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Give step-by-step solution with explanation and final answer:3. (a) Machines in a repair station arrive at rate of forty per hour and workman can cope with sixty per hour for an eight-hour day. You are required to calculate. (i) The average time in the system. (ii) The average length of the queue of machines that form from time to time. (iti) The implied cost of machine’s time if owner of the system has rejected a faster service which could raise cost by Rs. 800 per day and would raise the service to eighty per hour. (2+2+4=8)

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Give step-by-step solution with explanation and final answer:Uploaded Image3. (a) Machines in a repair station arrive at rate of forty per hour and workman can cope with sixty per hour for an eight-hour day. You are required to calculate. (i) The average time in the system. (ii) The average length of the queue of machines that form from time to time. (iti) The implied cost of machine’s time if owner of the system has rejected a faster service which could raise cost by Rs. 800 per day and would raise the service to eighty per hour. (2+2+4=8)

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Asked by: Test User 100171
Created at: 2025-05-13 20:19:49

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Let's solve each part step by step. --- ### **Given Data** - Arrival rate (\(\lambda\)) = 40 per hour - Service rate (\(\mu\)) = 60 per hour - System works for 8 hours a day This is a classic **M/M/1 queue** (single server, Poisson arrivals, exponential service time). --- ## **1. The average time in the system (\(W\))** For an M/M/1 system: \[ W = \frac{1}{\mu - \lambda} \] Substitute values: \[ \mu = 60, \quad \lambda = 40 \] \[ W = \frac{1}{60 - 40} = \frac{1}{20} \text{ hours} = 3 \text{ minutes} \] **Answer:** **Average time in the system = 3 minutes** --- ## **2. The average length of the queue (\(L_q\))** \[ L_q = \frac{\lambda^2}{\mu(\mu - \lambda)} \] Substitute values: \[ L_q = \frac{40^2}{60 \times (60-40)} = \frac{1600}{60 \times 20} = \frac{1600}{1200} = 1.33 \] **Answer:** **Average length of the queue = 1.33 machines** --- ## **3. Implied cost of machine's time (Cost analysis if faster service is rejected)** If owner rejects faster service (from 60/hr to 80/hr) which would cost Rs. 800 more per day, what is the implied cost per machine per hour? ### **Step 1: Calculate Average Time Saved per Machine with Faster Service** - New service rate, \(\mu' = 80\) per hour \[ W' = \frac{1}{\mu' - \lambda} = \frac{1}{80-40} = \frac{1}{40} \text{ hours} = 1.5 \text{ minutes} \] - Old average time in system = 3 minutes (from above) - New average time in system = 1.5 minutes (from above) - **Time saved per machine** = \(3 - 1.5 = 1.5\) minutes = \(0.025\) hours ### **Step 2: Find Total Time Saved in a Day** - Number of arrivals per day = \(40 \times 8 = 320\) machines - **Total time saved per day:** \[ 320 \times 0.025 = 8 \text{ hours} \] ### **Step 3: Implied Cost per Hour** - Cost to save 8 hours = Rs. 800 - **Implied cost per hour:** \[ \text{Implied cost per hour} = \frac{800}{8} = Rs. 100 \text{ per hour} \] --- ## **Final Answers (Summary)** 1. **Average time in the system:** \(3\) minutes 2. **Average length of the queue:** \(1.33\) machines 3. **Implied cost of machine's time:** **Rs. 100 per hour** --- **Let me know if you want detailed formulas or further explanations!**

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