Question

Time​ (days)						Immediate		Time​ (days)						Immediate
Activity	a		m		b		​Predecessor(s)	Activity	a		m		b		​Predecessor(s)
A	55		55		77		long dash—	H	44		44		66		​E, F
B	11		22		55		long dash—	I	22		77		1010		​G, H
C	55		55		55		A	J	22		44		77		I
D	44		88		1313		A	K	66		1010		1313		I
E	11		1010		1717		​B, C	L	22		66		66		J
F	11		55		77		D	M	22		22		33		K
G	22		66		99		D	N	77		77		1212		​L, M

Number of days that would result in​ 99% probability of completion​

Answer 1

For each activity we have got 3 time estimates

• a=Optimistic time
• m= Most probable time
• b=Pessimistic time

Using these we can calculate the expected time (t) for each activity as below

for example for activity A, the expected time is

Next we need to find the variance of each activity given by the formula

For example for activity A, the variance is

The expected time and variances are given below

The values are

Next we need to find the critical path pf the project, using the expected time as fixed activity time.

The following is the PERT/CPM with the expected activity times.

We take the earliest start as the earliest end of the predecessor. For example activity D has A as the predecessor. Hence the earliest start of D is earliest end of A = 58.7. The earliest end is got by adding the activity time to the earliest start. When there are more than one predecessors, we take the maximum of the earliest end times. For example Activity E has B,C as predecessors. The end time of B is 25.7 and C is 113.7. Hence the earliest start of E is 113.7

Updated chart with earliest start and end dates, by moving from left to right

Finally we move from the right to left and fill in the latest start/end times

The latest end of any activity is the latest start of its successor. If there are more than one successor the latest end is the minimum of the latest starts. For example activity I has J and K as successors and their latest starts are 1346 and 2168.5. Hence the latest end for I is 1346.0

The final chart is below

The critical path (shown in green) are the activities with zero slack (ealiest time= latest time)

The critical path is A,C,E,H,I,K,M,N

The expected time of the project is the sum of expected times of the activities in the critical path.

The variance for the project completion is the sum of the variances of the critical path activities

Let T indicate the time taken to complete the project. T has normal distribution with mean and

standard deviation

Number of days that would result is 99% probability of completion is given as

the probability that a project would complete in less than t days is 0.99

That is P(T<t) = 0.99

We need to find the value of t.

The standardized value of T is

Now let us find the value of z for which P(Z<z) = 0.99.

If we use the standard normal table we can see that for z=2.33 we get P(Z<2.33) = 0.5+0.4901=0.99

We equate this to the standardized value of T and get

That means for 3546.66 days, the probability of completion is 99%