Problem $18-10$

Two operators handle adjustments for a group of $10$ machines. Adjustment time is exponentially distributed and has a mean of $14$ minutes per machine. The machines operate for an average of $86$ minutes between adjustments. While running, each machine can turn out $50$ pieces per hour. Find: Use Table $1,$ Table $2$ and Table $3.$

a$.$ The probability that a machine will have to wait for an adjustment. $($Round your answer to $3$ decimal places.$)$

Probability

b$.$ The average number of machines waiting for adjustment. $($Round your answer to $2$ decimal places.$)$

Average number of machines waiting $$ machines

c$.$ The average number of machines being serviced. $($Round your answer to $2$ decimal places.$)$

Average number of machines $$ machines

d$.$ The expected hourly output of each machine, taking adjustments into account. $($Round your answer to $2$ decimal places.$)$

Expected hourly output $$ pieces

e$.$ Machine downtime represents a cost of $$70$ per hour; operator cost $($including salary and fringe benefits$)$ is $$15$ per hour. What is the optimum number of operators?

$1$ operator

$2$ operators

$3$ operators

$4$ operators