crashing presentation final

8/7/2019 crashing presentation final

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By :-

Anubha Jain

Kshitij Pareek



Satisifying unlimited demandby using limited resources



` T o ensure equitable distribution

` T o prevent day to day flucations in theActivities

` T o obtain uniform resource requirement



D

elibrate reduction of theactivity times by putting inextra efforts is called ascrashing



` Activity cost: -it refers to a reduction in theduration of an activity and a consequentialincrease in the cost

` Crash time:-it refers to the minimum activityduration to which an activity can be compressed byusing resources.

` Crash cost:-it is that cost of a project at which anactivity can be completed in minimum duration



` Normal cost:-it is that cost which is generallyincreased towards the completion of an activity

` Normal time:-it is the minimum duration for execution of an activity determined on the basis of norma l time.

` Cost slope:-it refers to an increase in direct cost per unit time for reducing the duration for completion of an activity.



Chapter 9 Resource Allocation 7

Slope = crash cost ± normal costnormal time ± crash time

Where: slope = cost per day of crashing a project

W

hen slope is negative : indicate the time requiredfor a project is decreased, the cost is increased



Pre ±feaibility stage

` Bureaucratic delays` Delay in financial assistance

Evaluation stage` Inadequate project study` Wrong selection of location

Technology selection and engineering stage

` Selection of poor and outdated technology



Construction stage

` Indecisions or delayed decision` Low productivity of contractors

Start up stage

` Defects in erection and installtations` Failures of certain equipments and parts





Step I:-Prepare the network diagram, find the normal critical

path and determine critical activities.

Step II:-Calculate cost slope,for different activities with the help

of formulae .Step III:-

Identify those activities on critical path which can becrashed.



Step IV:-Crash the least expensive activity i.e. the activity with

least slope

Step V:-By crashing the activity of critical path if other paths

also become critical ,a situation of parallel criticalpath is obtained which implies that the duration of

the project can further be reduced throughsimultaneous crashing of the activities in the parallelcritical paths



Step VI:-

Calculate the total cost.



` Crash only critical activities on the critical path

` Crashing non critical activities will result in

reduction of project duration

` First, activities with the lowest crashing cost per unitof time should be crashed and the process should berepeated until desired duration of projet is achieved

` In case of parallel critical paths, each of parallel pathsmust be compressed



Activity Crash

Time,Cost

Normal

Time,Cost

crashing

a 3,60 3,60 No

b 6,80 7,30 Yes

c 4,90 5,50 Yesd 5,50 6,30 Yes

e 2,100 4,40 Yes

1

2

3

4

a

be

c

d



` Paths Normal duration Crash duration1-2-4 981-2-3-4 127

1-3-4 11

1-2-3-4 is the critical path with highest duration of 12 days.

Project needs to be crashed from 12 ² 8 days.



Activity Cost slopeA 0

B 50

C 40

D 20

E 30



Determining the lowest cost slope of critical pathactivities.

Activitiy cost1-2 2-3

3-4 30Since the lowest cost is 30 therefore activity3-4 will be crashed by 1 day.



Activity Normal duration

1-2-4

1-2-3-4 1-3-4 ______________________________________________

ActivityN

ormal time Crash time3-4



Again, activity 3-4 will be crashed due to low T herefore,

Activity Normal duration1-2-4 1-2-3-4 1-3-4

Crashing cost :- Rs.30



Activity Normal time Crash tim3-4 4,3,2

Now, activity 3-4 can not be crashedfurther. Since normal time=crash ti



Further activity 2-3 with next lowest crash slope willbe crashed.i.e. 40

T herefore,

Activity Duration1-2-4

1-2-3-4 1-3-4



Possible cost options,

i) 1-2 - ii)1-2 - iii)1-3 2-4 20 2-3 40

3-4 -

Crashing cost=20+40+50=110



Path Duration1-2-4 1-2-3-4 1-3-4



T otal crashing cost=30+30+40+110=rs.210



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crashing presentation final