Problem involving Earliest Start Times in this Activity Network

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chwala
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Homework Statement
Kindly consider the attached diagram and its solution.
Relevant Equations
decision maths
1741388694875.png

Now the steps seem to be pretty clear on this ms below;
1741388749493.png



My challenge is on the highlighted, in short i note that EST is the earliest time that an activity can start. No challenge on that and by that we have to consider all dependencies. Now i can note that we have,
##B(7) + D(4) + E(5) + J(7) = 23##
##23 + K(5) =28##

that aspect is clear with me as also applies to other nodes. Now my question is the ##30## how to arrive at that? this is a new area to me but i have taken the challenge to self study.
Cheers.


aaaaaah late event time...i should be able to master that! from final node i can see that going backwards we have ##33-3=30## ...if that is correct then i will proceed.
 
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chwala said:
from final node i can see that going backwards we have ##33-3=30## ...if that is correct then i will proceed.

Yes, that is the correct way to enter the late finish times, but why have you entered 11 as the early start time in the top left box? There is a different path that allows this event to start at t = 5. That is not the only error you have made.
 
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That is not my working but a copy of mark scheme from one of the international exam boards...
 
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chwala said:
That is not my working but a copy of mark scheme from one of the international exam boards...

Ah, I have misinterpreted the question, let me have another look.
 
Sorry, I was thrown by the format. The answer is correct, as is your comment
chwala said:
from final node i can see that going backwards we have ##33-3=30## ...if that is correct then i will proceed.

Silly me o:)
 
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...just to get this clear. The project will take 33 days, correct. Now to my question, Calculate the maximum number of days by which activity H could be delayed without lengthening the completion time of the project?

My thinking; Earliest start ##9 +4 = 13##

Latest start ##27-4 = 23##

Therefore, Float = ##23-13=10##

on the contrary, if the question had asked for maximum number of days for activity H, where delayed in completing project is allowed, then how would we proceed? I guess ##79##
 
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chwala said:
...just to get this clear. The project will take 33 days, correct.
Correct.

chwala said:
My thinking; Earliest start ##9 +4 = 13##
No, the early start (I prefer the terms 'late' and 'early' rather than 'latest' and 'earliest') for H is 9 (the top number in its predecessor node): 13 is the early finish for H.

chwala said:
Latest start ##27-4 = 23##

No, the late start for H is 23 - 4 = 19 (the late finish (the bottom number) for its successor node - H's duration).

I am not a fan of "activity on arrow" networks, particularly when times are shown on nodes. An "activity on node" (AON) network with ES, EF, LF and LS shown for each task is much easier to work with and to easily see the critical path(s) and the float for each activity.

chwala said:
on the contrary, if the question had asked for maximum number of days for activity H, where delayed in completing project is allowed, then how would we proceed? I guess ##79##
I don't understand what you are asking here.