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CONS0009 Construction Planning and Control
LOB Exercise - Solution
Question 1.
A delivery rate of completed units in a housing project comprising 100 similar residential units may
be specified in the contract as at least 2 units/per week over a period of 50 weeks.
Assuming that the first housing unit will be fully completed on week 10, the contractor will then
need to delivery approximately 2.5 units per week over the remaining 40 weeks.
A) How many houses should be completed by Week 40?
B) How many Plumbing packages should be completed by Week 10?
C) How many Window packages should be completed by Week 20?
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Solution;
Footings Brickwork Roof Windows Painting Plaster Plumbing Clean- up 0
1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1
0 Join Elect
This means, Leading time for Footings is 8, Leading time for Brickwork is 6, Leading time for
Plumping is 3, Leading time for Roof and Windows is 5, Leading time for Plaster is 4, leading
time for Join and Elect is 3, Leading time for Painting is 1, and Leading time for Clean-up is
0.
0 20 40 60 80 100 0 10 20 30 40 50 Nu m be r o f U ni ts Time (Weeks) 76 9
38
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A) How many houses should be completed by Week 40?
At week 40, 75-77 units are completed
B) How many Plumbing packages should be completed by Week 10?
At week 10 we should add lead time for Plumbing that is 3, thus we look for week 13 (10+3).
Thus, at week 13, 9-11 Plumbing packages are completed
C) How many Window packages should be completed by Week 20?
At week 20 we should add lead time for window that is 5, thus we look for week 25 (20+5).
Thus, at week 25, 38-39 Window packages are completed
Question 2.
The following precedence diagram shows all the activities involved in the construction of one floor
of a high-rise building. The duration of each activity is also given (in days).
There are 30 typical
floors to be constructed for this high-rise building. Calculate the Start and Finish date for each
activity and fill in the following table.
A
3
B
2
E
3
C
1
D
5
F
3
G
2
4
Solution:
Activity
number
Duration
(days)
Total duration of
packages
(days)
Start of work
packages
(days)
Finish of work
packages
(days)
A 3 (3x30)= 90 0 90
B 2 (2x30)= 60 32 92
C 1 (1x30)= 30 63 93
D 5 (5x30)= 150 34 (32+2) 184
E 3 (3x30)= 90 64 (63+1) 154 (64+90)
F 3 (3x30)= 90 97 187
G 2 (2x30)= 60 129 189
Package (B) influenced by the completion of package (A). Package (B) duration per cycle is
faster than (A). Thus (B) will be scheduled from the finish of last (A) activity, therefore finish
of last activity (B) = 92 (90+2, duration of activity B). Start day of (B) is 92-60=32.
Package (C) influenced by the completion of package (B). Package (C) duration per cycle is
faster than (B). Thus (C) will be scheduled from the finish of last (B) activity, therefore finish
of last activity (C) = 93 (92+1, duration of activity C). Start day of (C) is 93-30=63.
Package (D) influenced by (B) which is longer per cycle than (B), then its start will be as
soon as first activity in package B is completed, 34 (32+2, duration of Activity B). the D
package finish day is 184 (34+150).
Package (E) influenced by (C), which is longer per cycle than (C), then its start will be 64
(63+1, duration of Activity C). the package finish day is 154 (64+90).
Package (F) influenced by (D) and (E). Let's take (E), since (F) duration per cycle is same as
(E), then its start will be 67 (64+3, duration of activity E). Let's take D, since (F) duration per
cycle is faster than (D), its finish of last activity will be scheduled on 187 (184+3, duration of
F activity). Thus, its start of work will be (187-90) = 97. Thus, start of work package (F) will
be on 97 (the larger).
Package (G) influenced by the completion of package (F). Package (G) duration per cycle is
faster than (F). Thus (G) will be scheduled from the finish of last (F) activity, therefore finish
of last activity (G) = 189 (187+2, duration of activity G). Start day of (G) is 189-60=129.
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Question 3.
50 similar houses will be built in a new development area. Each house involves 9 trade packages
given in the following diagram. Calculate the Start and Finish date (in weeks) for each trade
package and fill in the following table.
Solution:
ctivity number Duration Total duration Start Finish
A 2 (2x 50)=100 0 100
B 1 (1x50)= 50 51 101
C 2 (2x50)= 100 52 152
D 3 (3x50)= 150 52 202
E 3 (3x50)= 150 52 202
F 2 (2x50)= 100 104 204
G 2 (2x50)=100 104 204
H 2 (2x50)= 100 106 206
K 1 (1x50)= 50 157 207
Package (B) influenced by the completion of package (A). Package (B) duration per cycle is
faster than (A). Thus (B) will be scheduled from the finish of last (A) activity, therefore finish
of last activity (B) = 101 (100+1, duration of activity B). Start day of (B) is 101-50=51.
Package (C) influenced by the completion of package (B). Package (C) duration per cycle is
longer per cycle than (B), then its start will be as soon as first activity in package B is
completed, 52 (51+1, duration of Activity B). the C package finish day is 152 (52+100).
B
1
E
3
F
2
C
2
G
2
H
2
A
2
K
1
D
3
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Package (D) influenced by the completion of package (B). Package (D) duration per cycle is
longer per cycle than (B), then its start will be as soon as first activity in package B is
completed, 52 (51+1, duration of Activity B). the D package finish day is 202 (52+150).
Package (E) influenced by the completion of package (B). Package (E) duration per cycle is
longer per cycle than (B), then its start will be as soon as first activity in package B is
completed, 52 (51+1, duration of Activity B). the E package finish day is 202 (52+150).
Package (F) influenced by (C) and (D). Let's take (C), since (F) duration per cycle is same as
(C), then its start will be 54(52+2, duration of activity C). Let's take D, since (F) duration per
cycle is faster than (D), its finish of last activity will be scheduled on 204 (202+2, duration of
F activity). Thus, its start of work will be (204-100) = 104. Thus, start of work package (F)
will be on 104 (the larger).
Package (G) influenced by the completion of package (E). Package (G) duration per cycle is
faster than (E). Thus (G) will be scheduled from the finish of last (E) activity, therefore finish
of last activity (B) = 204 (202+2, duration of activity G). Start day of (G) is 204-100=104.
Package (H) influenced by (F) and (G). Since (H) duration per cycle is same as (F) and (G),
and both activities F and G starts same day, then its start will be 106 (104 +2, duration of
activity H).
Package (K) influenced by the completion of package (H). Package (K) duration per cycle is
faster than (H). Thus (K) will be scheduled from the finish of last (H) activity, therefore
finish of last activity (H) = 207 (206+1, duration of activity K). Start day of (K) is 207- 50=157.
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