Superior Wire Co$.$ produces three types, Product I, Product II$,$ and Product III, of cable for industrial purposes. Three machines, Machine A$,$ Machine B and Machine C$,$ are utilized to achieve three processes, weaving, coating, and coiling, respectively.

The firm operates throughout the year except for $121$ closure dates. The plant operates three $8-$hour shifts daily.

All machines require varying intervals of calibration, cleaning, and maintenance and are conducted in accordance with the following schedule in Table $1.$

The abovementioned processes differ in duration to produce three products, Product I, Product II$,$ and Product III. The durations in hours are listed in Table $2.$

Table $1$

$*$For example, Machine A will undergo a $40-$minute calibration $24$ times during the year. During the calibration event $($occurrence$),$ Process I, which is executed by Machina A$,$ will be paused.

$1.$ What is the design capacity of the machines? Specifically, what are the available hours of operation when only considering days of closure?

Design Capacity$=($Number of days in operation$-$closure days$)$ x number of shifts per day x hours per shift

Days in operation is $365$ days.

Closure days is $121$ days.

The number of shifts per day is $3$ shifts.

Hours per shift is $8$ hours.

Design Capacity$=(365-121)$ x $3$ x $8=9,552$ hours

$2.$ What is the effective capacity in hours of each, A$,$ B and C$,$ machine type after accounting for calibration, cleaning, and maintenance?

Effective Capacity $=$ Actual Output $/$ Effective Capacity x $100$

For Machine A:

Calibration $=40$ minutes x $24$ occurrences $=960$ minutes

Cleaning $=12$ minutes x $24$ occurrences $=288$ minutes

Maintenance $=40$ minutes x $24$ occurrences $=960$ minutes

Total time for Machine A $=960+288+960=2,208$ minutes

Effective Capacity $=$ Actual Output $/$ Effective Capacity x $100$

Actual Output $=9,552$ hours

Effective Capacity $=9,552/2,208$ x $100=44\mathrm{.}2\%$

For Machine B:

Calibration $=40$ minutes x $24$ occurrences $=960$ minutes

Cleaning $=12$ minutes x $24$ occurrences $=288$ minutes

Maintenance $=40$ minutes x $24$ occurrences $=960$ minutes

Total time for Machine B $=960+288+960=2,208$ minutes

Effective Capacity $=$ Actual Output $/$ Effective Capacity x $100$

Actual Output $=9,552$ hours

Effective Capacity $=9,552/2,208$ x $100=44\mathrm{.}2\%$

For Machine C:

Calibration $=40$ minutes x $24$ occurrences $=960$ minutes

Cleaning $=12$ minutes x $24$ occurrences $=288$ minutes

Maintenance $=40$ minutes x $24$ occurrences $=960$ minutes

Total time for Machine C $=960+288+960=2,208$ minutes

Effective Capacity $=$ Actual Output $/$ Effective Capacity x $100$

Actual Output $=9,552$ hours

Effective Capacity $=9,552/2,208$ x $100=44\mathrm{.}2\%$

Final answer: Therefore, the effective capacity in hours of Machine A is $44\mathrm{.}2\%,$ Machine B is $44\mathrm{.}2\%,$ and Machine C is $44\mathrm{.}2\%.$

Three product types, Product I, Product II and Product III, require different processing intervals during production. Specifically, Table $2$ depicts the processing time in hours for each product$/$process combination:

Table $2$

$*$For example, $1,000$ hours of processing time is required to complete the weaving of a single unit of Product I.

To determine the number of each machine type required to fulfill the demand for $70,62,$ and $40$ units of Product I, Product II$,$ and Product III, respectively, we can use the processing time data provided in Table $2.$

$3.$ How many of each machine type is required to fulfill $70,62,$ and $40$ units of Product I, Product II$,$ and Product III, respectively?

Units of Capacity Needed$=$Processing Time Needed$/$Processing Time Capacity Per Unit