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CPM with mason resource aggregation chart

Short answer

Project duration: 20 days, critical path: 1–2–4–7–9–11–12. With early start schedule, mason requirement peaks at 12 on days 8–9; total mason‑days = 101.

Draw the network diagram of the given project having the following activities. Obtain the project duration, critical path, TF, FF and interfering float. Prepare the resource aggregation chart and allocate the mason using early start schedule.

ActivityDuration (days)Mason (per day)
1-231
2-332
2-444
2-522
3-1032
4-623
4-743
5-944
6-822
7-941
8-932
9-1134
10-1122
11-1221

The activities are given by their event numbers (i-j), so each activity is an arrow from event i to event j.

Network diagram

(1) --1-2(3)--> (2)
(2) --2-3(3)--> (3)
(2) --2-4(4)--> (4)
(2) --2-5(2)--> (5)
(3) --3-10(3)--> (10)
(4) --4-6(2)--> (6)
(4) --4-7(4)--> (7)
(5) --5-9(4)--> (9)
(6) --6-8(2)--> (8)
(7) --7-9(4)--> (9)
(8) --8-9(3)--> (9)
(9) --9-11(3)--> (11)
(10) --10-11(2)--> (11)
(11) --11-12(2)--> (12)

Event times

TE from the forward pass, TL from the backward pass (days).

Event123456789101112
TE03675911111591820
TL031371110111215161820

Activity times and floats

TF=TLj−TEi−DTF = TL_j - TE_i - D, FF=TEj−TEi−DFF = TE_j - TE_i - D, IntF=TF−FFIntF = TF - FF.

ActivityDMasons/dayESTEFTLSTLFTTFFFInt.F
1-2310303000
2-332361013707
2-4443737000
2-52235911606
3-1032691316707
4-62379810101
4-743711711000
5-944591115660
6-8229111012101
7-94111151115000
8-93211141215110
9-113415181518000
10-11229111618770
11-122118201820000

Project duration and critical path

  • Critical activities (TF = 0): 1-2, 2-4, 4-7, 7-9, 9-11, 11-12
  • Critical path: 1-2-4-7-9-11-12 (3 + 4 + 4 + 4 + 3 + 2 = 20 days)

Answer: project duration = 20 days; critical path = 1-2-4-7-9-11-12.

Resource aggregation chart (early start schedule)

Each activity starts at its EST and needs the given number of masons every day until its EFT. The masons of all activities working on a given day are added.

Day12345678910
Masons11188101012127
Day11121314151617181920
Masons7333144411
 12 |              # #                       
 11 |              # #                       
 10 |          # # # #                       
  9 |          # # # #                       
  8 |      # # # # # #                       
  7 |      # # # # # # # #                   
  6 |      # # # # # # # #                   
  5 |      # # # # # # # #                   
  4 |      # # # # # # # #         # # #     
  3 |      # # # # # # # # # # #   # # #     
  2 |      # # # # # # # # # # #   # # #     
  1 |# # # # # # # # # # # # # # # # # # # # 
    +----------------------------------------
     1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0  day (last digit)
  • Peak requirement = 12 masons (days 8 and 9).
  • Total mason-days = 101 (check: the sum of duration x masons over all activities = 101).
  • Average = 101/20 = 5.05, i.e. about 5 masons per day.

Allocation of masons

The early start allocation needs only 1 mason on days 1-3, then 8 to 12 masons on days 4-9, and then 7, 3, 1, 4 and 1 masons on the later days. The demand is very uneven. Because the non-critical activities 2-3, 2-5, 3-10, 5-9 and 10-11 have a total float of 6-7 days, they can be delayed to cut the peak (for example, by starting 5-9, which needs 4 masons, later) without extending the 20 days. This is resource smoothing.

✓ Answer checked by Claude Opus 5.5 (high effort)

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