Expected time of contractors (PERT)

Expected time $t_e = \dfrac{t_o + 4t_m + t_p}{6}$; standard deviation $\sigma = \dfrac{t_p - t_o}{6}$ (the unit of time is as given).

Contractor A ($5, 7, 13$): $t_e = \dfrac{5 + 28 + 13}{6} = \dfrac{46}{6} = 7.67$, $\sigma = \dfrac{13 - 5}{6} = 1.33$, variance $= 1.78$

Contractor B ($6, 11, 12$): $t_e = \dfrac{6 + 44 + 12}{6} = \dfrac{62}{6} = 10.33$, $\sigma = \dfrac{12 - 6}{6} = 1.00$, variance $= 1.00$

Contractor C ($3, 5, 7$): $t_e = \dfrac{3 + 20 + 7}{6} = \dfrac{30}{6} = 5.00$, $\sigma = \dfrac{7 - 3}{6} = 0.67$, variance $= 0.44$

Choice of contractor

Contractor C is preferred because:

• it has the shortest expected time (5.00 against 7.67 and 10.33);
• it also has the smallest standard deviation (0.67) and variance (0.44), so its completion time is the most certain, and the risk of delay is the least.

(Cost and quality are assumed equal; if they differ, they must be considered along with time.)

Answer: $t_e$ = 7.67 (A), 10.33 (B), 5.00 (C); choose contractor C.