Speedy copy center located on Capitol Hill in Washington DC

Speedy copy center located on Capitol Hill in Washington D.C. has three coin operated copying machines used primarily by U.S. Senators to make copies of their confidential diaries and illegal agreements with foreign governments and corporations. The owner, Hugh Makeham, is a former Xerox repairman and can fix a broken machine within 20 minutes on the average. Many times the machines are only jammed and he can get them working again quickly so the exponential will describe his repair time distribution. The time-to-breakdown of a freshly repaired copier averages 60 minutes and is also exponentially distributed.

His revenue when all machines are working averages $80/hr per machine. When one machine is broken his revenue drops to $60/hr per working machine because of balking. When two machines are broken his revenue from the remaining machine decreases to $50/hr. When all machines are broken his revenue naturally drops to zero.

A.Compute his average hourly revenue.

B. If he employs a second, equally competent, repairman, what will be his total expected hourly income?

C. what is the largest hourly wage he can pay the employee without decreasing the expected hourly rate he earned when working alone?

Solution

dP(3)/dt = 3 P(2) - 3 P(3)

dP(2)/dt = 3 P(1) + 3 P(3) - 5 P(2)

dP(1)/dt = 2 P(2) + 3 P(0) - 4 P(1)

dP(0)/dt = P(1) - 3 P(0)

Of course, P(0) + P(1) + P(2) + P(3) = 1

From the top equation, we get that P(2) = P(3)

From the bottom equation, we get that P(1) = 3 P(0)

Then, as  2 P(2) + 3 P(0) - 4 P(1) = 0

2 P(2) + 3 P(0) - 4* 3 P(0) = 0

Then, 2P(2) = 9 P(0)

Then, P(2) = 9/2 P(0)

Then, P(0) + 3 P(0) + 9/2 P(0) + 9/2 P(0) = 1

13 P(0) = 1

P(0) = 1/13

P(1) = 3/13

P(2) = 9/26

P(3) = 9/26

Then, revenue is 9/26*3*80 + 9/26*2*60+3/13*1*50 = 1770/13 = $136.15

If we hire a second repairman, we have

dP(3)/dt = 3 P(2) - 3 P(3) (if 2 are working, only one repairman is working)

dP(2)/dt = 6 P(1) + 3 P(3) - 5 P(2)

dP(1)/dt = 2 P(2) + 6 P(0) - 7 P(1)

dP(0)/dt = P(1) - 6 P(0)

Of course, P(0) + P(1) + P(2) + P(3) = 1

From the top equation, we get that P(2) = P(3)

From the bottom equation, we get that P(1) = 6 P(0)

Then, as  2 P(2) + 6 P(0) - 7 P(1) = 0, we have

2P(2) + 6P(0) - 7 (6 P(0)) = 0

2P(2) = 36P(0)

P(2) = 18P(0)

Then, as P(0) + P(1) + P(2) + P(3) = 1

P(0) + 6 P(0) + 18P(0) + 18P(0) = 1

43P(0) = 1

P(0) = 1/43

P(1) = 6/43

P(2) = 18/43

P(3) = 18/43

Then, revenue is 18/43*3*80 + 18/43*2*60+6/43*1*50 = 6780/43 = $157.67

Then, $157.67-$136.15 = $21.52 This is how much we can pay the second repairman.