Statistics In Probabilistic Pert Analysis With Three Time Estimates

I. Estimation:

Let t_i denotes the duration of activity i. We assume that t_i is distributed according to the modified beta distribution. We make this assumption because the modified beta distribution is defined for non-negative values only and duration cannot be negative, and because the shape of the distribution is flexible it may be asymmetric and skewed to the left or to the right.

We denote and assume, the mean of t_I as,

,

where, a and b are the most optimistic and most pessimistic judgments respectively, and m is the most likely (mode) judgment.

The mean is identical to the one obtained for a discrete distribution with its greatest mass in the center with weights of 1/6 for the extreme optimistic(a) and pessimistic (b) values, and 4/6 for the mode (m).

And the standard deviation of t_I, as,

.

In going from the center 3 std deviations to the left and 3 to the right we capture most of the distribution of t_i. We thus assume that the distance between b and a covers the distance of 6 standard deviations.

The variance of t_I is clearly,

.

Let T denotes the project’s duration. We are making the heroic assumption that it depends only on the critical path.

In the book’s example CP= a, b, d, e, i, k, l.

This assumption is often not valid.

Hence,

Since, T is a sum of the random variables t_{i ,} the mean of T is denoted and obtained as,

= 5+2 +12+10+1+3+9 = 42

And assuming statistical independence among the t_is, variance T is,

= .444 + .444 + 7.111+1 + 0 +.25 +.444 = 9.693

and,

=3.11

The complete arguments are as follows:

The mean of a sum of random variables is always the sum of the means of the random variables regardless of statistical dependence.

The variance of the sum is the sum of all variances and covariances of the added random variables. However, if the random variables are statistically independent, the covariances vanish.

As a sum of random variables with standard deviations, according to the C.L.T (Central Limit Theorem), T is distributed approximately normal.

II. Inference.

Example 1: P = Pr(T<50)

P = Pr(T<50) =

By algebra of inequalities (division of both sides of the inequality by a positive number would not alter the direction of the inequalities. Subtraction of any number positive or negative on both side of the inequality is allowed. The inequality would have identical set of feasible solutions in T.

Let,

z is a standard random variable. It is also normal as a linear transformation of the normal variable T.

P=

=.5+ .4949 = .9949.

Example 2: P= Pr(T>40)

P = Pr(T>40) =

By algebra of inequalities.

Let,

z is a standard random variable. It is also normal as a linear transformation of the normal variable T.

P=

By symmetry

Pr(-.64 < z <0) = Pr(0 < z <.64)=.2389

Hence,

P = .2389 +.5 =.7389

Example 3: (Not in the book). P = Pr(40 < T < 50).

P = Pr(40<T<50) =

By algebra of inequalities

Let,

z is a standard random variable. It is also normal as a linear transformation of the normal variable T.

P=

By symmetry

Pr(-.64 < z <0) = Pr(0 < z <.64)=.2389

And

Pr (0 < z < 2.57) = .4949.

Hence,

P= .2389 +.4949 = .7338

Please notice. The yellow material is not needed for a computational question. It may be needed for a question requiring explanation of arguments.

Notice also that a computational answer would also requires the graphs of the distributions of T and Z as needed for determining the requested probability.