exercise answers - thomas uher

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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 Press2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN

Simultaneously published in the USA and Canada by Spon Press270 Madison Avenue, New York, NY 10016

Spon Press is an imprint of the Taylor & Francis Group, an informa business

© 2011 Thomas E Uher and Adam Zantis

The right of Thomas E Uher and Adam Zantis to be identified as authors of thiswork has been asserted by them in accordance with sections 77 and 78 of the

Copyright, Designs and Patents Act 1988.

All rights reserved. [No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, nowknown or hereafter invented, including photocopying and recording, or in anyinformation storage or retrieval system, without permission in writing from the

publishers.

The publisher makes no representation, express or implied, with regard to the

accuracy of the information contained in this book and cannot accept any legalresponsibility or liability for any errors or omissions that may be made.



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 toimprove 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. Wetrust you will find Answers to Exercises a useful supplement to the textbook. Weare confident that it will improve your understanding of the programming andscheduling techniques introduced in the textbook, and make your study mucheasier 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

K

G

E J L

N

M



3


Precedence schedule


Precedence schedule



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 varyingit or by splitting it.



8

Cumulative Hoist Lifting Schedule

WeekNo.

Hoist time/weekin hours

Cumulative hoist timein hours

1 28 28

2 36 643 2 66

4 8 745 4 78

6 16 947 26 120

8 10 130

9 1 13110 2 133

11 11 14412 9 153

13 9 16214 7 169

15 169016 169

17 16918 169

19 16920 169

21 141

22 10523 103

24 9525 91

26 7527 49

28 3929 38

30 3631 25

32 1633 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 3weeks to complete 18 days 18 days 18 days

4. The structure will take 34 weeks to complete once the jump form isinstalled. There are approximately 329 load lifts required per floor withthe 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 scaleTake 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 days6. The roof plantroom is to be constructed from structural steel with the

largest steel member weighing 5 tonnes and being located 45 m from thegoods lift shaft. OK OK

NOT

OK

Test Crane load = tonne/metre * metre. Final load to be confirmed bycrane 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 thegoods 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 t8. 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 bereviewed and a crane selected based on the crane speed, maximum reach, capacity at themaximum 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 therequirements.



11

ANSWERES TO EXERCISES IN

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




12




13




14


CHAPTER 6 (pp. )




15



16



17




18



19




20



21



22



23



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 $10,0002 $10,000 $5,000 $4,000 $4,000 $10,0003 $10,000 $5,000 $5,000 $4,000 $8,0004 $10,000 $5,000 $6,000 $4,000 $6,6675 $10,000 $5,000 $4,000 $5,000 $12,5006 $10,000 $5,000 $6,000 $5,000 $8,3337 $10,000 $5,000 $4,000 $6,000 $15,0008 $10,000 $5,000 $5,000 $6,000 $12,0009 $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 $7,50012 $10,000 $5,000 $7,000 $6,000 $8,57113 $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 TradesA, 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 2time units. Discontinuity in the use of the committed resources continues in Trades Band C.

A clearer picture of the use of resources can be deduced by converting a critical pathschedule to a MAC schedule.



26

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



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 alsochanges substantially from one level to the next.



30



31


The first truck will be unloaded in 54 minutes when the 8 th 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 iscalculated 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 40 th 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 the1st 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 160B 4 160 4 164C 6 240 8 248D 3 120 47 167E 8 320 14 334F 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 zonesActivity Time duration

in days

Total

construction

time in days

Start of activity

in days

Finish of

activity in

days

A 4 160 0 160B 4 160 10 170C 6 240 20 260D 3 120 59 179E 8 320 32 352F 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 onday 206.



37

The LOB schedule in the form of a graph



39

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

in weeks

Productionrate per

week

Number

of crews

Totalduration

in weeks

Start of

activity

Finish of

activity

A 3 0.33 6 51 0 51B 3 0.33 6 51 4 55C 2 0.50 4 50 4 54D 3 0.33 6 51 4 55E 2 0.50 4 50 8 58F 4 0.25 8 50 11 61G 2 0.50 4 50 14 64H 4 0.25 8 50 19 69I 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.



41


The MAC schedule shows that the present method of work that uses 1 m3 kibble candeliver 1 m3 of concrete every 4.5 minutes. Consequently, the delivery of 130 m3 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 m3 of concreteto the working floor would be 5.5 minutes. The delivery of 130 m 3 of concrete wouldthen take ((130 / 1.5) x 5.5 min.) / 60 = 7.9 hours, which is within the limit of deliveryhours.



42


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


S = (1.77 + 0.69 + 0.25 + 0.69) = 3.4S = 1.84

Ts – Te 25 – 23.5z = = = 0.815

S 1.84

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


Precedence schedule

exercise answers - thomas uher

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