line diagrams

LINE DIAGRAMS

Learning objectivesAfter this lecture you will be able to:• Prepare and use line diagrams• Demonstrate the effect that making activities ‘continuous’ has on a project• Describe why this method is particularly suitable for repetitive work

Quantity

Activity A Activity B80%

20%

Time Day X

Example

Project = To build six partly prefabricated houses where each house consists of the following four activities.

A B C D

A = Excavation/Services takes 2 weeks/houseB = Footings takes 3 weeks/houseC = Superstructure takes 4 weeks/houseD = 2nd Fix takes 2 weeks/house

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 House 1. A B C DHouse 2. A B C DHouse 3. A B C DHouse 4. A B C DHouse 5. A B C DHouse 6. A 31 weeks B C D

0 2 2 5 5 9 9 11House 1 A/2 B/3 C/4 D/2 0 2 2 5 5 9 11 13 2 4 5 8 9 13 13 15House 2 A/2 B/3 C/4 D/2 4 6 6 9 9 13 15 17 4 6 8 11 13 17 17 19House 3 A/2 B/3 C/4 D/2 8 10 10 13 13 17 19 21 6 8 11 14 17 21 21 23House 4 A/2 B/3 C/4 D/2 12 14 14 17 17 21 23 25 8 10 14 17 21 25 25 27House 5 A/2 B/3 C/4 D/2 16 18 18 21 21 25 27 29 10 12 17 20 25 29 29 31 = Project completion timeHouse 6 A/2 B/3 C/4 D/2

20 22 22 25 25 29 29 31

House No. 12 20 29 31 (Project compl.) 5 10 15 20 25 30 Weeks6

5

4 A B C D alt. D

3

2

1

2 5 9 19

ExampleThese are two alternative schedules for building six identical houses. In Schedule 1 it takes 15 weeks to build one house whilst in Schedule 2 it takes 20 weeks to build one house. Which alternative would result in the quickest project completion time if all activities must be continuous?

Schedule 1 3 2 4 2 4 Cycle = 15 weeks6 houses

Schedule 2 4 4 4 4 4 Cycle = 20 weeks6 houses

Schedule 1House No. 18 20 34 36 50 5 10 15 20 25 30 35 40 45 50 wks6

5

4 (3) (2) (4) (2) (4)3

2

1

8 10 24 26

Schedule 2House No. 24 28 32 36 40 5 10 15 20 25 30 35 40 45 50 wks6

5

4 (4) (4) (4) (4) (4)3

2

1

4 8 12 16

Demonstration of affect of continuous activities

Say we have a project consisting of six identical houses with each house having the three activities A, B and C.

Starting with all activities having a duration of 5 we will check the result of making each activity both longer and shorter.

A/5 B/5 C/5 Cycle time = 15 weeks

6 Houses

All equal: A/5 B/5 C/5 Cycle = 15 weeks

5 10 15 20 25 30 35 40 456 5

4

3

2

1

5 10

Say, we reduce Activity A to only 4 weeks on each of the six houses.

What affect will that have on the project completion time?

‘A’ reduced to 4: A/4 B/5 C/5 Cycle = 14 weeks

5 10 15 20 24 30 34 39 456 5

4

3 A saving of only ONE week

2

1

4 9

Say, we increase Activity A to 6 weeks on each of the six houses.


‘A’ increased to 6: A/6 B/5 C/5 Cycle = 16 weeks

5 10 15 20 25 30 36 41 466 5

4

3

2 A delay of SIX weeks

1

11 16

Say, we reduce Activity B to 4 weeks on each of the six houses.


‘B’ reduced to 4: A/5 B/4 C/5 Cycle = 14 weeks

5 10 15 20 25 30 34 40 44 6 5

4

3

2 A DELAY of FOUR weeks!

1

10 14

Say, we increase Activity B to 6 weeks on each of the six houses.


‘B’ increased to 6: A/5 B/6 C/5 Cycle = 16 weeks

5 10 15 20 25 30 35 41 46 6 5

4

3

2 A delay of SIX weeks!

1

5 16

Say, we reduce Activity C to 4 weeks on each of the six houses.


‘C’ reduced to 4: A/5 B/5 C/4 Cycle = 14 weeks

5 10 15 20 25 30 35 39 6 5

4

3

2 A saving of ONE week!

1

5 15

Finally, say we increase Activity C to 6 weeks on each of the six houses.


‘C’ increased to 6: A/5 B/5 C/6 Cycle = 16 weeks

5 10 15 20 25 30 35 46 6 5

4

3

2 A delay of SIX weeks!

1

5 10

SUMMARYReduce ‘A’ by one week/house: Saving of ONE week for projectExtend ‘A’ by one week/house: Delay of SIX weeks for projectReduce ‘B’ by one week/house: Delay of FOUR weeks for projectExtend ‘B’ by one week/house: Delay of SIX weeks for projectReduce ‘C’ by one week/house: Saving of ONE week for projectExtend ‘C’ by one week/house: Delay of SIX weeks for project

CONCLUSIONIt is difficult to predict the result of changing activity durations

when activities must be continuous and it is efficient to have parallel activities.

Project Evaluation and Review Technique (PERT)

Similar to the Critical Path Method, but an attempt to estimate the probability of meeting certain deadlines or milestones.

EXAMPLE

A B C

D E F

For each activity we select three possible activity durationsO = Optimistic durationM = Most likely durationP = Pessimistic duration (within reason e.g. not incl WW3 or similar)

Normal distribution

Activity O M P A 1 2 3 (weeks) B 2 4 7 C 1 2 5 D 1 2 3 E 2 4 6 F 2 3 7

Expected activity duration

te = O + 4xM + P

6

In our example we get these te-valuesA. te = (1 + 4*2 + 3)/6 = 2.00B. te = (2 + 4*4 + 7)/6 = 4.17C. te = ( 1 + 4*2 + 5)/6 = 2.33D. te = (1 + 4*2 + 3)/6 = 2.00E. te = (2 + 4*4 + 6)/6 = 4.00F. te = (2 + 4*3 + 7)/6 = 3.50

Use the te-values as activity durations and analyse the network A/2.00 B/4.17 C/2.33 D/2.00 E/4.00 F/3.50

Use the te-values as activity durations and analyse the network 0 2.00 2.00 6.17 6.17 8.50 A/2.00 B/4.17 C/2.33 0 2.00 2.00 6.17 6.17 8.50 2.00 4.00 4.00 8.00 8.50 12.00 D/2.00 E/4.00 F/3.50 2.50 4.50 4.50 8.50 8.50 12.00

CP = A – B – C – F Expected project completion time is 12.00 weeks

STATISTICSIf a series of Variables t1, t2, t3,…, tn

have theMean Times te1, te2, te3,…, ten and the

Standard Deviations S1, S2, S3,…, Sn, then the expected total time

T = t1 + t2 + t3 + …+tn follows a Normal Distribution with a

Mean Value Te = te1 + te2 + te3 +…+ ten and the

Variance V = S^2 = S1^2 + S2^2 + S3^2 +…Sn^2

Say: te1 +/- S1 te2 +/- S2 te3 +/- S3

Then: Te = te1 + te2 + te3 +/- S1^2 + S2^2 + S3^2

So we’ll use the following formula to determine the expected project completion time

Te = te +/- S2

To calculate the Standard Deviation for each activity we useThis formula S = P – O 6In our example we getA. S = (3 – 1)/6 = 0.33B. S = ( 7 – 2)/6 = 0.83C. S = (5 - 1)/6 = 0.67 D. S = (3 – 1)/6 = 0.33E. S = (6 – 2)/6 = 0.67 F. S = (7 – 2)/6 = 0.83

For our project we use the critical path activities to calculate the expected project completion time with its standard deviation.

Te = 2 + 4.17 + 2.33 + 3.50 +/- 0.332 + 0.832 + 0.672 + 0.832 = 12.00 +/- 1.39 weeks

1 SD = 1.39 weeks

12.00 weeksNext we need to look at a Normal Distribution table in order to

determine the probability of completing the project with a certain time.

SD .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

.1 .5398 .5435 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7703 .7734 .7764 .7794 .7823 .7852

NORMAL DISTRIBUTION TABLE

Read the whole number and the first decimal of the number of Standard Deviations down the left hand column and read the second decimal across to the right. The number at that intersection represents the probability. For example 0.56 SD has a probability of 0.7123 or ~71%

SD .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

.1 .5398 .5435 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7703 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .90147

1.3 .90320 .90490 .90658 .90824 .90988 .91149 .91309 .91466 .91621 .91774

1.4 .91924 .92073 .92220 .92364 .92507 .92647 .92785 .92922 .93056 .93189

1.5 .93319 .93448 .93574 .93699 .93822 .93943 .94062 .94179 .94295 .94408

1.6 .94520 .94630 .94738 .94845 .94950 .95053 .95154 .95254 .95352 .95449

1.7 .95543 .95637 .95728 .95818 .95907 .95994 .96080 .96164 .96246 .96327

1.8 .96407 .96485 .96562 .96638 .96712 .96784 .96856 .96926 .96995 .97062

1.9 .97128 .97193 .97257 .97320 .97381 .97441 .97500 .97558 .97615 .97670

2.0 .97725 .97778 .97831 .97882 .97932 .97982 .98030 .98077 .98124 .98169

2.1 .98214 .98257 .98300 .98341 .98382 .98422 .98461 .98500 .98537 .98574

2.2 .98610 .98645 .98679 .98713 .98745 .98778 .98809 .98840 .98870 .98899

2.3 .98928 .98956 .98983 .99010 .99036 .99061 .99086 .99110 .99134 .99158

2.4 .99180 .99202 .99224 .99245 .99266 .99286 .99305 .99324 .99343 .99361

2.5 .99379 .99396 .99413 .99430 .99446 .99461 .99477 .99420 .99506 .99520

2.6 .99534 .99547 .99560 .99573 .99586 .99598 .99609 .99621 .99632 .99643

2.7 .99653 .99664 .99674 .99683 .99693 .99702 .99711 .99720 .99728 .99736

2.8 .99744 .99752 .99760 .99767 .99774 .99781 .99788 .99795 .99801 .99804

2.9 .99813 .99819 .99825 .99830 .99836 .99841 .99846 .99851 .99856 .99860

3.0 .99865 .99869 .99874 .99878 .99882 .99886 .99889 .99893 .99896 .99900

3.1 .99903 .99906 .99910 .99913 .99915 .99918 .99921 .99924 .99926 .99929

3.2 .99931 .99934 .99936 .99938 .99940 .99942 .99944 .99946 .99948 .99950

3.3 .99952 .99953 .99955 .99957 .99958 .99960 .99961 .99962 .99964 .99965

3.4 .99966 .99968 .99969 .99970 .99971 .99972 .99973 .99974 .99975 .99976

3.5 .99977 .99978 .99978 .99979 .99980 .99981 .99981 .99982 .99983 .99983

What is the probability of completing the project in 12 weeks plus one Standard Deviation, i.e. 13.39 weeks?


1.00 SD = 0.8413 = 84%

Note that completion times longer than the mean value will have higher than 50% probability whilst shorter completion times will have probabilities of less than 50%. The table is only made up for 50% and greater probabilities.


1.00 SD = 0.8413 = 84%

and within 12 weeks plus two SDs, i.e. 14.78 weeks?

What is the probability of completing the project I 12 weeks plus one Standard Deviation, i.e. 13.39 weeks?

1.00 SD = 0.8413 = 84%

and within 12 weeks plus two SDs, i.e. 14.78 weeks?2.00 SD = 0.97725 = 98%


1.00 SD = 0.8413 = 84%


What about 12 weeks minus one SD, i.e. 10.61 weeks?


1.00 SD = 0.8413 = 84%


What about 12 weeks minus one SD, i.e. 10.61 weeks?The table is only set up for durations longer than the 12 weeks

i.e. with probabilities higher than 50% so we need to take 1.0 minus the probability we read in the table, so;

1.00 SD = 0.8413 but 1.0 – 0.8413 = 0.1587 = 16%

To find the probability of completion within a certain number of weeks we need to determine how many Standard Deviations that time represents.

To finish the project within 12 weeks represents zero Standard Deviations from the Mean value of 12.00.

0.00 SD = 0.500 = 50%

To find the probability of completion within a certain number of weeks we need to determine how many Standard Deviations that time represents. We had T = 12 +/- 1.39 weeks.


0.00 SD = 0.500 = 50%

To finish within 13 weeks represents (13-12)/1.39 = 0.72 SD0.72 SD = 0.7642 = 76%

To find the probability of completion within a certain number of weeks we need to determine how many Standard Deviations that time represents. We had T = 12 +/- 1.39 weeks.


0.00 SD = 0.500 = 50%

To finish within 13 weeks represents (13-12)/1.39 = 0.72 SD0.72 SD = 0.7642 = 76%

To finish within 11 weeks represents (12-11)/1.39 = 0.72 SD which has a probability of 0.7642 (as above) but since 11 is less than 12 the probability will be less than 50% so we get

1 – 0.7642 = 0.2358 = 23%

Tutorial example: Line DiagramsOne of your colleagues has handed you a schedule for a housing project consisting of 48 identical, partly prefabricated, houses. The schedule for each house is linear and contains the following activities with their respective duration (in days).

A/3 B/2 C/1 D/4 E/2 F/6 •Determine the shortest possible completion time for the whole project, as it has been scheduled.

•It is important that you manage to keep the workforce for the duration of the whole project. It is therefore essential that you can offer the workers continuous employment. How long would it take to complete the project if that condition was applied?

•You now realise that activities C and D utilise the same type of trade people, which means that you can distribute the workers between those two activities as you see fit. Assume that after the first 24 houses, you move some workers from activity C to activity D. This has the effect of changing the activity duration for the following 24 houses so that C takes 2 days and D takes 3 days. Determine the effect this has on the completion time for the whole project.

•Assume that all the workers, in all activities, were multi-skilled so that you could decide how to distribute them between all activities and thereby control the duration of each activity. The total number of working days per house must remain the same as before. How would you suggest that each activity’s duration should be modified in order to complete the project in the shortest possible time, whilst maintaining continuity? What would that completion time be?

Tutorial: PERTThe optimistic (O), most likely (M) and pessimistic (P) activity durations have been given for the project described by the network below. Use the scheduling technique PERT to;

•Calculate the probability of the whole project being completed within 15 weeks.

•Calculate which project completion time would give you at least a 90% probability of completion (in a whole number of weeks).

•Calculate the probability of activity 7-9 being able to start in week 11.

Activity O M P1-2 3 4 51-3 4 5 72-5 2 3 5 3 4 82-6 2 4 83-4 2 3 44-7 1 1 1 1 2 5 7 94-8 3 4 65-7 2 2 36-7 2 3 4 67-9 3 4 58-9 2 2 2

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