6- Time Cost Trade Off

8/10/2019 6- Time Cost Trade Off

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Tutorial # 6

Time-Cost Trade-Off

(Time Shortening)

King Saud University

College of Engineering

GE 404 Engineering Management



Basic Steps for crash calculation

1. Compute the crash cost per time period,considering crash costs are linear over time

2. Using current activity times, find the critical path and identify the

critical activities

3. Select the activity on the critical path of which (a) it can still be crashed,

and (b) it has the smallest crash cost per period.

o If there is more than one critical path, then select one activity from each

critical path such that (a) each selected activity can still be crashed, and (b)the total crash cost of all selected activities is the smallest.

o Note that the same activity may be common to more than one critical path

4. Update all activity times. If the desired due date has been reached, stop. If not,

return to Step 2





Network Compression Algorithm Update the project network

When a new Critical path is formed:

- Shorten the activity which Falls on Both Critical Paths, OR

- Shorten one activity from each of the critical paths. Use the combined cost

of shortening both activities when determining if it is cost effective to

shorten the project.

At each shortening cycle, compute the new project duration and project cost

Continue until no further shortening is possible

Use the total project cost-time curve to find the optimum time.

Tabulate and Plot the Indirect project Cost on the same time-cost graph

Add direct and indirect cost to find the project cost at each duration.



Example1:

The below network shows theactivities of a small engineering

project. Data of the project is given inthe below table. The indirect cost is

estimated to be SR 90/day . Determinethe optimum contract duration.

Activity

Code

Time (day) Cost (SR)

Normal Crash Normal Crash

A 7 3 1500 2300

B 9 5 1400 1840

C 8 5 1280 1430

D 14 9 1100 1400

E 4 2 1200 1500

F 9 7 1500 1680

G 7 3 1200 1520

H 11 7 1750 1910

I 11 7 1500 1720



1- Develop AON, calculate times, find critical path

0 7 7 7 8 15 23 4 27

A 0 C 8 E 2

10 10 17 17 10 25 25 2 29

0 0

0 0 0

0 9 9 9 14 23 18 7 25 29 11 40

B 0 D 0 G 4 I

0 0 9 11 2 25 22 4 29 29 0 40

9 9 18 18 11 29

F 0 H 0

9 0 18 18 0 29

START

Project completion time = 40 working days

Critical Path: B, F, H, I.

ES EF LS LF TF FF CP

A 0 7 10 17 10 0 N

B 0 9 0 9 0 0 YC 7 15 17 25 17 8 N

D 9 23 11 25 2 0 N

E 23 27 25 29 2 2 N

F 9 18 9 18 0 0 Y

G 18 29 22 29 4 4 N

H 18 29 18 29 0 0 YI 29 40 29 40 0 0 Y



Activity

Code

Time (day) Cost (SR) Time can be

crashed

Slope,

SR/week

Total Crash

Cost, SRNormal Crash Normal Crash

A 7 3 1500 2300 4 200 800

B 9 5 1400 1840 4 110 440C 8 5 1280 1430 3 50 150

D 14 9 1100 1400 5 60 300

E 4 2 1200 1500 2 150 300

F 9 7 1500 1680 2 90 180

G 7 3 1200 1520 4 80 320

H 11 7 1750 1910 4 40 160

I 11 7 1500 1720 4 55 220

∑ 12430 15300 ∑ 2870

2- Calculate cost slope

onash duratiation - Cr Normal dur

t t - Normal CrashCost Slope

coscos



3- prepare data for crash analysis

Add the crash limit on node

diagram and table of iteration

ES D LS

Act FF

EF TF LF

Crash limit

(d @ cost)

0 7 7 7 8 15 23 4 27

A 0 C 8 E 2

10 10 17 17 10 25 25 2 29

0 0 4@200 3@50 2@150

0 0 0

0 9 9 9 14 23 18 7 25 29 11 40

B 0 D 0 G 4 I

0 0 9 11 2 25 22 4 29 29 0 40

4@110 5@60 4@80 4@55

9 9 18 18 11 29

F 0 H 0

9 0 18 18 0 292@90 4@40

START

Cycle #Activity to

Shorten

Can Be

ShortenedNIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project

Duration

0 - 12430 40

1

2



4- Start to Short Critical Activities beginning with activity

having lowest cost activity

a) Find compression time to reduce activity (H) time

Activity (H) can be reduced by 4 days, but non critical activity (E) on

other paths has only FF=2 Then Nil is 2 day

b) Calculate solution and organize data in the table of iteration

table of cycle reduction

Cycle #

Activity to

Shorten

Can Be

Shortened NIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project

Duration0 - 12430 40

1 H 42 (Due FF of

E)2 40 80 12510 38

2

Cycle1: reduce critical Activity (H) time which has lowcost



5- Update the project AON

Cycle 1: Modified AON After Time reduction of Activity (H) by 2 days

After reduction other critical Path is developed: B, D, E, I

0 7 7 7 8 15 23 4 27

A 0 C 8 E 0

8 8 15 15 8 23 23 0 27

0 0 4@200 3@50 2@150

0 0 0

0 9 9 9 14 23 18 7 25 27 11 38

B 0 D 0 G 2 I

0 0 9 9 0 23 22 2 27 27 0 38

4@110 5@60 4@80 4@55

9 9 18 18 9 27

F 0 H 0

9 0 18 18 0 27

2@90 2@40

START





a) Find compression time to reduce activity (I) time

Activity (I) can be reduced by 4 days, without any constraint for FF,

Then NIL =4



Cycle #Activity to

Shorten

Can Be

ShortenedNIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project


1 H 42 (Due FF

of E)2 40 80 12510 38

2 I 4 4 4 55 220 12730 34

3

Cycle2: reduce critical Activity (I) time which has low cost




Cycle 2: Modified AON After Time reduction of Activity (I) by 4 days

0 7 7 7 8 15 23 4 27

A 0 C 8 E 0

8 8 15 15 8 23 23 0 27

0 0 4@200 3@50 2@150

0 0 0

0 9 9 9 14 23 18 7 25 27 7 34B 0 D 0 G 2 I

0 0 9 9 0 23 22 2 27 27 0 34

4@110 5@60 4@80 0

9 9 18 18 9 27

F 0 H 09 0 18 18 0 27

2@90 2@40

START





a) Find compression time to reduce activity (H,D) time



Cycle #Activity to

Shorten

Can Be

ShortenedNIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project


1 H 42 (Due FF

of E)2 40 80 12510 38

2 I 4 4 4 55 220 12730 34

3 H,D (2,2)2(Due FF

of G)2 (40,60) 200 12930 32

Cycle3: reduce critical Activities (H,D) times which have

low cost and satisfy the Critical Paths

Activity (H,D) can be reduced by 2 days. Also, Activity (G) has

FF=2, Then NIL =2




Cycle 3: Modified AON After Time reduction of Activity (H,D) by 2 days

After reduction other critical Path is developed: B, F, G, I

0 7 7 7 8 15 21 4 25A 0 C 6 E 0

6 6 13 13 6 21 21 0 25

0 0 4@200 3@50 2@150

0 0 0

0 9 9 9 12 21 18 7 25 25 7 32

B 0 D 0 G 0 I0 0 9 9 0 21 18 0 25 25 0 32

4@110 3@60 4@80 0

9 9 18 18 7 25

F 0 H 0

9 0 18 18 0 252@90 0

START





a) Find compression time to reduce activity (B) time



Cycle #Activity to

Shorten

Can Be

ShortenedNIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project


1 H 42 (Due FF

of E)2 40 80 12510 38

2 I 4 4 4 55 220 12730 34

3 H,D (2,2)2(Due FF

of G)2 (40,60) 200 12930 32

4 B 4 4 4 110 440 13370 28

Cycle4: reduce critical Activity (B) time which has low cost

and satisfies the Critical Paths

Activity (B) can be reduced by 4 days. Then NIL=4




Cycle 4: Modified AON After Time reduction of Activity (B) by 4 days





a) Find compression time to reduce activities (D,F) times

b) Calculate solution and organize data in the table of iterationtable of cycle reduction

Cycle #Activity to

Shorten

Can Be

ShortenedNIL

Week

Shortened

Cost / week,

SR

Cost / Cycle,

SR

Total Direct

Cost, SR

Project

Duration

0 - 12430 40

1 H 4 2 (Due FFof E) 2 40 80 12510 38

2 I 4 4 4 55 220 12730 34

3 H,D (2,2)2(Due FF

of G)2 (40,60) 200 12930 32

4 B 4 4 4 110 440 13370 28

5 D,F (2,2) 2 2 (60,90) 300 13670 26

Cycle5: reduce critical Activities (D,F) times which have low cost and

satisfy the Critical Paths

Activities (D,F) can be reduced by 2 days, Since the FF of activity C

is 6, Then NIL=2




Cycle 5: Modified AON After Time reduction of Activity (B) by 4 days

After reduction other critical Path is developed: A, C, E, I

0 7 7 7 8 15 15 4 19

A 0 C 4 E 0

0 4 7 7 4 15 15 0 19

0 0 4@200 3@50 2@150

0 0 0

0 5 5 5 10 15 12 7 19 19 7 26

B 0 D 0 G 0 I

0 0 5 5 0 15 12 0 19 19 0 26

0 1@60 4@80 0

5 7 12 12 7 19

F 0 H 0

5 0 12 12 0 19

0 0

START

Remark:

All Activity on network are critical and no further reduction is possible



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6- Time Cost Trade Off