A car repair shop has two hoists where cars can be lifted for repair work. Currently customers come in at the rate of 4 per hour and are processed at a similar rate. On average 8 cars are waiting to be processed, 4 needing routine repairs and 4 needing major repairs. People are served on a first come first serve basis. Now: The repair shop owner feels that he is losing many customers needing routine repair because of the long wait. He dedicates one hoist for routine repair and one for major repairs. A study indicates that routine repairs are processed at the rate of 3 per hour and major repairs at the rate of 1 per hour. There are now 5 people waiting on average for routine repairs and 3 waiting on average for major repairs. With the new system, what is the average waiting time over all customers

At the routine repair hoist, 5 people waiting on average hence the Inventory (I) = 5 cars. The cars are processed at a rate of 3 per hour, hence the Throughput (R) = 3 cars per hour.

Therefore the Flow time (T) = I/R = 5/3 = 1.67 hours.

The Cars wait an average of 1.67 hours before being served at routine repairs.

At the major repair hoist, 3 people waiting on average hence the Inventory (I) = 3 cars. The cars are processed at a rate of 1 per hour, hence the Throughput (R) = 1 cars per hour.

Therefore the Flow time (T) = I/R = 3/1 = 3 hours.

The Cars wait an average of 3 hours before being served at major repairs.

TOPEADENIRAN2GO TOPEADENIRAN2GO · Answer 2

The cars will wait an average of 1.67 hours before being served at routine repairs while they'll wait an average of 3 hours before being served at major repairs.

From the information given, at the routine repair hoist, 5 people waiting on average and the cars are processed at a rate of 3 per hour, therefore the flow time (T) will be:

= I/R = 5/3 = 1.67 hours.

Also, at the major repair hoist, 3 people wait on average and the cars are processed at a rate of 1 per hour. Therefore, the Flow time (T) will be:

= I/R = 3/1 = 3 hours.

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