Question: Given the project network and time (in days) and cost information listed below, what is the lowest possible cost after expediting the project by 3
- Given the project network and time (in days) and cost information listed below, what is the lowest possible cost after expediting the project by 3 days? (CBC = Cannot be crashed.
| Activity | Normal Time | Crash Time | Normal Cost | Crash Cost |
| A | 1 | CBC | $400 | CBC |
| B | 3 | 2 | $600 | $700 |
| C | 5 | 2 | $300 | $1200 |
| D | 6 | 4 | $200 | $1000 |
| E | 4 | 2 | $500 | $800 |
| F | 2 | 1 | $600 | $800 |
A. $2900 B. $3100 C. $3300 D. $3400 E. $3600
- In a G/G/1 queue, the average arrival rate is 10 customers per hour and the standard deviation of inter-arrival times is 12 minutes. The service time is 3 minutes per customer and its standard deviation is zero. The average queue length is:
A. 1/3 B. 1/2 C. 1 D. 2 E. None of the above
- Consider a process consisting of two activities with different unit loads shown below (s for second):
To double the output per hour, you can
A. Buy one more Machine 1 for Activity 1 (working in parallel)
B. Replace Machine 2 with a faster machine which cuts the unit load by half
C. Buy one more Machine 2 for Activity 2 (working in parallel)
D. Both B and C; E. None of the above
AT > Activity 1 Machine 1 >Activity 2 Machine 2 10s 85Step by Step Solution
There are 3 Steps involved in it
Get step-by-step solutions from verified subject matter experts
