PERT-CPM OPERATIONS RESEARCH CONTINUED AND MORE

Part CPM

1. PERT-CPM CONTINUED The concepts that you will learn next are: Finding the earliest and latest event times of a network We will find the Earliest Start times, Earliest Finish times, Latest Start times, Latest Finish times and various float times for all the activities. Using concept of probability in order to apply a sort of sensitivity analysis for a network

2. Consider the following network: Activity Immediate Predecessor Duration (Days) A -- 2 B -- 3 C A 1 D B 2 E C 4 F D 3 G E, F 5 H G 4 I G 6 J G 2 K H 7 L I 2 M J 8 N K, L, M 10

3. The event times are found in this way: Each of the events has been divided into 2 halves – upper circle and lower circle. The lower circle has been further subdivided into 2 halves. The left half will show the earliest times of events and the right half will show the latest time of the event. The earliest times of the events will be found out in the forward direction, that is, starting from event 1 and ending at the last event, event 12. This way of finding the earliest times of the events is known as the forward Pass method. The latest event times will found out in the backward direction, that is, starting from the last event, event 12 and finishing at the first event, event 1. This way of finding the latest time is called Backward Pass method.

4. At first, event 1 starts at time 0. Time 0 may represent any starting time such as , 10 AM, for example. However, activity A takes 2 time units and thus the earliest time when event 2 can start is (0+2)=2 time units, as shown in the left half of the lower circle of event 2. Similarly, activity B takes 3 time units and thus the earliest time for event 3 is (0+3) = 3 time units. Activity C takes 1 unit of time. Thus, the earliest event time for event 4 = earliest time of its previous event, event 2 + duration of activity C = 2+1=3 time units. Duration of activity D is 2 time units and thus the earliest time for event 5 is = earliest time for event 3 + duration of activity D = 3+2=5 time units. For event 6, it can be observed that, two different paths are reaching event 6 – one is through event 4 and another is through event 5. Event 6 cannot start unless both the activities E and F finish. Thus the earliest time for event 6 = maximum of time through activity E and time through activity F = maximum(earliest time of event 4 + duration of activity E, earliest time of event 5 + duration of activity F) = maximum(3+4,5+3) =maximum(7,8) = 8 time units.

5. Next, duration of activity G is 5 time units. Thus, earliest time of event 7 = earliest time of previous event 6 + duration of activity G = 8+5 = 13 time units. Durations of activities H, I, and J are 4 time units, 6 time units, and 2 time units respectively. Thus the earliest time of event 8 = earliest time of event 7 + duration of activity H = 13+4=17 time units. The earliest time of event 9 = earliest time of event 7 + duration of activity I = 13+6=19 time units. The earliest time of event 10= earliest time of event 7 + duration of activity J = 13+2=15 time units. For event 11, it can be observed that, 3 different paths are reaching event 11– first one is through event 8, 2 nd one is through event 9 and 3 rd one is through event 10. Event 11 cannot start unless the activities K, L and M finish. Thus the earliest time for event 11 = maximum of time through activity K, time through activity L and time through activity M = maximum(earliest time of event 8 + duration of activity K, earliest time of event 9 + duration of activity L, earliest time for event 10 + duration of activity M) = maximum(17+7,19+2,15+8) =maximum(24,21,23) = 24 time units. Duration of activity N is 10 time units. Thus, earliest time for event 12 is = earliest time of previous event 11 + duration of activity N = 24+10 = 34 time units. Thus now, all the earliest times for all the events have been calculated.

6. Next, the event times will be calculated by Backward Pass method, starting from event 12. The latest time for the last event 12 = earliest time of the event = 34 time units. Instead of adding the activity duration to the earliest time as in forward pass method, this time, the duration of the previous activity will be subtracted from the latest event time of an event in order to find the latest time of the previous event. Thus, the latest time of event 11 = latest time of event 12 – duration of activity N = 34 – 10 = 24 time units, as shown on the lower right half of the lower half circle of event 11. Duration of activity K is 7 time units. Thus, latest time of event 8 = latest time of event 11 – duration of activity K = 24 – 7 = 17 time units. Similarly, latest time of event 9 = latest time of event 11 – duration of activity L = 24 – 2 = 22 time units and latest time of event 10 = latest time of event 11 – duration of activity M = 24 – 8 = 16 time units.

7. For event 7, we can reach event 7 in backward pass method, through 3 different paths – through event 8, through event 9 and through event 10. Unlike before for forward pas s method, we will take minimum of the times through these 3 paths, not the maximum one as we did in forward pass method. Thus the latest time for event 7 = minimum(latest time of event 8 + duration of activity H, latest time of event 9 + duration of activity I, latest time of event 10 + duration of activity J) = minimum(17 – 4, 22 – 6, 16 – 2) = minimum(13,16,14) =13 time units. Duration of activity G is 5 time units. Thus, latest time of event 6 = latest time of event 7 – duration of activity G = 13 – 5 =8 time units. Duration of activity E is 4 time units. Thus, latest time of event 4 = latest time of event 6 – duration of activity E = 8 – 4 =4 time units. Duration of activity F is 3 time units. Thus, latest time of event 5 = latest time of event 6 – duration of activity F = 8 – 3 =5 time units. Again, duration of activity C is 1 time unit. Thus, latest time of event 2 = latest time of event 4 – duration of activity C = 4 – 1 =3 time units. Duration of activity D is 2 time units. Thus, latest time of event 3 = latest time of event 5 – duration of activity D = 5 – 2 =3 time units.

8. For event 1, we can reach event 1 in backward pass method, through 2 different paths – through event 2, and through event 3. Thus the latest time of event 1 = minimum(latest time of event 2 – duration of activity A, latest time of event 3 – duration of activity B) = minimum(3 – 2, 3 – 3) = minimum(1,0) =0 time unit, as shown in the right half of the lower half circle of event 1. Now, based on the above earliest and latest event times, we can find out the project length or project duration which is the length of the longest path. This longest path is known as CRITICAL PATH. The critical path of a network is that path on which each of the events have equal earliest and latest event times. In the above network, such event as observed are: events 1, 3, 5, 6, 7, 8, 11, and 12. Thus the critical path of this network is: B – D – F – G – H – K – N, as shown by the double arrow in the above network. The project duration is the length of this path which is also equal to the latest time of the last event, that is, 34 time units.

9. Calculate the earliest and latest event times of the above network. HOMEWORK PLEASE COMPLETE THE HOMEWORK AS GIVEN ON THE NEXT SLIDE AND SEND THE ANSWER TO ME BY MAY 25, 2020