Problem 18-10 (Algo)

Two operators handle adjustments for a group of 10 machines. Adjustment time is exponentially distributed and has a mean of 16 minutes per machine. The machines operate for an average of 84 minutes between adjustments. While running, each machine can turn out 54 pieces per hour. Use Table 1, Table 2 and Table 3. a. Find the probability that a machine will have to wait for an adjustment. (Round your answer to 3 decimal places.)

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

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

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

e. Machine downtime represents a cost of $68 per hour; operator cost (including salary and fringe benefits) is $15 per hour. What is the optimum number of operators?

Answer is complete but not entirely correct. \begin{tabular}{|l|r|l|} \hline \multicolumn{3}{|c|}{ Answer is complete but not entirely correct. } \\ \hline Average number of machines waiting & 0.00 & machines \\ \hline \end{tabular} \begin{tabular}{|l|r|l|} \hline \multicolumn{3}{|c|}{ Answer is complete but not entirely correct. } \\ \hline \hline Average number of machines & 0.19 & machines \\ \hline \end{tabular} \begin{tabular}{|l|l|l|} \hline \multicolumn{2}{|c|}{ Answer is complete but not entirely correct. } \\ \hline \hline Expected hourly output & 45.25 & pieces \\ \hline \end{tabular} 1 operator 2 operators 3 operators 4 operators