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a ., t. 29, , Figure 5.20, d on the assumption, , time for each activity and the corresponding variance 1S calculate, Beta. Thus using the formulae:, , , , , , that the di :, hat the distribution of time in the production process 1s, , Best Estimated 1 ‘ : a+4ni+b ‘em, Ls st Estimated Time t., = oe qu -Lu), a=, c, i : b-a), And Variance of the Estimate =| —— Le, 6 ¥, , One get the time estimates in given table:, , ! Activity to” | Variance of to, , ee, , 1-3, LA, | __0.12, , , , , , , , , , , , , , , , PION TP 20s 1h) |s8%e)

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242, best estimated times, we can calculate the critical path, which comes out to be:, , Using the, , With an estimated time of the project = [43.2+2+10+21.2]=76.5 days., With variange=16.0+0.12+0.45+3.35=19.92 Cah, Ve, 1 a, , The confidence limits for the completion time of the project are given by, , 68 days approximately., The calculation of probability can be done with the help of normal distribution shown in figure §, , , , , tet, , Due Date, , , , , , ~Bo' nto as 0! gene ae, , Duration, , Manna £94 or, , ee Be A ng

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Due Date, , , , , , abe 23 tore o 20, , , , Duration, Here due date T; = 60 days Figure 5.21, Te = 76.5 days Z, and o = V19.92, : 3 ary a, | _One get, z=18 ig O83. 2 ne, a o 419.92 = Ph, , From the tables of normal distribution, we find- the renin, approximately; which is very small., , , , Example 22: A project manager has obtained the following optimistic, pessimistic ar 02 ay Acre eX), , ee