum of the above is 138, which is the LF for job G.
LSG = LFG – tG = 138 – 30 = 108.
For job F: LFF = LSM – SF SM + tF = 180 – 7 + 90 = 263
LFF = LFM – FF FM = 195 – 7 = 188
LFF = LFG – FF FG = 138 – 2 = 136
The minimum of the above is 136, which is the LF for job F.
LSF = LFF – tF = 136 – 90 = 46.
For job E: LFE = LSF – SE SF + tE = 46 – 28 + 80 = 98
LFE = LFF – FE FF = 136 – 28 = 108
Hence, LF for job E is 98.
LSE = LFE – tE = 98 – 80 = 18
For job B: LFB = LSF – SB SF + tB = 46 – 3 + 18 = 61
LFB = LFF – FB FF = 136 – 3 = 133
Hence, LFB = 61 and LSB = LFB – tB = 61 – 18 = 43
For job A: LFA = LSB – SA SB + tA = 43 – 3 + 43 = 83

10. 10
LFA = LFB – FA FB = 61 – 2 = 59
Hence, LFA = 59 and LSA = 59 – 43 = 16
For job D: LFD = LSE – SD SE + tD = 18 – 2 + 56 = 72
LFD = LFE – FD FE = 98 – 0 = 98
The minimum of the above is 72, which is the LF for job D.
LSD = LFD – tD = 72 – 56 = 16.
For job C: LFC = LSE – FC SE = 18 – 0 = 18
LSC = LFC – tC = 18 – 18 = 0
For job P: LFP = LSN – SP SN + tP = 77 – 4 + 80 =153
LFP = LFN – FP FN = 167 – 4 = 163
The minimum of the above is 153, which is the LF for job P.
LSP = LFP – tP = 153 – 80 = 73.
A table of Early Start (ES) and Late Start (LS) for the different jobs is furnished
below. The total slack (LSj – ESj) is also noted.
JOB ES LS TOTAL SLACK
(LS – ES)
A 0 16 16
B 27 43 16
C 0 0 0
D 0 16 16
P 0 73 73
E 18 18 0
N 4 77 73
F 46 46 0
G 108 108 0
H 110 110 0
I 110 113 3
J 115 115 0
M 128 180 52
K 195 195 0
L 215 215 0

11. 11
CHAPTER 34 & 35: Project Management – I & II
Objective Questions
1. The problem with Social Cost Benefit Analysis is that :
a. some benefits and costs are not measurable in monetary terms.
b. it is difficult to allocate the expected benefits amongst different
projects.
c. the time preference rate of interest is an assumption.
√d. all of the above.
2. Social Cost Benefit Analysis is useful for :
a. Government or public sector projects.
b. Private sector projects.
√c. all of the above.
d. none of the above.
3. In CPM, a ‘critical activity’ is defined as one whose :
√a. total float is zero.
b. free float is zero.
c. duration is the longest.
d. none of the above.
4. Activity sequencing depends upon :
a. technological constraints
b. normal availability of resources
√c. all of the above
d. none of the above
5. Three time estimates are used by :
√a. PERT
b. CPM
c. Precedence Network
d. None of the above.
6. CPM uses :
√a. 1 time estimate
b. 2 time estimates
c. 3 times estimates
d. none of the above
7. The main difference between PERT and CPM is :
a. PERT uses Activity on Node diagram, whereas CPM uses Activity on
Arrow diagram.

12. 12
b. PERT is event-oriented whereas CPM is activity-oriented.
c. PERT lays emphasis on cost, whereas CPM lays emphasis on time.
√d. PERT is probabilistic, while CPM is deterministic.
8. PERT / CPM network can be used for :
a. Project appraisal
b. Project implementation
c. Project monitoring and control
√d. all of the above
9. An activity can become critical by consuming its :
√a. Total float fully.
b. Free float partially.
c. Any or all of the above.
d. None of the above.
10. Free float is always :
a. equal to or more than Independent Float.
b. equal to or less than Total Float.
√c. a & b
d. none of the above.
11.The 3 time estimates in network analysis assume that :
√a. The differences in time are attributable to difficulties inherent in
performing the tasks.
b. The differences in time arise due to the limits in resource availability.
c. a & b
d. none of the above.
12. For projects in the ‘service’ sector :
a. PERT can be used.
b. CPM can be used.
√c. all of the above.
d. none of the above.
Check the following network. Activities A & B each have a time duration of 7
and C & D each have 9.
B O
7
A D
O O O
7 9
C
9

13. 13
13. The dummy activity in the above diagram has a total float of :
a. 0 (Zero)
√b. 2
c. 4
d. none of the above
14. The dummy activity has a free float of :
a. 0 (Zero)
√b. 2
c. 4
d. none of the above
15. The dummy activity in the above diagram is required for :
√a. showing correct sequencing relationship.
b. nomenclature purpose.
c. a & b
d. none of the above.
16. ‘Index of criticality’ is encountered in :
a. CPM
√b. PERT
c. Precedence Network
d. all of the above
17. Early Start Schedule and Late Start Schedule can differ in their :
a. available floats
b. required resources
√ c. a & b
d. none of the above
18. Work Breakdown Structure is used in arriving at :
a. the daily resource requirements.
√b. the list of activities.
c. a reschedule of the project.
d. none of the above.
19. Ford and Fulkerson’s Method is used in :
a. resource leveling
b. resource smoothing
c. computing criticality index
√d. numbering the events

14. 14
20. Under PERT, the activity duration is assumed to follow a :
a. Poisson distribution
b. Normal distribution
√ c. Beta distribution
d. Negative exponential distribution
21. When job interruptions are allowed :
a. the different segments of an activity will have different slacks.
b. the critical path may consist of activity segments.
√ c. a & b
d. none of the above.