Question

What is the minimum cost of crashing the following project that Roger Solano manages at Slippery Rock University by 4 days? Normal Crash Time Normal Activity Time (days) (days) Cost $800 $250 $600 $750 $1,200 Uw > A Awarw Total Cost with Crashing Immediate Predecessor(s) $1,100 $600 $650 $1,500 $1,650 OD By how many days should each activity be crashed to reduce the project completion time by 4 days? Fill in the table below. (Enter your responses as whole numbers.) Each Activity Should be Reduced BY (days) Activity The total cost of crashing the project by 4 days is $ . (Enter your response as a whole number.)

Answer 1

The step by step solution of the problem is given below:

1.

y Port, drawing the diagram (А - yday 6 days days → The different paths and total days for completing the projāt : Activity days AD 4+6 = 10 В 7= CE 4+7 = 1 citical Path (It is the path including activities which * The project can be completed in Ildays & take maximum time) we have to reduce it to 7. → Updating the table table: (I-1) (I-I) (I) TIN (I-II) (I) Maximum - Noumal Acuity Total Normal Time cuaching Crash lane crashing days cost cost with coot Crashing ч - А $800 1 $1100 $300 1 5 2 4250 4.600 4350 $600 $650 $50 4750 $1500 $ 750 $1200 1650 4450 ua w is mos

2.

- now we will start veashing ie decreasing the day • Frist we will crash the time of writical path le lldays Here we can crash Activity core. now, cost of crashing c is 450 and Eis $450, In order to minimise the cost we will go for C. Thus, Crash C by Iday and cost is $50. now, we have A,D 4+6 = 10days B 7 z 7days CE 347 z lodays Now we be can crash A, D, C, E activities as all of them have lodays. The minimum cost of maching is of activity with $50. But maximum crashing time is (day which we have already done. So, the next minimum coot is of activity A which is $300 Thus, crash A by Iday and coat is $300 Now, we have A, D 376 - 9 days B 7. - 1 days GE 347 z lodays . Now we have to deal with c&e. we cannot crash C so we will crash & with a cost of 9450 Now we have AD St 629 day B 727 days GE 3+629 days

3.

• Now we to have two vitical path again, (A,D) & (CE) out of them we cannot chash A & C as Iday washing limit has been used. So, we will crash & as it has minimum coot. crash & with a Now we have cost AD of $450 3+6=9 727 34528 AD • Now we d will deal with bath (AD) and crash D Cost=$750 Then, 3+5=8 7=7 GE 3+5=8 • Ag. As we have to reduce project days by u ie- bring it to I days, we need to crash both I and E by 1 day each. Also, we have crashing time available of both actrities Dkg Thus total cost of cashing - D 28750 E- $450 $1200 Now, at last we will he able to reduce the days by 4 and thue it will be completed in 7days A,A 344 = 7 days B 1= 7 days GERT 3442 7 days

4.

Therefore Days my each to actority has to reduced: pelivity Each Activity should we reduced by I day Iday 2 days 3 days Total caut of crashing the perfect by 4 days 'ü $3200. In order to calculate this take the sum of crashing cost of each activity that we have crashed. A + C + E + E + D & D TE de 50+300F 450+450+70+ 150+450