Question

Consider the following PERT network:



The optimistic time, most likely time and pessimistic time of all the activities are given in the table below:

Activity	Optimistic time (days)	Most likely time (days)	Pessimistic time (days)
1 - 2	1	2	3
1 - 3	5	6	7
1 - 4	3	5	7
2 - 5	5	7	9
3 - 5	2	4	6
5 - 6	4	5	6
4 - 7	4	6	8
6 - 7	2	3	4

The critical path duration of the network (in days) is

• A. 11
• B. 14
• C. 18
• D. 17

Answer 1

The correct option is C 18

Activities	Duration $t_e=\frac{t_o+4t_m+t_p}{6}$	EST	LST	EFT	LFT	LST - EST = Float	Variance ${\sigma }^2{\left(\frac{t_p-t_o}{6}\right)}^2$
1 - 2	$\frac{1+4\times 2+3}{6}=2$	0	0	2	3	0	${\left(\frac{3-1}{6}\right)}^2=\frac{4}{36}$
1 - 3	$\frac{5+6\times 4+7}{6}=6$	0	0	6	6	0	${\left(\frac{7-5}{6}\right)}^2=\frac{4}{36}$
1 - 4	$\frac{3+4\times 5+7}{6}=5$	0	0	5	12	7	${\left(\frac{7-3}{6}\right)}^2=\frac{16}{36}$
2 - 5	$\frac{5+4\times 7+9}{6}=7$	2	3	10	10	1	${\left(\frac{9-5}{6}\right)}^2=\frac{16}{36}$
3 - 5	$\frac{2+4\times 4+6}{6}=4$	6	6	10	10	0	${\left(\frac{6-2}{6}\right)}^2=\frac{16}{36}$
5 - 6	$\frac{4+4\times 5+6}{6}=5$	10	10	15	15	0	${\left(\frac{6-4}{6}\right)}^2=\frac{4}{36}$
4 - 7	$\frac{4+4\times 6+8}{6}=6$	5	12	18	18	7	${\left(\frac{8-4}{6}\right)}^2=\frac{16}{36}$
6 - 7	$\frac{2+4\times 3+4}{6}=3$	15	15	18	18	0	${\left(\frac{4-2}{6}\right)}^2=\frac{4}{36}$

$\text{Critical path}=1\to 3\to 5\to 6\to 7$
(Marked by double lines)

$\text{Critical path duration of the network = 6 + 4 + 5 + 3 = 18 days}$