Crashing

COST ANALYSIS AND

CRASHING OF ACTIVITIESCRITICAL PATH MEASURES EXPECTED DURATION OF PROJECT AND IDENTIFIES ACTIVITIES WHICH NEED ATTENTION IN COMPLETING THE PROJECT

TO CUT DOWN TOTAL PROJECT TIME AS PER CRITICAL PATH, PROJECT CRASHING IS ADOPTED

OBJECTIVES

•TO COMPLETE IMPORTANT PROJECT IN LEAST POSSIBLE TIME WHERE TIME IS MORE IMPORTANT THAN COST•TO EFFECT COST ECONOMY WHILE CRASHING•TO EXPEDITE WHEN EARLIER ACTIVITIES TOOK LONGER TIME

•TO REDUCE IDLE TIME FOR NON CRITICAL ACTIVITIES AND ENSURE UNIFORM RESOURCES UTILISATION•TO RELEASE FACILITY/T&P QUICKLY AND TRANSFER TO OTHER PROFITABLE PROJECTS•TO ENHANCE REPUTATION OF FIRM

MEANS OF CRASHING

•ALLOCATING MORE MAN POWER AND MACHINES

•WORKING ADDITIONAL SHIFTS

•OVERTIME

•MAKING LOCAL PURCHASES

•ENGAGING COSTLIER AGENCIES

•RELAXING TECHNICAL SPECIFICATIONS

PROJECT COST

DIRECT INDIRECT

DIRECT COST

•LABOUR COST•MATERIAL COST•EQUIPMENT COST DIRECT COST INCREASES WITH DECREASE IN PROJECT PERIOD

INDIRECT COST

•ADMINISTRATIVE/ ESTABLISHMENT / OVERHEADS

•SUPERVISION

•LOSS OF BUSINESS/ REVENUE / PROFITS

•PENALITIES

•INTEREST PAYABLE ON CAPITALS

Indirect cost falls with decrease in project duration

C C

C n

t Ct nTIME

COST

LINEAR APPROXIMATION

CRASH TIME & DIRECT COST RELATIONSHIP

PROJECT DIRECT COST

CRASH DIRECT COST > NORMAL DIRECT COST

C C

C n

t Ct nTIME

COST

CRASH TIME & INDIRECT COST RELATIONSHIP

PROJECT INDIRECT COST

CRASH INDIRECT COST < NORMAL INDIRECT COST

INDIRECT COST CURVE

TOTAL COST CURVE

DIRECT COST CURVE

CO

ST

DURATION

CRASH OPTIMUM NORMAL

tC to tn

TOTAL COST CURVE

TERMINOLOGY:-

• NORMAL TIME (tn) OF AN ACTIVITY REPRESENTS EXPECTED DURATION OF THE ACTIVITY

• CRASH TIME (tc) OF AN ACTIVITY REPRESENTS THE IRREDUCIBLE MINIMUM TIME REQUIRED (BELOW WHICH IT CAN NOT BE REDUCED) TO COMPLETE AN ACTIVITY

• NORMAL COST (Cn) OF AN ACTIVITY REPRESENTS LOWEST POSSIBLE DIRECT COST TO COMPLETE THE ACTIVITY WITHIN ITS NORMAL TIME

• CRASH COST (CC) OF AN ACTIVITY REPRESENTS LOWEST POSSIBLE DIRECT COST REQUIRED TO COMPLETE THE ACTIVITY WITHIN ITS CRASHED TIME

• COST SLOPE (Δc/Δt) OF AN ACTIVITY REPRESENTS ADDITIONAL DIRECT COST TO BE INCURRED TO REDUCE THE DURATION OF AN ACTIVITY BY ONE UNIT OF TIME

Δ c / Δ t = {Cc - Cn }/ {tn - tc}

• EXPECTED DURATION OF THE PROJECT: SUM OF NORMAL TIMES OF ALL ACTIVITIES ON CRITICAL PATH

• CRASH DURATION OF THE PROJECT: IRREDUCIBLE MINIMUM DURATION OF THE PROJECT WHEN ALL CRICTICAL ACTIVITIES HAVE BEEN CRASHED

BASIC STEPS IN CRASHINGSTEP 1:

CALCULATE COST SLOPE OF ALL THE ACTIVITIES OF THE PROJECT

Δc / Δt = {Cc - Cn } / {tn - tc}

STEP 2:

FIND FOR THE PROJECT:

CRITICAL PATH (S) EXPECTED DURATIONASSOCIATED PROJECT COST

STEP 3:

SELECT AN ACTIVITY ON CRITICAL PATH WITH LEAST COST SLOPE

IN CASE OF MORE THAN ONE CRITICAL PATHS, SELECT COMBINATIONS OF ACTIVITIES - ONE ON EACH CRICITAL PATH, SUCH THAT TOTAL COST OF CRASHING IS MINIMUM AMONG ALL SUCH COMBINATIONS

STEP 4: REDUCE ACTIVITY TIME OF SELECTED ACTIVITIES PROGRESSIVELY TILL:

CRASHED TIME IS REACHED OR

NON-CRITICAL PARALLEL PATH BECOMES CRITICAL

STEP 5: REPEAT STEPS 2 TO 4 UNTIL THERE IS AT LEAST ONE CRITICAL PATH ON WHICH NO FURTHER ACTIVITY CAN BE CRASHED. (i.e. ALL ACTIVITES ON ORIGINAL CRITICAL PATH OR ON PREVIOUS NON-CRITICAL PATH HAVE BEEN CRASHED)

STEP 6: PREPARE PROJECT TIME-COST MATRIX TO SELECT OPTIMAL DURATION OF PROJECT

1 2 3A B

8 / 5 4 / 20 8 12

1280

ACT Cn Cc Tn Tc CS

A 500 650 8 5 50

B 400 450 4 2 25

INDIRECT COST RS 30 /- DAY

ACTIVITY DURATION COST

1-2 8 5002-3 4 400DC 900IC (12X30) 360TOTAL 12 1260


1-2 8 5002-3 3 425DC 925IC (11X30) 330TOTAL 11 1255

ACTIVITY DURATION COST1-2 8 5002-3 2 450DC 950IC (10X30) 300TOTAL 10 1250


1-2 7 5502-3 2 450DC 1000IC (9X30) 270TOTAL 9 1270

DURATION 12 11 10 9 8 7DIRECT COST 900 925 950 1000 1050 1100INDIRECT COST 360 330 300 270 240 210TOTAL PROJECT COST 1260 1255 1250 1270 1290 1310

PROBLEM

ACTIVITY NORMAL DURATION

CRASH DURATION

NORMALCOST

CRASH COST

1-2 A 6 DAYS 4 DAYS Rs 400 Rs 540

1-3 B 4 DAYS 2 DAYS Rs 500 Rs 660

2-4 C 3 DAYS 1 DAY Rs 130 Rs 250

3-5 D 8 DAYS 6 DAYS Rs 600 Rs 780

4-5 E 7 DAYS 4 DAYS Rs 500 Rs 620

3-4 DUMMY ACTIVITY

Indirect Cost is Rs.70/- per day

1

4

3

2

5

A6/4

B

4/2

C3/1

D

8/6

E

7/4

0/0d1

ACTIVITY NORMAL DURATIO

N

CRASH DURATION

NORMALCOST

CRASH COST

Δ c Δ t COST SLOPE

Δ c / Δ t

1-2 A 6 DAYS 4 DAYS Rs 400 Rs 540 140 2 70

1-3 B 4 DAYS 2 DAYS Rs 500 Rs 660 160 2 80

2-4 C 3 DAYS 1 DAYS Rs 130 Rs 250 120 2 60

3-4 d1 0 0 0 0 0 0 -

3-5 D 8 DAYS 6 DAYS Rs 600 Rs 780 180 2 90

4-5 E 7 DAYS 4 DAYS Rs 500 Rs 620 120 3 40


Step 1

1

4

3

2

5

A6/4/70

B

4/2/80

C3/1/60

D

8/6/90

E

7/4/40

Project duration = 16 Days

d1 0/0/--

1

4

3

2

5

A6/4/70

B

4/2/80

C3/1/60

D

8/6/90

E

7/4/40

0

6

8

9

16

16

9

4

6

0

Project duration = 16 Days

Step 2 i/ii

d1 0/0/--

ACTIVITY

NORMAL DURATIO

N

CRASH DURATION

NORMALCOST

CRASH COST

Δ c Δ t COST SLOPE

Δ c / Δ t

1-2 A 6 DAYS 4 DAYS Rs 400 Rs 540 140 2 70

1-3 B 4 DAYS 2 DAYS Rs 500 Rs 660 160 2 80

2-4 C 3 DAYS 1 DAYS Rs 130 Rs 250 120 2 60

3-4 d1 0 0 0 0 0 0 -3-5 D 8 DAYS 6 DAYS Rs 600 Rs 780 180 2 90

4-5 E 7 DAYS 4 DAYS Rs 500 Rs 620 120 3 40

Step 2 iii Rs 2130


1

4

3

2

5

A6/4/70

B

4/2/80

C3/1/60

D

8/6/90

E

6/4/40

0

6

4

9

15

15

9

8

6

0

Crashed Project duration = 15 Days

Step 3,4

0/0/--d1

1

4

3

2

5

A6/4/70

B

4/2/80

C3/1/60

D

8/6/90

E

5/4/40

0

6

4

9

14

14

9

8

6

0


Step 3,4

0/0/--d1

.

1

4

3

2

5

A6/4/70

B

4/2/80

C3/1/60

D

8/6/90

E

4/4/40

0

6

4

9

13

13

9

8

6

0


Step 3,4

0/0/--d1

1

4

3

2

5

A6/4/70

B

4/2/80

C2/1/60

D

8/6/90

E

4/4/40

0

6

4

9

12

12

9

8

6

0


Step 5

0/0/--d1

1

4

3

2

5

A6/4/70

B

3/2/80

C1/1/60

D

8/6/90

E

4/4/40

0

6

4

9

11

11

9

8

6

0


0/0/--d1

.

1

4

3

2

5

A5/4/70

B

2/2/80

C1/1/60

D

8/6/90

E

4/4/40

0

6

4

9

10

10

9

8

6

0


0/0/--d1

.

1

4

3

2

5

A4/4/70

B

2/2/80

C1/1/60

D

7/6/90

E

4/4/40

0

6

4

9

9

9

9

8

6

0


0/0/--d1

ActivityCost Slope Days

Direct Cost

Addl DC

Indirect Cost TOTAL

i ii iii iv v vi vii ivxv ix=vix70 x xi

1

A-C-E =16 B-d1- E= 11 B- D =12 Nil Nil Nil 16 Days 2130 Nil 1120 3250

2

A-C-E =13 B-d1- E=8 B- D =12 E 40/- 3 Days 13 Days 2130 120 910 3160

3

A-C-E =12 B-d1- E= 8 B- D =12 C 60/- 1 Day 12 Days 2250 60 840 3150

4

A-C-E =11 B-d1-E= 7 B- D =11 B,C

60/-+ 80/- 1 Day 11 Days 2310 140 770 3220

5

A-C-E =10 B-d1-E=6 B- D =10 A,B

70/-+80/- 1 Day 10 Days 2450 150 700 3300

6

A-C-E =9 B-d1-E= 6 B- D =9 A,D

70/-+90/- 1 Day 9 Days 2600 160 630 3390

Path & Duration

Selected for Crashing Crashed Project Cost Crashed Project Duration

Sl No

No further crashing except 'D' which do not further minimise proj durationMinimum duration = 9 days. Corresponding project cost = Rs.3390.Minimum Cost = Rs. 3150. Corresponding duration = 12 days.

4th Crash

5th Crash

Remarks

No Crash

1st Crash

2nd Crash

3rd Crash

630 700 770 840 9101120

27602600

24502310 2250 2130

3390 3300 3220 3150 3160 3250

0

500

1000

1500

2000

2500

3000

3500

4000

8 9 10 11 12 13 14 15 16

INDIRECT COST DIRECT COST TOTAL COST

Step 6

From the given data calculate normal duration & corresponding total cost of the ProjectIf duration is to be compressed by 5 days, find activities to be compressed & corresponding total minimum costIndirect cost = Rs 600/day SL NO

Act Code

Act Descrip

Duration (In days)

Crash Duration (In days)

Normal Cost (Rs)

Crash Cost (Rs)

1 1-2 A 2 2 2000 20002 1-3 B 6 3 3000 42003 1-4 C 4 2 1600 20004 2-5 D 5 2 4000 55005 3-4 d1 - - - -6 3-7 E 4 1 1000 22007 4-6 F 8 3 5600 81008 5-7 G 3 2 1200 17009 6-7 H 4 2 2400 3000

PROBLEM

1

2

3

4

5

7

6

A

D

G

B E

F

C H

2,2

5,2

4,1

8,3

3,2

4,2

6,3

4,2d1

SL NO

Act Code

Activity Descrip

Normal Dur (In days)

Crash Dur (In days)

Normal Cost (Rs)

Crash Cost (Rs)

Cost Slope (Rs/Day)

1 1-2 A 2 2 2000 2000 -2 1-3 B 6 3 3000 4200 4003 1-4 C 4 2 1600 2000 2004 2-5 D 5 2 4000 5500 5005 3-4 d1 - - - - -6 3-7 E 4 1 1000 2200 4007 4-6 F 8 3 5600 8100 5008 5-7 G 3 2 1200 1700 5009 6-7 H 4 2 2400 3000 300

Step 1

Step 2 20800

1

2

3

4

5

7

6

A

D

G

B E

F

C H

2,2,-

5,2,500

4,1,400

8,3,500

3,2,500

4,2,200

6,3,400

4,2,300d1

10

2

0

015

7

6

6

6

6

18

18

14

14

Step 2

Sl No

Path &Duration

Selected for CrashingCrashed Project

Dur

Crashed Project CostRem Activit

yCost Slope Days Direct

CostAddl DC

Indir Cost Total

1 A-D-G=10 B-E= 10 B-F-H =18 C-F-H=16

Nil Nil Nil 18 Days 20800 Nil 10800 31600 No

Crash

2 A-D-G=10 B-E= 10 B-F-H =16 C-F-H=14

H 300 2 Days 16 Days 20800 600 9600 31000 1st

Crash

3 A-D-G=10 B-E= 8 B-F-H =14 C-F-H=14

B 400 2 Days 14 Days 21400 800 8400 30600 2nd

Crash

4 A-D-G=10 B-E= 8 B-F-H =13 C-F-H=13

F 500 1 Day 13 Days 22200 500 7800 30500 3rd

Crash

Step 3,4,5

thanks

Sl No

Path &Duration

Selected for CrashingCrashed Project

Duration

Crashed Project CostRemar-

ksActivity Cost

Slope Days Direct Cost

Addl DC

Indir Cost Total

1 A-D-G=10 B-E= 10 B-F-H =18 C-F-H=16

Nil Nil Nil 18 Days 20800 Nil 10800 31600 No Crash

2 A-D-G=10 B-E= 10 B-F-H =16 C-F-H=14

H 300 2 Days 16 Days 20800 600 9600 31000 1st Crash

3 A-D-G=10 B-E= 10 B-F-H =16 C-F-H=14

B 400 2 Days 14 Days 21400 800 8400 30600 2nd Crash

4 A-D-G=10 B-E= 8 B-F-H =14 C-F-H=14

F 500 1 Day 13 Days 22200 500 7800 30500 3rd Crash

1

2

3

4

5

7

6

A D G

B E

FC H

2,2,- 5,2,500

4,1,400

8,3,500

3,2,500

4,2,200

6,3,400

4,2,300

Crashing

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