Question

Use this info for questions 9 through 13. Chuck's Custom Boats (CCB) builds customer luxury yachts. CCB has landed a contract with a mysterious New York real estate developer, Mr. T. Relevant data on the project are shown below. The complication is that Mr. T wants delivery in 30 weeks or he will impose a penalty of $250 for each week his yacht is late. Determine the lowest cost plan for crashing this project. Note: All numbers are in thousands, but keep them that way (i.e., don't enter the last three zeros). What is the expected project duration before crashing any activities? Activity # Possible Crashes Immediate Predecessor - Normal Time (weeks) 10 8 Crash Cost Per Week 175 1 K 2 2 45 60 м к 2 4 6 5 55 MN 265 70 P 35

2. What is the cost of the third crash?

3.What is the cost of the fifth crash?

4.What is the total cost of all crashes that should be done to minimize costs?

Answer 1

diagram Activity On Node te=34 L=34 Ez240 EM Istart 7 7 Il 2240 X Ek=1% 4K=18 EM=22 LM322 0 - 10 u Lsco 6 Eps 27 m5 Lpz2 I CH EESYO EEU O LE=40 EE15 L=16 EN=21 LN=22 EH=35 44240 while crashing the Network 2 things we always remember, 1) Crashing should be done in critical path only 11) among the critical activity , crash those - activity first , which have minimum crash cost per week, Here, critical pathis are J K -M P 9+2. 10 + 8 +4+5+ 7+6.40 weeks Among the critical to activity My has less crash cost per week Cie 2o) & Maximum crash possible is a week as given but if we Crash M by 2 weeks then critical path will change, Hence, possibilites of crash of activity am is only ONE week,

Now the new path will be JKM P.) asz lo +8+3+ 5+7 +6=39 weeks & another critical path is ) د N داد و 7 د و 10+5+6 + 5 + 7 + 6 = 39 weeley additional Crash cost due to activity Mis=20 Now here a critical path, Hence A crash those activity which is common in both the critical path or have to crash Simultaneously 2 diffrent different activity from the both the path, & also which should have less crash lost slope. Among the to both the critical path activity a is common in both the path and also have les crash cost slope.lice 70) & possibilites of crashing is 1 week. Now the New path will be J -KM P -Q- - 10+8+3 + 5+ 6+6 = 38 weeks J→LANA PSQ2 10+5+6+5+6+6=38 weeks addtional crash cost due to activity & is = 70 Now again re will repeat same thing Among the critical path Now simultaneoryly o we can crash Men by I weeks

So, that New tofpath will be JKM porsz= lo + 8 + 2 +5+6+6 = 37 weeks JLAN+ Pa+2 lo +5+5+5+6+6 = {F weeks additional cost due to MAN =(20+55) = 75 if wet we crash (K&L) = (45+60) zlos if we erash [K&NE (45+55)=100 but I we crash z then additional - cost is = 95 & maximum possibilities of crashing is I week. Hence next a zis crashed by I week New path JakAMAPOZ 10+8+2+5 +6+5=36 weeks JLANA Pye+ 10+5+5+5+6+5=36 weeks additional cost=.95 Next will crash (K&N) which has total crash Cost possibilites (45+55)=100& of crash is by I week only Now critical path is JJK +M > Paz 10 +7+2 + 5+6+5 = 35 weely 7 رده م دله ډا د ل 10+5+4+5+6 +5 = 35 weeks additional cost is (45+55) 2100

Next crash to the activity K&L by one week & additional cost is (45+60)=105 JAKAMAP a» 10+6+2 +5+6 +5=34 weeks JLNA PAG-2 to +4+4+5+6+5 =34 weeks, Now the only possibility is to crash is I which has maximum possible of week and additional cost is 175 Hence New & final critical path's will be J KAMP oyz 9 +6+2 +5 +6 +5 = 33 weeks JALAN> P 79» Z = 33 weeke g tu+4+5+6+5 after this crashing is not possible, if we crash the activity P, then crashing line cost of activity P is (ie 265) is more than the penalty cast of (250 $) so, we will not crash to the activity P. & Minimum possible project completion time is 33 weeks after crashing.

No. of crashing Crashing without (crashing activity week Gashed Cost 20 total=40 7 39 38 H 70 1 37 H I Man . 3 (20+55)=75 95 (45+55)=100 35 (45+60)=105 H KAN H K &L т 175 H a final so, from the above table duration we can give the of Arywer (33 weeks) - all question Arywer 2 > cost of 3rd crash is 75$ Argwer 3> cost of 5th crash is 100$ Argwer uy Total cost of all crashes = (20+70+75+95 Hou Ho5+175) = 640$ are Mence, 640$ is the total cost of all crashes that should be done to minimize cost
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