A simple project comprises of two start-to-end parallel paths, each with three activities in series, with no interpath dependencies. The a, m, b data (in days) for each activity are shown in the diagram. Assuming that three activities in series are enough for further computations, what will be the total project duration and its standard deviation?

Explanation:

Here two path – (I) ⇒ 1 – 2 – 4 – 6

(II) ⇒ 1 – 3 – 5 – 6

Calculation of expected time for both path – 

\({T_e} = \frac{{{T_0} + 4{T_m} + {T_p}}}{6}\)

For I 

\( \Rightarrow \frac{{2 + 4 \times 3 + 4}}{6} + \frac{{4 + 4 \times 6 + 8}}{6} + \frac{{5 \times 4 \times 8 + 11}}{6}\)

= 3 + 6 + 8

= 17 days

For II 

\( \Rightarrow \frac{{6 + 4 \times 7 + 8}}{6} + \frac{{12 + 4 \times 12 + 18}}{6} + \frac{{9 + 4 \times 15 + 18}}{6}\)

= 7 + 13 + 14.5

\( = 34\frac{1}{2}{\rm{\;days}}\)

Hence longest path II so it is critical standard deviation.

\(\delta d = \sqrt {\sum {{\left( {\frac{{{T_p} - {T_0}}}{6}} \right)}^2}} \)

\(\sigma = \sqrt {{{\left( {\frac{{8 - 6}}{6}} \right)}^2} + {{\left( {\frac{{18 - 12}}{6}} \right)}^2} + {{\left( {\frac{{18 - 9}}{6}} \right)}^2}} \)

\(\sigma = \sqrt {{{\left( {\frac{1}{3}} \right)}^2} + 1 + {{\left( {\frac{3}{2}} \right)}^2}} = \sqrt {\frac{1}{9} + 1 + \frac{9}{4}} = \sqrt {\frac{{121}}{{36}}} \)

\(\therefore \;\left[ {\sigma = \frac{{11}}{6}\;days} \right]\)