Question

Kimpel Products makes pizza ovens for commercial use. James Kimpel, CEO, is contemplating producing smaller ovens for use in high school and college kitchens. The activities necessary to build an experimental model and related data are listed in the following table: Activity A Normal Time (weeks) 3 Crash Time (weeks) 2 B 2 1 1 Normal Cost $1,200 $2,000 $400 $1,500 $950 $4,500 $1,600 Total Cost with Immediate Crashing Predecessor(s) $1,800 $2,800 $400 $1,820 A $1,250 B $5,500 с $2,300 1 7 3 3 E 6 F 2 1 G 4 2 D, E a) Based on the given information regarding the activities for the project, the project completion date = weeks.

Answer 1

Step 1: Draw Activity on Node network diagram as follows: Start E End F Step 2: Draw the table listing Max crash time and Crash cost per week Max Crash time = Normal time - Crash time Crash cost per week = (Crash Cost - Normal Cost) / Max Crash time Normal Crash Normal Activity Predecessor Crash Max Crash Time Time Cost Cost Time A 3 2 1,200 1,800 1 B 1 2,000 2,800 1 с 1 400 400 0 D A 3 1,500 1,820 4 E 6 3 950 1,250 3 F С 1 4,500 5,500 DE 2 1,600 2,300 Total normal cost = 12,150 Crash cost/week 600 800 0 1 7 80 B 1 100 1,000 350 G N

14 Step 3: Identify the alternative paths and Critical path. Critical path is the path of longest duration in the network A-D-G Critical path B-E-G 12 C-F 3 weeks Normal Time = Total Direct Cost = 14 12,150 Step 4: Project Crashing using least cost method 1. Identify and list the various alternative paths through the network and their respective duration. 2. Determine Max crash time (=Normal time - Crash time) and Crash cost per unit time =(Crash cost - Normal cost)/Max Crash Time for each activity. Also determine whether the crash cost rate remains constant as the activity is crashed, or it changes. Determine the variable crash costs by time, if they are variable. 3. Determine the critical path, i.e. the path of longest duration. Only activities along the critical path need to be crashed. 4. Identify the eligible activity on the critical path having the least crash cost per unit time, and reduce by 1 time unit at a time. See if the critical path changes as the activities along the critical path are crashed. 5. If there are multiple critical paths at any stage during the crashing. Then determine whether there is any of the activites common to the critical paths. Then calculate whether the crash cost of common activity is less than the combined crash costs of the least cost crash cost activities along the critical paths. Activity Crash Cost Cumulative Crash Cost Activity Crash Path 1 Path 2 Path 3 Project Crashing Crashed Time A-D-G B-E-G C-F Duration 0 14 12 3 14 1 D 2 12 12 3 12 D 2 10 12 3 12 2 E 2 10 10 3 10 3 G 2 8 8 3 8 1 7 8 3 4 E 1 7 7 3 7 Critical path times after each crashing, are highlighted in yellow color 160 160 200 700 600 100 160 320 520 1,220 1,820 1,920 TOTAL COST 12,150 12,310 12,470 12,670 13,370 13,970 14,070

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a)

Project completion date =14 weeks

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b)

To reduce the duration of the project to 10 weeks,

The first activity to crash is D by 2 weeks at a per week crashing cost of 80 dollars

Next, one should cash activity D and activity E by 2 weeks (each) for an additional total per week crashing cost of 80+100 = 180 dollars

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c)

To reduce the duration of the project to 7 weeks,

Next activity to crash in order is G by 2 weeks at a per week crashing cost of 350 dollars

Next, one should cash activity A and activity E by 1 week (each) for an additional total per week crashing cost of 600+100 = 700 dollars

Total cost of crashing the project to 7 weeks = 1920 dollars

After crashing the project to 7 weeks, total project cost, including normal time = 12150+1920 = 14070 dollars