Home > Operation Research calculators > Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic) example

3. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic) example ( Enter your problem )
 Example-1Example-2 Other related methods Network diagramCritical path, Total float, Free float, Independent floatProject scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)Project crashing to solve Time-Cost Trade-Off with fixed Indirect costProject crashing to solve Time-Cost Trade-Off with varying Indirect cost

1. Example-1

1. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)
 1-2 1 1 7 1-3 1 4 7 1-4 2 2 8 2-5 1 1 1 3-5 2 5 14 4-6 2 5 8 5-6 3 6 15

Solution:
Expected time of each activity,
 Activity t_o t_m t_p t_e=(t_o + 4*t_m + t_p)/6 sigma^2=((t_p - t_o)/6)^2 1-2 1 1 7 2 1 1-3 1 4 7 4 1 1-4 2 2 8 3 1 2-5 1 1 1 1 0 3-5 2 5 14 6 4 4-6 2 5 8 5 1 5-6 3 6 15 7 4

The earliest and latest expected time for each activity is calculated by considering the expected time t_e

The given problem is
 Activity Duration 1-2 2 1-3 4 1-4 3 2-5 1 3-5 6 4-6 5 5-6 7

Edge and it's preceded and succeeded node
 Edge Node1 -> Node2 A 1->2 B 1->3 C 1->4 D 2->5 E 3->5 F 4->6 G 5->6

The network diagram for the project, along with activity time, is
 2
 A(2) A : 1->2
 D(1) D : 2->5 E(6) E : 3->5
 5
 1
 B(4) B : 1->3
 3
 G(7) G : 5->6
 C(3) C : 1->4
 G(7) G : 5->6
 4
 F(5) F : 4->6
 6

Forward Pass Method
E_1=0

E_2=E_1 + t_(1,2) [t_(1,2) = A = 2]=0 + 2=2

E_3=E_1 + t_(1,3) [t_(1,3) = B = 4]=0 + 4=4

E_4=E_1 + t_(1,4) [t_(1,4) = C = 3]=0 + 3=3

E_5=Max{E_i + t_(i,5)} [i=2, 3]

=Max{E_2 + t_(2,5); E_3 + t_(3,5)}

=Max{2 + 1; 4 + 6}

=Max{3; 10}

=10

E_6=Max{E_i + t_(i,6)} [i=4, 5]

=Max{E_4 + t_(4,6); E_5 + t_(5,6)}

=Max{3 + 5; 10 + 7}

=Max{8; 17}

=17

Backward Pass Method
L_6=E_6=17

L_5=L_6 - t_(5,6) [t_(5,6) = G = 7]=17 - 7=10

L_4=L_6 - t_(4,6) [t_(4,6) = F = 5]=17 - 5=12

L_3=L_5 - t_(3,5) [t_(3,5) = E = 6]=10 - 6=4

L_2=L_5 - t_(2,5) [t_(2,5) = D = 1]=10 - 1=9

L_1=text{Min}{L_j - t_(1,j)} [j=4, 3, 2]

=text{Min}{L_4 - t_(1,4); L_3 - t_(1,3); L_2 - t_(1,2)}

=text{Min}{12 - 3; 4 - 4; 9 - 2}

=text{Min}{9; 0; 7}

=0

(b) The critical path in the network diagram has been shown. This has been done by double lines by joining all those events where E-values and L-values are equal.
The critical path of the project is : 1-3-5-6 and critical activities are B,E,G

The total project time is 17
The network diagram for the project, along with E-values and L-values, is
 2 E_(2)=2L_(2)=9
 E_(5)=10L_(5)=10
 A(2) A : 1->2
 E_(2)=2L_(2)=9
 D(1) D : 2->5 E(6) E : 3->5
 5 E_(5)=10L_(5)=10
 1 E_(1)=0L_(1)=0
 B(4) B : 1->3
 3 E_(3)=4L_(3)=4
 G(7) G : 5->6
 E_(1)=0L_(1)=0
 C(3) C : 1->4
 E_(3)=4L_(3)=4
 G(7) G : 5->6
 E_(4)=3L_(4)=12
 4 E_(4)=3L_(4)=12
 F(5) F : 4->6
 6 E_(6)=17L_(6)=17
 E_(6)=17L_(6)=17

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here

Dear user,