11-28 on page 418, MODIFIED You are NOT required to use Linear Programming, you

Question

11-28 on page 418, MODIFIED You are NOT required to use Linear Programming, you may use Excel OM or compute manually. Bowman Builders manufactures steel storage sheds for commercial use. Joe Bowman, president of Bow- man Builders, is contemplating producing sheds for home use. The activities necessary to build an experimental model and related data are given in the following table Immediate predecessors Activity N2ormal time Crash timeN Normal cost Crash Cost 1,000 2,000 300 1,300 1,600 2,700 300 1,600 1,000 5,000 2,000 4,000 1,500 D, E 11-28a: What is expected project complete time (before crashing)? 11-28b: Which activities should be crashed and by how much? 11-28c: What is the additional project cost to crash to 10 weeks?

Explanation / Answer

Slope for any activity will be defined as :

Slope = ( Crashed cost – Normal cost ) / ( Normal duration – Crashed duration )

Accordingly ,

Slope for Task A =( 1600 – 1000) / ( 3 -2 ) = 600

Slope for Task B = ( 2700 – 2000) / ( 2 -1 ) = 700

Slope for Task C = 0 ( Cannot be crashed as normal time = crash time )

Slope for Task D =( 1600 – 1300) / ( 7 – 3 ) = 300/ 4 =75

Slope for Task E = ( 1000 – 850 ) / ( 6 – 3 ) = 150 / 3 = 50

Slope for Task F = ( 5000 – 4000) / ( 2 -1 ) = 1000

Slope for Task G = ( 2000 – 1500) / ( 4 – 2 ) = 500/2 = 250

While crashing , we must first start with thetask with least slope and then go for higher ones. Thetask with least slop is Task E followed by Task D, Task G , Task A , Rask B and Task F

The precedence diagram as follows :

A

B

C

D

E

                                 F

                                                                 G

The parallel paths and their corresponding normal durations as follows :

A-D-G = 3 + 7 + 4 = 14

B-E-G = 2 + 6 + 4 = 12

C-F = 1 + 2 = 3

Since A-D-G has the longest duration, it forms the critical path .

Project duration ( which is same as duration of critical path ) is 14

EXPECTED PROJECT COMPLETION TIME BEFORE CRASHING = 14

For the project to crash to 10 weeks ,

a)A-D-G must be crashed to 10 weeks

As well as

b)B-E-G must be crashed to 10 weeks

For A-D-G to crash to 10 weeks : CRASH activity D by 4 weeks from 7 to 3 weeks at an additional crash cost of = $1600 - $1300 = $ 300

For B-E-G to crash to 10 weeks :

Crash activity E by 2 weeks from 6 weeks to 4 weeks at an additional cost of = 2 x $50 = $100

Therefore , total additional cost of crashing = $300 + $100 = $400

CRASH ACTIVITY D BY 4 WEEKS AND ACTIVITY E BY 2 WEEKS

ADDITIONAL PROJECT COST TO CRASH TO 10 WEEKS = $400

A

B

C

D

E

                                 F

                                                                 G

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