Hi - I just want to be sure I did this problem correctly.
There is a copy center with 3 copiers. The owner can fix a broken machine within 20 minutes and the time to breakdown of a freshly repaired machine is 60 minutes. Everything is exponentially distributed. Revenue with all 3 machines working is 80/hr, 60/hr with 2 machines, 50/hr with 1 and obviously 0 for 0 machines. What is his average hourly revenue? What is the highest hourly wage he can pay a second, equally good, repairman and not lose money?
To find the revenue I found the average number of working machines/hour. Using lambda = 1 breakdown/hour, k = 3 machines, R = 1 repairman, and mu = 3 repairs/hour with p=1/3
pi1 = (3!/1!2!)*(1/3)*1!*pi0/1!*1^-1= pi0
pi2=0.667pi0
pi3=0.222pi3
pi0*(1+1+.667+.222) = 1
pi0 = 0.346
so pi1 = 0.346, pi2 = 0.2309, pi3 = 0.077
expected # of machines in good condition = K-sumjpij = 3-[0*.346+1*.346+2*.2309+3*.077] = 3-1.04 = 1.96 machines
0.96*60+0.04*50= 59.60/hour = average hourly revenue
If you repeat all of the calculations for R=2 you end up with 2.46 = expected # of machines in good condition
.46*80+.54*60= 69.2/hour = average hourly revenue with second repair man
69.2-59.6 = 9.60/hour = highest hourly wage he can afford for a second repairman without losing money.
Does this seem correct? Thanks for any help!
There is a copy center with 3 copiers. The owner can fix a broken machine within 20 minutes and the time to breakdown of a freshly repaired machine is 60 minutes. Everything is exponentially distributed. Revenue with all 3 machines working is 80/hr, 60/hr with 2 machines, 50/hr with 1 and obviously 0 for 0 machines. What is his average hourly revenue? What is the highest hourly wage he can pay a second, equally good, repairman and not lose money?
To find the revenue I found the average number of working machines/hour. Using lambda = 1 breakdown/hour, k = 3 machines, R = 1 repairman, and mu = 3 repairs/hour with p=1/3
pi1 = (3!/1!2!)*(1/3)*1!*pi0/1!*1^-1= pi0
pi2=0.667pi0
pi3=0.222pi3
pi0*(1+1+.667+.222) = 1
pi0 = 0.346
so pi1 = 0.346, pi2 = 0.2309, pi3 = 0.077
expected # of machines in good condition = K-sumjpij = 3-[0*.346+1*.346+2*.2309+3*.077] = 3-1.04 = 1.96 machines
0.96*60+0.04*50= 59.60/hour = average hourly revenue
If you repeat all of the calculations for R=2 you end up with 2.46 = expected # of machines in good condition
.46*80+.54*60= 69.2/hour = average hourly revenue with second repair man
69.2-59.6 = 9.60/hour = highest hourly wage he can afford for a second repairman without losing money.
Does this seem correct? Thanks for any help!