lec11 crashing-b

CHAPTER 5: SCHEDULE COMPRESSION AND TIME-COST TRADE OFF

THE CRASHING PROCEDURE.Estimates of activity times for projects usually are made for some

level of resources.

It is possible to reduce the length of a project by injecting additional resources.

The desire to shorten the length of a project merely reflects an attempt to reduce the indirect costs associated with running the project and increasing direct expenses by using additional funds to support additional personnel or more efficient equipment.

The goal of time-cost trade-offs is to identify activities that will reduce the sum of the indirect and direct costs (the total cost).

DIRECT AND INDIRECT COSTS

The contractor’s main expenses are as follows:

I- Direct Costs1. Labor, particularly hourly workers, for whom a labor expenses

can be directly linked to a particular work item

2. Materials, such as concrete, lumber, nails, paints, steel, and installed equipment, such as elevators, air-conditioning units, and kitchen equipment.

3. Equipment, particularly construction equipment (bulldozers, excavators, cranes, pumps….)

4. Other costs, such as government permits and fees , and fees for lawyers and consultants hired for specific task in a project.

DIRECT AND INDIRECT COSTS

II- Indirect Costs

A. Project overhead (or job overhead), such as the following:

1. Project staff (project manager, project engineer, secretary, clerk,…..)

2. Office trailer and other temporary structures 3. Cars and trucks assigned to the project team. 4. Office equipment (copying machine, fax machine,

computers…) 5. Temporary utilities (electricity, water, drinking

water, telephone, cell phone, portable toilets…)

II- Indirect Costs B. General overhead , such as the following: 1. Main office expenses (rent, maintenance,…) 2. Main office personnel. 3. Main office equipment and vehicles. 4. Other main office expenses, such as advertising.

Project total cost = Direct costs + Indirect costs

EFFECT OF CRASHING ON INDIRECT COST

EFFECT OF CRASHING ON DIRECT COST

EFFECT OF CRASHING ON TOTAL COST

Totalcost

Shorten

Shorten

Cumulativecost of crashing

Expected indirect costs

Optimum

CRASH

19 20 22 24 260

500010000150002000025000300003500040000

cost

cost

Example1:Using the following information, develop the optimal time-cost solution . Normal cost of the

project $6000. Indirect costs are $1,000 per day.

Activity predecessors Normal time

Crash time

Crash potential

Cost slope

abcdef

---a---cdb, e

6105492

684171

------21321

---$500 300 700 600 800

6a

4d

5c

10b

9 e

2f

1- Determine which activities are on the critical path: Critical path :c-d-e-f………………..length 20 days.2- Rank the critical path activities in of lowest crashing cost

and determine the number of days each can be crashed.

SOLUTION

path duration

Crash 1 Crash 2 Crash 3

A-b-f 18 18 18 17C-d-e-f 20 19 18 17

SOLUTION 1-Crash C by 1 day @ $300 (cost slope value of C) 2-Crash E by 1 day @ $600 3-Crash B&E by 1 day @ $500+ $600=

$1100>1000

Note that both paths are critical that is why we had to crash both B&E at the same time

FIND THE OPTIMAL SOLUTION??Crashed activity

Project duration

Direct cost

Indirect cost

Total cost

None 20 $6000 20x 1000 $26,000C 19 6300 19x 1000 25300E 18 6900 18x1000 24900

optimumB, E 17 8000 17x1000 25,000

CAN U FIND A BETTER SOLUTION?3- Begin shortening the project, and check after

each reduction to see which path is critical. a- shorten activity c one day at $300. b- shorten activity e one day at a cost of $600.

the length of path c-d-e-f is now 18 days the same as the length of path a-b-f.

c- the paths are now both critical. d- crash f by one day at $800. e- At this point no further improvement is

feasible.

Crashing b and e at $1,100 would exceed the indirect project costs of $1,000 per day.

SOLUTION OF SECOND METHOD 1-Crash C by 1 day @ $300 (cost slope value of C) 2-Crash E by 1 day @ $600 3-Crash F by 1 day @ $800 4-Crash B&E by 1 day @ $500+ $600=

$1100>1000

Note that both paths are critical that is why we had to crash both B&E at the same time

MORE OPTIMAL SOLUTION

Crashed activity

Project duration

Direct cost

Indirect cost

Total cost

None 20 $6000 20x 1000 $26,000C 19 6300 19x 1000 25300E 18 6900 18x1000 24900 F 17 7700 17x1000 24,700 optimumB,E 16 8800 16x1000 24,800

Example2:Using the following information, develop the optimal time-cost solution . Indirect costs are $120

per day.

Activity Immediate predecessors

Normal time (days)

Crash time(days)

Normal cost $

Crash cost $

ABCDEFGH

---AAAB, CCD, FE, F

578116476

45574455

5003508001200600500700300

60050092014007005001000420

Solution: First calculate cost slope Second sum up the total direct cost=$4950

Activity Immediate predecessors

Normal time (days)

Crash time(days)

Normal cost $

Crash cost $

Cost slope

ABCDEFGH

---AAAB, CCD, FE, F

578116476

45574455

5003508001200600500700300

60050092014007005001000420

10075405050----150120

SOLUTION path duration 1 2 3ABEH 24 24 23ACEH 25 24 23ACFH 23 22 22ACFG 24 23 23ADG 23 23 23

A

H

G

F

EB

C

D

SOLUTION: STRATEGY 1 Crash activity C by 1 day @ 40$ Crash activity E by 1 day @ 50$ Crash activity A by 1 day @ 100$ Crash activity B,C,D by 1 day @ 165$

SOLUTION: STRATEGY 1

Cashed activity

Project duration

Direct cost

Indirect cost Total cost

None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

E 23 5040 23x120= 2760 7800

A 22 5140 22x120= 2640 7780

B,C,D 21 5305 21x120= 2520 7825


Cashed activity

Project duration

Direct cost


None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

E 23 5040 23x120= 2760 7800

A 22 5140 22x120= 2640 7780

H,G 21 5410 21x120= 2520 7930


Cashed activity

Project duration

Direct cost


None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

E 23 5040 23x120= 2760 7800

A 22 5140 22x120= 2640 7780

E,G 21 5340 21x120= 2520 7860


Cashed activity

Project duration

Direct cost


None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

E 23 5040 23x120= 2760 7800

E,G 22 5240 22x120= 2640 7880


Cashed activity

Project duration

Direct cost


None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

B,C 23 5080 23x120= 2760 7840

A 22 5180 22x120= 2640 7820

E,D 21 5380 2520 7800

E,G 20 5580 2400 7980

SOLUTION: STRATEGY 6 (NOT-PREFERRED)

Cashed activity

Project duration

Direct cost


None 25 $4950 25x 120=3000 $7950

C 24 4990 24x 120=2880 7870

B,C 23 5105 23x120= 2760 7865

E,C,D 22 5245 22x120= 2640 7885

STRATEGY 5 (NOT-PREFERRED) SINCE PATH ACEH IS REDUCED BY 2 DAYS IN ONE STEP

EXAMPLE 3 Using the following information, develop the optimal

time-cost solution . Indirect costs are $12,000 per week. Direct cost is assumed to be $21,000 Suppose direct cost is not given what should you do?

Activity Normal duration

Immediate predecessors

Crashing potential (weeks)

Cost per week to crash

A B

C D E F

10 14

13 6 15 8

-- A

-- C -- E

3 3 2 1 3 1

$11,000 3,0001st week,$4,000

others 6,000 1,000 6,000 2,000

SOLUTION path duratio

n1 2 3 4 5

A-B 24 23 22

21 20 19

C-D 19 19 19

19 19 19

E-F 23 23 22

21 20 19

Start End

F,8

D,6

B,14A,10

C,13

E,15

SOLUTION 1-Crash B by 1 Week @ $3000 (cost slope value ) 2-Crash B&F by 1 Week @ $4000+ $2000=

$6000 3-Crash B&E by 1 Week @ $4000+ $6000=

$10,000 4-Crash A&E by 1 Week @ $11000+

$6000=$17000 5-Crash A&E by 1 Week @ $11000+

$6000=$17000

SOLUTION

Crashed activity

Project duration

Direct cost


None 24 $21,000 24x 12000=288,000

$309,000

B 23 24,000 23x 12000=276,000

300,000

B,F 22 30,000 22x12000=264,000

294,000

B,E 21 40,000 21x12000=252,000

292,000 optimum

A,E 20 57,000 20x12000=240,000

297,000

A,E 19 74,000 19x12000=228,000

302,000

SOLUTION WHEN DIRECT COST IS NOT GIVEN

Crashed activity

Project duration

Direct cost


None 24 X 24x 12000=288,000

288,000 +X

B 23 X+3000 23x 12000=276,000

279,000 +X

B,F 22 X+9000 22x12000=264,000

273,000 +X

B,E 21 X+19,000

21x12000=252,000

271,000 +X optimum

A,E 20 X+36,000

20x12000=240,000

276,000 +X

lec11 crashing-b

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