A project consist of the following activities and estimated time as shown below.
a.) Construct a network diagram
b.) Determine the project completion time and critical path
Activity
LEGEND:
A - 5 6 18 8 ES
Earliest Start
B A 8 8 19 10 EF
Eartliest Finish
C A,B 3 5 12 6 LS
Latest Start
D B,C 6 9 14 9 LF
Latest Finish
E D 3 4 10 5
F E 3 3 9 4 ACT
START A
Te = (To+4Tm+Tp)/6 8
Te
Activity A A = (5 + 4x6 +18) / 6 = 7.83 or 8
Activity B B = (8 + 4x8 +19) / 6 = 9.83 or 10 ACTIVITY ES LS
Activity C C = (3 + 4x5 +12) / 6 = 5.83 or 6 A0 0
Activity D D = (6 + 4x9 +14) / 6 = 9.33 or 9 B8 8
Activity E E = (3 + 4x9 +10) / 6 = 4.83 or 5 C18 18
Activity F F = (3 + 4x3 + 9) / 6 = 4.00 or 4 D24 24
E33 33
F38 38
CRITICAL PATHS
A
A
A
Intermediate
Predecessor
Optimistic
Time (To)
Most
Probable
Time (Tm)
Pessimistic
Time (Tp)
Expected
Time (Te)
Solution Notes:
A.) For the first Activity or Activity A, the earliest start (ES) is always equal
to 0
B.) The estimated times (Te) have been plotted below the Activity label.
Notice that activity A has a Te of 8, which was computed using the formula
shown in slide number 3.
C.) Earliest finish (EF) for Activity A is Te + ES, thus, 8+0 = 8
E.) For Activity B, the ES is the EF of the preceeding activity A, thus, 8
F.) For Activity B, the EF is Te+ES, thus, 10+8 = 18
G.) For Activity C, the ES is 18 which is the EF of Activity B. Notice that
Activity C is connected to both A and B. In cases like this, and with a
forward pass, the rule is to select the higher EF. EF of A is 8 while EF of B is
18. The higher value is 18, and it will therefore be the ES of Activity C. In
cases like this, and with a backward pass, the rule is reversed. Thus , the
lower value is selected between or among those items connected to the
activity being processed.
H.) Using the same procedure, the computed EF for F is 42. It will also be
used as the LF for F and begin the backward pass by reversing the
procedure, thus, subtracting the LS by the Te to get the EF. For F, it is
computed as 42-4=38.
I. ) The LS of F is the LS of G, thus 38. The procedure is similar for all the
other backward pass.
