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A barbershop has two barbers: an experienced owner and an apprentice. The owner cuts hair at the rate of 4 customers/hour, while the apprentice can only do 2 customers/hour. The owner and the apprentice work simultaneously, however, any new customer will always go first to the owner if the latter is available. The barbershop has a waiting room for only 1 customer (in case both barbers are busy), and any additional customers are turned away. Suppose customers walk by the barbershop at the rate of 6 customers/hour. Find the long-run proportion of time the apprentice is busy cutting hair. (Hint: try first to construct a continuous-time Markov chain to represent the status of the barbershop.)

User Wenger
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Final answer:

Modeling the barbershop scenario with a continuous-time Markov chain allows for the calculation of the apprentice's busy time by finding the steady-state probabilities of the system's states.

Step-by-step explanation:

The student is asking about the long-run proportion of time the apprentice in a barbershop is busy cutting hair when customers arrive at a given rate and when there are two barbers with different service rates. To solve this, one can model the situation using a continuous-time Markov chain. The owner cuts at a rate of 4 customers per hour, the apprentice at 2 customers per hour, and customers walk in at a rate of 6 customers per hour. Since the apprentice is only engaged when the owner is busy, and there's a capacity for one waiting customer, we need to calculate the probability of the states where the apprentice is actively cutting hair.

When modeling the states, state 0 represents both barbers being free, state 1 represents only the owner being busy, and state 2 represents both the owner and the apprentice being busy. Using the arrival rates and service rates, the transition rates between these states can be determined. The steady-state probabilities are then calculated, providing the long-term proportion of time each state occurs. Finally, the apprentice's busy proportion is the steady-state probability of state 2, as this is the only state where the apprentice works.

User Brad Gardner
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