1
QUANTITATIVE ANALYSIS
QuantitativeAnalysis
QuantitativeAnalysis

Company A

The expected completion time for the project:
Theexpected duration and variance for each activity can be calculatedfrom the following formula:
Expectedduration, d_{e}= (d_{o}+ 4d_{m}+)/6
Variance,σ= [(– do)/6)]^{2}
Expectedtime of completion variance

d_{e } = [2 + 4(3) + 4]/6 = 3 σ^{2} = [(4 – 2)/6]^{2} = 0.11

de=[5 + 4(6) + 13]/6 = 7 σ^{2} = [(13 – 5)/6]^{2} = 1.78

de = [3 + 4(4) + 8]/6 = 4.5 σ^{2} = [(8 – 3)/6]^{2} = 0.69

de = [10 + 4(11) + 15]/6 = 11.5 σ^{2}= [(15 – 10)/6]^{2} = 0.69

de = [4 + 4(5) + 6]/6 = 5 σ^{2}= [(6 – 4)/6]^{2} = 0.11

de = [8 + 4(10) + 12]/6 = 10 σ^{2}= [(12 – 8)/6]^{2} = 0.44

de = [4 + 4(6) + 11]/6 = 6.5 σ^{2}= [(11 – 4)/6]^{2} = 1.36

de = [8 + 4(10) + 18]/6 = 11 σ^{2}= [(18 – 8)/6]^{2} = 2.78

de = [3 + 4(6) + 12]/6 = 6.5 σ^{2}= [(12 – 3)/6]^{2} = 2.25

de = [2 + 4(3) + 7]/6 = 3.5 σ^{2}= [(7 – 2)/6]^{2} = 0.69
[Formulaefrom [ CITATION ICM10 l 1033 ]]
2.
0 3
A, 3
0 2
3 7.5
C, 4.5
2 6.5
26626.6.56.5
7.512.5
E, 5
6.5 19
18.529.5
H, 30
19 30
0 0
START
0 0
3033.5
J, 3.5
30 33.5
0 7
B , 7
0 7
7 18.5
D, 11.5
7.5 19
7 17
F, 10
7 17
1723.5
G, 6.5
1723.5
23.5 30
I, 6.5
23.5 30
23.5 23.5
FINISH
23.5 23.5
Figure1, showing the PERT chart.
Theabove chart has been prepared using the method of activity on node(AON)
Key:ES – Earliest start, EF – Earliest finish, LS – Latest start,LF – Latest finish
Fromthe calculations above, the expected duration for each of the tenproject activities is:
Expectedduration for
A= 3 weeks B = 7 weeks C = 4.5 weeks D = 11.5 weeks E = 5 weeks F= 10 weeks G = 6.5 weeks H = 11 weeks I = 6.5 weeks J = 3.5weeks.
ES EF
TASK
LS LF
Theresults in A.1 above can be summarized in the table below:
Task 
Preceding Activity 
Optimistic Time to Complete(weeks) 
Probable Time to Complete(weeks) 
Pessimistic Time to Complete(weeks) 
Expected Time to Complete (weeks) 
Variance(weeks) 
START 
  
  
  
  
  
  
A 
START 
2 
3 
4 
 3 
0.111  
B 
START 
5 
6 
13 
 7 
 1.778 
C 
A 
3 
4 
8 
 4.5 
 0.694 
D 
B 
10 
11 
15 
 11.5 
 0.694 
E 
C 
4 
5 
6 
 5 
 0.111 
F 
B 
8 
10 
12 
 10 
 0.444 
G 
F 
4 
6 
11 
 6.5 
 1.361 
H 
D, E 
8 
10 
18 
 11 
2.778  
I 
G 
3 
6 
12 
 6.5 
 2.25 
J 
H, I 
2 
3 
7 
 3.5 
 0.694 
Definitions
Theseare definition according to Evans(1972):

Earliest time of start event (ES)
Thisis the earliest possible time a task (activity) can probably start.

Latest time of finish event (LF)
Thisrefers to the latest time an activity (task) can be done withoutdelaying the completion of the project.

Latest time of start event (LS)
Thisrefers to the latest time a preceding activity (task) may finish.

Earliest time of finish event (EF)
Thisrefers to the earliest time that a succeeding activity (task) mayprobably start.
Therefore,from the PERT chart, one can determine the critical activities. Theprimary activities refer to the activities, which have no float, andwhose earliest and latest times of start event coincide. Therefore,their latest and earliest times of finish event coincide (are equal)and the time difference between start event and complete developmentequals to the duration of the activity.
Floatis the amounts of time by which an activity can be delayed orextended and still not impede with the project’s deadline (Evans1972):
Criticalpath is B – F – G – I – J
3(a) From the PERT chart, the expected duration of the whole projectis equal to the sum of the duration of the activities along thecritical path = 7+10+6.5+6.5+3.5 = 33.5.
Note:Slack time is the time differential between the necessary date tomeet the critical path and the scheduled deadline.
(b)Slack for project task A = Difference between the latest time offinish of event A and the earliest time of finish of event A = 3 –2 = 1 week
(c)Slack for project task H = Difference between latest time of finishof event H and the earliest time of finish of event H= 30 29.5 = 0.5 week
(d)The starting week for project taskF is given by the earliest of startof event F = 7^{th}week. Note that we take the earliest time of start event (ES) forthis task other than the latest time of start event (LS) to reducedelays, since executing works in the shortest time possible is moreeconomical.
(e)The finishing time of the project task I, is given by the latest timeof finish of project task I = 30^{th}week.Though the earliest time of finish would give the time of finish ofthis task, we take the latest time of finish because we have worstcase i.e. when work does not progress as planned due tocontingencies.
4(a) Calculating the probability of finishing the project within agiven time is done but if the project activities are statisticallyindependent of each other. The activities are assumed to be normallydistributed.
LetX equal to the duration within which the project can be completed.
Theprobability of completing the project in 34 weeks is given by:
P(X<=34) = P (T<= T_{1}),where T_{1}= (Target duration – Expected duration)/Standard deviation
Targetduration = Total project time = 33.5
Expectedduration = required time of completion given = 34
TheStandard deviation, σ =(sum variance of critical activities)^{0.5}
Variance,σ^{2}= 1.778+0.444+1.361+2.25+0.694 = 6.527
σ= 2.555
P(X<=34)= (33.4 – 34)/2.555 = 0.196
Fromthe normal distribution table, P(X<=34) = 0.5 – 0.0777 = 0.4223
Hencethe chance of completing the project in 34 weeks = 0.9223 x100% =92.23%
Upondetermining this probability, one can estimate to what extent we aresure of completing the project within a given time.
B.Company B
1.Maximum reduction in time is given by the maximum crash time (weeks).From the table, the time reduction the activities are:
A– 2 weeks B = 5 weeks C = 3 weeks D = 7 weeks E = 7 weeks F =3 weeks G = 5 weeks
H= 8 weeks I= 4.5 weeks J = 3 weeks.
2.Projectcrushing involves reducing on the completion time planned earlier onso that the work is completed earlier than expected. Since theproject completion duration depends on the term of the criticalactivities, reducing the length of the activities along the criticalpath does crushing. It is worth noting that as one carries outcrashing, the cost implications should be borne in mind. On thisbasis, therefore, to decide which activity should have its durationreduced the most one should determine the crash cost per week of allactivities (Taggart1996).
Crashcost per week = (Crash cost – normal cost)/ (Normal time – crashtime)
Crashcost per week for:
TaskA = (11200 – 8400)/ (3 – 2) = $2,800 Task C = (24,000 –18000)/ (4.5 – 3) = $ 4,000 Task E = (16,800 – 14000)/(5 – 3)= $1,400
TaskB = (40,000 – 28000)/ (7 – 5) = $6,000 Task D = (36,400 –44,800)/ (11.5 – 7) = $1,778 Task F = (24,000 – 20,000)/(10 –6) = $1,000
TaskG = (28,000 – 18200)/(6.5 – 5) = $6533 Task H = (64,000 –44,000)/(11 – 8) = $6,667 Task I = (36,000 – 26,000)/(6.5 –4.5) = $5000
TaskJ = (36,000 – 21,000)/(3.5 – 3) = $30,000
Task 
Expected Time to Complete 
  
`Normal cost 
Crash Cost 
Maximum Reduction in Time 
Crash Cost per week 
Normal (weeks) 
Crash (weeks) 

START 
  
  
  
  
  
  
A 
3 
2 
$ 8,400 
$ 11,200 
2 
$ 2,800 
B 
7 
5 
$ 28,000 
$ 40,000 
5 
$ 6,000 
C 
4.5 
3 
$ 18,000 
$ 24,000 
3 
$ 4,000 
D 
11.5 
7 
$ 36,800 
$ 44,800 
7 
$ 1,778 
E 
5 
3 
$ 14,000 
$ 16,800 
3 
$ 1,400 
F 
10 
6 
$ 20,000 
$ 24,000 
6 
$ 1,000 
G 
6.5 
5 
$ 18,200 
$ 28,000 
5 
$ 6,533 
H 
11 
8 
$ 44,000 
$ 64,000 
8 
$ 6,667 
I 
6.5 
4.5 
$ 26,000 
$ 36,000 
4.5 
$ 5,000 
J 
3.5 
3 
$ 21,000 
$ 36,000 
3 
$ 30,000 

Identify the critical activity with the lowest crash cost per week. Reduce the duration for this activity as much as possible.
Fromthe above table, the critical activity with the lowest crash cost perweek is task F. Therefore we are going to reduce the duration fortask F as much as possible. Since we are required to reduce theproject completion duration from 33.5 to 30, by 3.5 weeks, task F canbe crashed by a maximum of 4 weeks without another path becomingcritical:
Crashfor task F = 10 – 6 = 4 weeks.
Sinceit’s possible to crash task F by a maximum of 4 weeks, and thiswill give the lowest increase in cost, we will apply all the 3.5weeks crash to task F.
Therefore,the duration for task F will be reduced by 3.5 weeks to meet thedeadline with the lowest possible increase in cost.
Theadditional crashing cost for the project is going to be $(24,000 –20,000) = $4,000
Accordingto Marshall(2010), it is clear that Networkanalysis offers the all the advantages of being able to manipulatethe planning data by holding the data in computer files. The planningdata in a network is linked through the logic that defines therelationship between the activities.
Otherterminologies to note:

Total float
Isthe total amount of time by which an activity could be delayed orextended and still not interfere with project end date.

Free float
Thisis the amount of time by which an activity could be extended ordelayed without interfering with the succeeding activity.
References
Evans,G. C. (1972). Thequantitative analysis of plant growth.Berkeley: University of California Press.
Marshall,P. (2010). Quantitativeanalysis of cognitive radio and network performance.Norwood, Mass.
Taggart,R. A. (1996). Quantitativeanalysis for investment management.Upper Saddle River, NJ: Prentice Hall.