A computer technician arrives at work at 9 AM and finds a department manager delivering three malfunctioning laptop computers. The computer technician begins to repair the problems as soon as possible. The technician begins on the first computer and does not start the second computer until the first is repaired. The technician works on only one computer at a time. The technician takes a lunch break from 12-1 PM.

The first computer is repaired by 11 AM

The second computer is repaired by 1:30 PM

The third computer is repaired by 4 PM.

What is the total process velocity?

The total process velocity = 2.25

Explanation:

The first computer is repaired by 11 am:-

Total time taken from the initiating the repair work = 11 am-9 am = 2 hours.

Total time spent in repairing the first computer = 11 am-9 am = 2 hours.

The second computer is repaired by 1:30 pm:-

Total time taken from the initiating the repair work = 1:30 pm - 9 am = 4:5 hours

Total time spent in repairing the first computer = 1:30 pm - 11 am = 2.5 hour

The third computer is repaired by 4 pm:-

Total time taken from the initiating the repair work = 4 pm - 9 am = 7 hours

Total time spent in repairing the first computer = 4 pm-1:30 pm = 2.5 hour

As there is a break of 1 hour, thus effective time in repairing the third computer = 2.5 hours - 1 hour = 1.5 hour

The total process velocity = The total time taken from the beginning / the total time taken in repairing each computer

The total process velocity = (2+4.5+7) / (2.0+2.5+1.5)

The total process velocity=13.5/6

The total process velocity=2.25