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ANSWERS TO EXERCISES

IN

PROGRAMMING AND SCHEDULING

TECHNIQUES 2

nd edition

Thomas E. Uher Adam Zantis

This edition published 2011

by Spon Press

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Simultaneously published in the USA and Canada

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© 2011 Thomas E Uher and Adam Zantis

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ANSWERS TO EXERCISES

IN PROGRAMMING AND SCHEDULING

TECHNIQUES

2nd

edition

Thomas E. Uher

Adam Zantis

FOREWORD

This Answers to Exercises document supplements the Programming and

Scheduling Techniques textbook. It contains worked solutions to exercises set out

in most chapters of the textbook. The exercises have been carefully formulated to

improve your comprehension of important topics explained in the textbook and

to enable you to self-test your knowledge. Upon accessing Answers to Exercises

on the Spon Press website, you may peruse this document, download it or even

print it free of charge.

The most effective way of using Answers to Exercises is for you to solve or

attempt to solve individual problems first before looking up the answers. We

trust you will find Answers to Exercises a useful supplement to the textbook. We

are confident that it will improve your understanding of the programming and

scheduling techniques introduced in the textbook, and make your study much

easier and more enjoyable.

T.E. Uher

A. Zantis

CONTENTS

ANSWERS TO EXERCISES IN CHAPTER 3 5








1

ANSWERS TO EXERCISES IN

CHAPTER 3 (pp )

Solution to exercise 3.1

Precedence schedule

E

A

J

C

D

G

F

L

H

K

M

N

O

P

B

2


Precedence schedule


Precedence schedule

E

A

B M

C

D

G

H

J

F

L

K

N

O

Q

P

A

F

B

C D

H

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G

E J L

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M

3


Precedence schedule


Precedence schedule

4


5


CHAPTER 4 (pp...........)

Solution to exercise 4.1 (a)

Solution to exercise 4.1 (b)

6

Solution to Exercise 4.2 (a)

Solution to exercise 4.2 (b)

A more even distribution of the total daily labour resource may be achieved by varying

it or by splitting it.

7


Hoist Lifting Table

Trade

Contract

Activity No. of loads/

floor

Cycle/

floor (min.)

Activity time/

floor

(hrs)

Total time/

floor

(hrs)

Cumulat.

time (hrs)

1 Formwork 100 15 25 28 28

Contingency 10% 3

2 Reinforcement 40 15 10 36 64

Concrete 170 7 20

Conduits & cables 5 30 3

Contingency 10% 3

3 Handrails 6 15 2 2 66

Contingency 10% 0

4 A/C ducts 20 15 5 8 74

Sprinkler pipes 10 15 3

Contingency 10% 1

5 Plumbing stock 5 30 3 4 78

Lift rails 3 30 2

Contingency 10% 0

6 Bricks 15 15 4 16 94

Mortar 10 15 3

Windows 7 60 7

Door frames 3 30 2

Contingency 10% 1

7 Electrical 8 60 8 26 120

Plaster 30 15 8

Glazing 8 60 8

Contingency 10% 2

8 Ceiling frames 4 30 2 10 130

Wall & floor tiles 20 20 7

Contingency 10% 1

9 Toilet partitions 2 30 1 1 131

Contingency 10% 0

10 Plumbing fixtures 2 60 2 2 133

Contingency 10% 0

11 Ceiling tiles 8 30 4 11 144

Lights 6 60 6

Contingency 10% 1

12 Lift doors 17 30 9 9 153

Contingency 10% 1

13 Doors 2 30 1 9 162

Vanity units 3 60 3

Venetian blinds 1 60 1

Mirrors 3 60 3

Contingency 10% 1

14 Induction units 2 30 1 7 169

Lift lobby finish 12 20 4

Door hardware 4 15 1

Contingency 10% 1

8

Cumulative Hoist Lifting Schedule

Week No.

Hoist time/week in hours

Cumulative hoist time in hours

1 28 28

2 36 64

3 2 66

4 8 74

5 4 78

6 16 94

7 26 120

8 10 130

9 1 131

10 2 133

11 11 144

12 9 153

13 9 162

14 7 169

15 169

016 169

17 169

18 169

19 169

20 169

21 141

22 105

23 103

24 95

25 91

26 75

27 49

28 39

29 38

30 36

31 25

32 16

33 7

34 0

9

Hoist Lifting Schedule


Crane 1 Crane 2 Crane 3

1. The lift shaft will be built to level 3 (3 storeys) prior to installation of

crane and will take 4 weeks to complete 28 days 28 days 28 days

2. The crane will be installed within the only Goods Lift 5 days 3 days 3 days

3. The jump form system will be installed using the crane and will take 3

weeks to complete 18 days 18 days 18 days

4. The structure will take 34 weeks to complete once the jump form is

installed. There are approximately 329 load lifts required per floor with

the average load weighing 4 tonnes and distance of 200 m NOT

OK OK OK

Test Crane Speed = Loads/floor * no. of floors * cycle time per load =

x, then convert to time scale

Take Crane 1 for example = ((329 loads/floor * 34 floors * 12 min/load)

/ (60 min x 8 hrs))/ 6 days 47 weeks 35 weeks 31 weeks

5. The jump form system removal can take place after the structure is

complete and will take 3 weeks to complete. 18 days 18 days 18 days

6. The roof plantroom is to be constructed from structural steel with the

largest steel member weighing 5 tonnes and being located 45 m from the

goods lift shaft. OK OK

NOT

OK

Test Crane load = tonne/metre * metre. Final load to be confirmed by crane supplier and structural engineer.

Take Crane 1 for example = 8.25 tonne / 60 m * 45 m, then check with

structural engineer & crane supplier 6.2 t 5.4 t 4.5 t

7. Heaviest permanent plant weighs 7 tonnes and is located 40 m from the

goods lift shaft. OK OK

NOT

OK

Test Crane load = tonne/metre * metre. Final load to be confirmed by

crane supplier and structural engineer.

Take Crane 1 for example = 8.25 tonne / 60 metres * 40 m, then check

with structural engineer & crane supplier 5.5 t 4.8 t 4.0 t

8. The crane can be removed after the final piece of plant is lifted into

position and jump form removed. 6 days 4 days 4 days

10

In selecting the appropriate crane for the project, all project information needs to be

reviewed and a crane selected based on the crane speed, maximum reach, capacity at the

maximum reach & average cycle time per lift. The project particular information should

be tabulated as shown in the above table and each crane's ability to meet the project

particular information should be analysed. The crane that can carry all heavy loads at

the required distances and has the most efficient cycle time should then be selected.

Based on the requirements of in the above exercise, Crane 2 appears to meet the

requirements.

11

ANSWERES TO EXERCISES IN

CHAPTER 5 (pp...........)


12


13


14


CHAPTER 6 (pp. )


17


19


24

Solution to exercise 6.4 (adapted from Burke, 1999, p 213)

The EAC calculations are performed using the following equation:

EAC = (ACWP/BCWP) x BAC

Cases BAC BCWS BCWP ACWP EAC

1 $10,000 $5,000 $5,000 $5,000 $10,000 2 $10,000 $5,000 $4,000 $4,000 $10,000

3 $10,000 $5,000 $5,000 $4,000 $8,000

4 $10,000 $5,000 $6,000 $4,000 $6,667 5 $10,000 $5,000 $4,000 $5,000 $12,500

6 $10,000 $5,000 $6,000 $5,000 $8,333

7 $10,000 $5,000 $4,000 $6,000 $15,000

8 $10,000 $5,000 $5,000 $6,000 $12,000 9 $10,000 $5,000 $6,000 $6,000 $10,000

10 $10,000 $5,000 $3,000 $4,000 $13,333

11 $10,000 $5,000 $4,000 $3,000 $7,500 12 $10,000 $5,000 $7,000 $6,000 $8,571

13 $10,000 $5,000 $6,000 $7,000 $11,667

Case 1: The project is on schedule and within cost budget.

Case 2: The project is behind schedule but within cost budget.

Case 3: The project is on schedule and under cost budget.

Case 4: The project is ahead of schedule and under cost budget.

Case 5: The project is behind schedule and over cost budget.



Case 8: The project is on schedule but over cost budget.

Case 9: The project is ahead of schedule and within cost budget.


Case 11: The project is behind schedule but under cost budget.


Case 13: The project is ahead of schedule but over cost budget.

25


CHAPTER 9 (pp............)


The original schedule provides continuity of resource use for Trade A only (verify this

by examining the earliest start and finish dates). Trade B is discontinuous as is Trade C.

In Trade C, two activities ‘Level 4’ and ‘Level 5’ compete for the same resource.

With introduction of resource restraints, which ensure a logical progression of Trades

A, B and C through the structure, the overlap between the activities ‘Level 4’ and

‘Level 5’ in Trade C was eliminated. However, the project duration was extended by 2

time units. Discontinuity in the use of the committed resources continues in Trades B

and C.

A clearer picture of the use of resources can be deduced by converting a critical path

schedule to a MAC schedule.

26

The first three columns of MAC show the converted original schedule. The

discontinuity in the use of the committed resources is clearly apparent as is the overlap

between the activities ‘Level 4’ and ‘Level 5’ in Trade C.

The next three columns show the converted schedule with the resource restraints. The

planner can now adjust the schedule to eliminate or minimise discontinuity in the use of

resources. The adjusted MAC is shown in the last three columns.

27


Alternative 1

Quantities of materials per floor and crew sizes have been calculated and are given in

the table below.

Activities Quantities Production rates

Person hours

Activity duration in hours

Crew size

Setout ceiling grid 8 4

2

Ceiling hangers

@ 2m centres

16x11 = 176 0.1 person

hrs/hanger

say 20 10 2

Ceiling frame (31x20m) + (41x30m)

= 1,850m

50m/person

hr

37 say

20

2

Ceiling tiles 30mx20m = 600m2 12.2

m2/person

48 24 2

Sprinkler heads

@ 3m centres

9x6 = 54 3/person 18 say

10

2

Light fittings

@ 3m centres

9x6 = 54 6/person 9 say

10

1

Aircon. registers

@ 4m centres

7x4 = 28 3/person 10 10 1

In this alternative, crew sizes have been kept at the maximum of 2 persons per crew per

activity. To speed up ‘Ceiling fixing’, it is assumed that 2 crew of 2 persons each will

work at the same time. The following MAC shows the arrangement of work for all the

crews.

28

While the continuity of work of the ceiling fixing crews has been maintained, the other

crews work discontinuously. The pattern of work of the ceiling fixing crews also

changes substantially from one level to the next.

29

Alternative 2

The previous solution requires crews to move from floor to floor. Let’s try to work out a

better solution by changing crew sizes. The adjusted MAC now provides a more

satisfactory solution.

Activities Quantities Production rates

Person hours

Activity duration in hours

Crew size

Setout ceiling grid 8 4

2

Ceiling hangers

@ 2m centres

16x11 = 176 0.1 person

hrs/hanger

say 20 20 1

Ceiling frame (31x20m) + (41x30m)

= 1,850m

50m/person

hr

37 10 4

Ceiling tiles 30mx20m = 600m2 12.2

m2/person

48 12 4

Sprinkler heads

@ 3m centres

9x6 = 54 3/person 18 10 2

Light fittings

@ 3m centres

9x6 = 54 6/person 9 9 1

Aircon. registers

@ 4m centres

7x4 = 28 3/person 10 10 1

31


The first truck will be unloaded in 54 minutes when the 8th pallet has been loaded with

bricks. The cycle time of moving three pallets through the system is 21 minutes.

32


Precedence schedule

33

MAC schedule

Solution to Exercise 4.

34


Preliminary MAC schedule

35

Final MAC schedule

36


CHAPTER 10 (pp. ..............)


With only one crew per activity, the total construction time in days per activity is

calculated first. For example, the activity ‘A’ will take 4 days x 40 floors = 160 days.

Durations of the other activities are given in the following table in the third column.

Start and finish dates of the repetitive activities are given in columns 4 & 5. For

example, the activity ‘D’, which is faster per cycle of work than its preceding activity

‘B’ will be scheduled from the end of the 40th completed activity ‘B’, which is day 164.

The 40th activity ‘D’ will then be completed 3 days later or on day 167. The start of the

1st activity ‘D’ will then be 167 – 120 = 47.

The LOB table

Activity Time duration

in days

Total

construction

time in days

Start of activity

in days

Finish of

activity in

days

A 4 160 0 160

B 4 160 4 164

C 6 240 8 248

D 3 120 47 167

E 8 320 14 334 F 4 160 178 338

Let’s now add time buffer zones of 6 days and recalculate start and finish dates of the

activities.

The LOB table with buffer zones Activity Time duration

in days

Total

construction

time in days

Start of activity

in days

Finish of

activity in

days

A 4 160 0 160

B 4 160 10 170

C 6 240 20 260 D 3 120 59 179

E 8 320 32 352

F 4 160 202 362

The LOB schedule for the project in question is given below.

40 typical floors will be constructed in 362 days with the first floor fully completed on

day 206.

37

The LOB schedule in the form of a graph

38


The production rates per week are calculated first from activity durations. They are

given in the following LOB table in column 3. To achieve the required production

output of 2 service stations per week, multiple crews are introduced to each activity, see

column 4. Please remember that with multiple crews durations of activities per

repetitive cycle will remain the same. Crews in each activity will work concurrently.

For example, the activity ‘A’ will continue to take 3 weeks per service station, however,

there will be 6 of these activities built concurrently with 6 crews.

The total duration in weeks per activity will be calculated next, see column 5. Since the

crew numbers per activity are uneven, these durations may need to be adjusted later.

Star and finish dates of each activity will be determined in the same manner as in

Exercise 10.1, see columns 6 & 7.

The LOB table

Activity Time

duration

in weeks

Production

rate per

week

Number

of crews

Total

duration

in weeks

Start of

activity

Finish of

activity

A 3 0.33 6 51 0 51

B 3 0.33 6 51 3 54

C 2 0.50 4 50 3 53

D 3 0.33 6 51 3 54

E 2 0.50 4 50 6 56

F 4 0.25 8 50 6 56

G 2 0.50 4 50 8 58

H 4 0.25 8 50 10 60

I 2 0.50 4 50 12 62

Let’s examine the impact of having an uneven number of multiple crews of workers.

The activity ‘F’ requires 8 crews of workers while the preceding activities ‘C’ and ‘D’

only 4 & 6 respectively. It means that on week 6 only six crews of the activity ‘F’ will

be able to star. It will therefore be necessary to delay the start of the activity ‘F’ until all

of its crew could star, which will be on week 9. The activity ‘F’ will then be completed

on week 59, see the following table. The start and finish dates of the activities ‘G’ &

‘H’ will be recalculated as 11 & 61 and 13 & 63 respectively.

The adjusted LOB table

Activity Time

duration

in weeks

Production

rate per

week

Number

of crews

Total

duration

in weeks

Start of

activity

Finish of

activity

A 3 0.33 6 51 0 51

B 3 0.33 6 51 3 54

C 2 0.50 4 50 3 53

D 3 0.33 6 51 3 54

E 2 0.50 4 50 6 56

F 4 0.25 8 50 9 59

G 2 0.50 4 50 11 61

H 4 0.25 8 50 15 65

I 2 0.50 4 50 17 67

The activity ‘H’ requires eight crews of workers while the preceding activities ‘G’ and

‘E’ only four each. It means that on week 13 only four crews of the activity ‘H’ will be

39

able to start. By delaying the activity ‘H’ by two weeks, all of its crews will be able to

commence work. Therefore, the activity ‘H’ will start on week 15 and finish on week

65. The activity ‘I’ will then start on week 17 and finish on week 67.

The final start and finish dates of the activities are given in the following table.

The adjusted LOB table with buffer zones

Activity Time

duration

in weeks

Production

rate per

week

Number

of crews

Total

duration

in weeks

Start of

activity

Finish of

activity

A 3 0.33 6 51 0 51

B 3 0.33 6 51 4 55

C 2 0.50 4 50 4 54

D 3 0.33 6 51 4 55

E 2 0.50 4 50 8 58

F 4 0.25 8 50 11 61

G 2 0.50 4 50 14 64

H 4 0.25 8 50 19 69

I 2 0.50 4 50 22 72

The project will completed in 72 weeks with the first service station delivered on week

24. The crew sizes are sufficient to meet the contract requirements.

40


CHAPTER 11 (pp. .............)


Standard time = Basic time + Relaxation allowance + Contingency

Basic time = Observed time x

Utilisation =

Basic time = 6 min. x (120/100) = 7.2 minutes

Standard time = 7.2 + (7.2 x 50/100) = 10.8 minutes

Utilisation = ((10.8 x 160) / (4days x 8hrs x 60min)) x 100% = 90%

Assessed rating

Total standard time of work

Standard rating

Time of work available x 100%

41


The MAC schedule shows that the present method of work that uses 1 m3 kibble can

deliver 1 m3 of concrete every 4.5 minutes. Consequently, the delivery of 130 m

3 of

concrete will take (130 x 4.5 min.) / 60 or 9.8 hours. However, since the delivery of

materials is restricted to only 8.5 hours during the working day, this method of

distributing concrete is inadequate.

If the contractor engaged 1.5 m3 kibble, the cycle time of delivering 1.5 m

3 of concrete

to the working floor would be 5.5 minutes. The delivery of 130 m3 of concrete would

then take ((130 / 1.5) x 5.5 min.) / 60 = 7.9 hours, which is within the limit of delivery

hours.

42


CHAPTER 13 (pp............)


S = (1.77 + 0.69 + 0.25 + 0.69) = 3.4

S = 1.84

Ts – Te 25 – 23.5

z = = = 0.815

S 1.84

The probability of completing the project within 25 weeks is 79%.


Precedence schedule

43


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