The table below contains data for the installation of new equipment in a manufacturing process at Excello Corporation. Your company is responsible for the installation project. Indirect costs are $12,000 perweek, and a penalty cost of $6,000 per week will be incurred by your company for every week the project is delayed beyond week 12.ActivityABсDEFGImmediatePredecessor(s)ABActivityABсDACD, EENormal Time Crash Time(weeks) (weeks)3122134163Time according toschedule (weeks)1a. What is the shortest time duration for this project regardless of cost?The shortest project duration regardless of cost is 9 weeks. (Enter your response as a whole number.)b. What is the minimum total cost associated with completing the project in 12 weeks?The minimum total cost to complete the project in 12 weeks is $ 203500. (Enter your response as a whole number.)c. What is the total time of the minimum-cost schedule? (Enter your responses as whole numbers.)The total project's duration according to the minimum-cost schedule is 10 week(s). The minimum-cost schedule is as follows:Time according toschedule (weeks)3142ActivityEFGNormal Cost($)7,0003,00012,00012,0008,0005,0009,0001Crash Cost($)14,0003,00040,00022,0008,000313,00016,000

Indirect costs (per week) $12,000 Penalty cost (per week) beyond week 12 $6,000 Crash cost is the cost of shortening the length of the project. 1- Draw the network diagram and calculate the paths. As per the normal time estimates, Path A-C-E-F has the highest duration (13 weeks), and this is the critical path. 2- Calculate the Maximum crash time and crash cost per week. Activity B and E cannot be crashed as both activities have equal Normal and Crash times. 3- Crash one activity at a time by 1 week each time and see if the critical path changes. If two critical paths are obtained, choose the one with which it is possible to reduce the maximum of critical path duration The result of the crashing process is as follows: Each activity is crashed step by step and accordingly path changes. Note: B and E cannot be crashed because normal and crash time are equal. We will not crash further as it will not help in reducing the critical path anymore. (B-D-F) is the new critical path of which B cannot be crashed, and the limit of crashing for D and F are exhausted. Observe the new critical path duration which is 9 weeks. This means that regardless of the crashing cost, the shortest time duration of completing the given project is 9 weeks. Thus, the Minimum total cost for project completion in 12 weeks is $203,500.In this step, we will crash the activities considering the crash cost per week (slope) of each activity. This will result in following schedule: At stage 2, there are two critical paths. In such a case, compare the cost of crashing a single activity with the cost of crashing a common activity. At Stage 2, A has the minimum crash cost per week i.e., $3500. However, if F (common activity) is crashed, we will be able to reduce the project length by 2 weeks. Thus, F should be crashed by 1 week in both paths. At stage 3: F cannot be crashed more. B & E cannot be crashed as normal and crash times are equal. Thus, A will be crashed. We will not choose activities C and D because of the higher crash cost per week. At stage 4: again we have two critical paths of which B & E cannot be crashed. F is exhausted. Thus, A will be crashed. Also, from path B-D-F activity D will be crashed. Again two critical paths are obtained and the only activity left for crashing is C. However, crashing this activity will not have any impact on the critical path. So, we will stop here. Total time of minimum cost schedule is as follows: