Lags in precedence networks

Lag Time

• When an activity is completed, and there is a delay or wait period before the second activity starts, this is called lag and the delay is known as the Lag Time.

• A lag specifies an offset or delay between an activity and its successor

• Lags can be used to constrain the start and finish of an activity

• Can be a positive or a negative value

Example

• Suppose you have to paint a newly constructed room.

• So the first activity would be applying the primer coating and then you will go for the final painting.

• However, after applying the primer coating, you must give it some time to dry properly. Once the primer coating dries, you can start final painting.

• The time given for coating to dry itself is called the lag time.

Lag Time

• FS +1:• For example, the time duration

for the first activity is 3 days, and for the second activity it is 2 days.

• After completing the first activity you wait for one day, and then you start second activity.

• Lag Time is one day• Finish to Start activity with one

day delay or lag.

Example 1

Answer

Example 2

Answer

Example 3

Answer

Example: a successor finishes before the predecessor

Excavate-1 Subbase-7 Lay pipe-4 Backfill-3 Compact-2

1 2 1 1

In this example, FF relationships ensure that no successor can finish before its predecessor.

Subbase cannot start till at least 1 day after excavate has started.Subbase cannot finish earlier than 1 day after the finish of excavate.

Excavate Subbase-7 Lay pipe-4 Backfill-3 Compact-2

1 2 1 1

1 1 2 1

CPM calculations for precedence networks

CPM calculations for precedence diagrams are similar to, but not the same as the CPM calculations discussed previously.

The following example illustrate the calculations procedure.

A-5 B-7 C-3 Figure6

0 5

0 7

0 7

0 7 74

0 7

In the forward pass, Activity A starts the projects. ES=0, EF=0+5=5Activity B is connected to A by 2 relationships: SS and FF. ES(B)= ES (A)+lag(if any) = 0, However the early finish of B is controlled by the max of: EF(B)=ES(B) +Dur = 0+7=7 or EF(B)= EF(A)+lag(if any) = 5 Thus, EF(B)=7

The same method applies for C.ES (C)=ES(B)+lag(if any) = 0, However the early finish of C is controlled by the max of: EF(C)=ES(C) +Dur = 0+3=3 or EF(C)= EF(B)+lag(if any) = 7 Thus, EF(C)=7

In the backward pass, Activity C must finish no earlier than day 7. LF(C)=7, LS=7-3=4

Activity B must finish no later than the LF(C). LF(B)= LF(C)-lag(if any) = 7, However the latest start of B is controlled by the min of: LS(B)=LF(B) - Dur = 7-7=0 or LS(B)= LS(C)-lag(if any) = 4 Thus, LS(B)=0

Activity A must finish no later than the LF(B).LF (A)=LF(B)-lag(if any) = 7, However the latest start of A is controlled by the min of: LS(A)=LF(A) - Dur = 7-5=2 or LS(B)= LS(B)-lag(if any) = 0 Thus, LS(A)=0

A-5 B-7 C-3 Figure6

0 5

0 7

0 7

0 7 74

0 7

In figure6, activity A must start at 0 otherwise activity B, which is critical, will be delayed. Activity A has 2 days of restricted float(start-restricted float).

Activity C has 4 days of float at its start. However, no matter when it starts, it must finish on day 7.C has 4 days restricted float (finish-start float).

Unlike the arrow and node diagrams, precedence networks may show the start or finish of an activity to be critical while the rest of the activity is not.

A-10 B-3 C-5 Figure7

0 10

0 10

2 10

2 10 105

5 102 3

In figure7, both A and C are completely critical. Activity B has a critical start and a critical finish but is not critical.

Activity B must start on day 2 and must finish on day 10. note that B has a duration of 3 days, but must fill an 8-day interval. Thus B has 5days double-restricted float(start-finish-restricted float)

Example:

Activity Duration (weeks)

Immediate predecessors

Type of relation

Lag

A 2 ----

B 6 A

C 11 A SS 1

D 7 B SS, 2

FF 0

E 4 B

C FF 5

F 8 ----

G 3 D,E

F FF 4

Start

C-11

A-2 B-6 D-7

F-8

E-4 G-3

8

1

4

0 2

0 7

1 12

1 12

4 11

10 17

17 20

17 20

0 8

2

16

2 8

7 13

8 17

13 17

Forward pass, Activity A: starts the projects. ES=0, EF=0+2=2Activity B: ES(B)= EF (A)= 2, EF(B)=2+6=7Activity C: ES(C) = ES(A)+lag=1+0=1 , EF(C)= 1+11=12Activity D: ES(D) = ES(B)+lag=2+2=4 , EF(D)= max{4+7=11 or 8}=11Activity E: ES(E)=8, EF(E)= MAX{8+4=12 or 12+5=17}=17Activity F: ES(F)=0, EF(F)= 0+8=8Activity G: ES(G)=max{EF(E)=17,EF(B)=11}=17 , EF(G)=max{ES(G)+Dur=17+3=20 or EF(F)+Lag=8+4=12}=20

Backward pass, Activity G: LF(G)=20, LS(G)=20-3=17Activity F: LF(F)=LF(G)-Lag=20-4=16,LS(F)=16-8=8Activity E: LF(E) =17, LS(E)= 17-4=13Activity D: LF(D) = 17, LS(D)=17-7=10Activity B: LF(B)=min{LF(D)=17 or LS(E)=13}=13, LS(B)=min{LF(B)-Dur=13-6=7 or LS(D)-Lag= 10-2=8}=7Activity C: LF(C)=LF(E)-Lag= 17-5=12, LS(C)= 12-11=1Activity A: LF(A)=LS(B)=7, LS(A)=min{LS(C) -Lag =1-1=0 or LS(A)-Dur=7-2=5}=0

Critical Path

• Start of A-C- end of E-G • Duration : lag 1+ 11+lag 5 +3= 20