ABS specializes in repair and maintenance of industrial machines.One repair technician has been assigned to five identical grinding machines for repair and maintenance.The machine breaks down after about 20 hours of use,and breakdowns have a Poisson probability distribution.It takes 6 hours to repair a machine,and repair times follow an exponential distribution.Machine downtime costs the company $250 an hour,and the technician is paid $50 an hour.Determine the cost savings of having two technicians as compared to one technician.
A)$320.50
B)$100.50
C)$162.50
D)$175.00

Question

Quizplus · Accepted Answer

Let λ be breakdowns per hour of each machine.
Poisson distribution will have mean and variance λt since t=20 hours and Poisson distribution is memoryless, which means probability distribution is not affected by time, which has passed until now. Hence,
λt=20 implies λ=1/20
The average number of failures per day on all 5 machines=5λ=1/4
Let μ be the number of machines repaired per hour.
Then, μ=1/6 since it takes 6 hours to repair one machine and repair times follow exponential distribution.
The average time to repair a machine is 1/μ=6 hours.
Let X be the total number of machines that breakdown per day.
X is a Poisson distribution with mean 1/4.
Let Y be the total number of machines repaired per day.
Y is a Poisson distribution with mean λtμ=5/2.
Let C1 be the total cost of one technician per day and C2 be the total cost of two technicians per day.
C1=250×24×E[X]+50×24×E[Y]=E[X]/4×250×24+5/2×50×24=5925
C2=2[250×12×E[X/2]+50×12×E[Y/2]]=2(250×12×E[X/2]+50×12×E[Y/2])
=2(250×12(E[X]-E[X/2])+50×12(E[Y]-E[Y/2]))=2(250×12×(E[X]-E[X]/2)+50×12×(E[Y]-E[Y]/2))
=2[250×12(E[X]/2)+50×12(E[Y]/2)]=2[250×12/2×E[X]+50×12/2×E[Y]]=12C1/5=14190
Cost savings=C1-C2=14190-5925=$8265/5=$1653/2=$162.50.

Therefore, the cost savings of having two technicians as compared to one technician is $162.50.

ABS Specializes in Repair and Maintenance of Industrial Machines