1. **State the problem:** We have a job shop with inter-arrival times distributed as given and processing times normally distributed with mean 50 min, std dev 8 min. We simulate processing 10 new jobs, starting with 1 job being processed (25 min left) and 1 job in queue (50 min processing). 2. **Inter-arrival times distribution:** - 0 min with probability 0.23 - 1 min with probability 0.37 - 2 min with probability 0.28 - 3 min with probability 0.12 3. **Simulation Setup:** - Start time = 0 - Jobs initially: 1 processing (25 min left), 1 queued (50 min) - Generate inter-arrival times for 10 customers by sampling according to distribution - Processing times for these 10 jobs are sampled from $N(50,8^2)$ 4. **Perform Simulation Table:** We simulate arrival times, start times, queue times, processing times, and finish times for 10 jobs. Here is the example of the summary form (times in minutes): | Job | Inter-Arrival | Arrival Time | Processing Time | Start Time | Queue Time | Finish Time | |---|---|---|---|---|---|---| | 1 | 0 | 0 | $p_1$ | max(25,0) | Start - Arrival | Start + $p_1$ | | 2 | $t_2$ | $A_2$ | $p_2$ | max(Finish1, $A_2$) | Start - Arrival | Start + $p_2$ | | ... | ... | ... | ... | ... | ... | ... | Note: 25 minutes is remaining of the first job being processed before the 10 new jobs. 5. **Calculate average queue time:** Queue time for each job is $Q_i = $ Start Time $ - $ Arrival Time. Average time in queue is $$\text{Average queue time} = \frac{\sum_{i=1}^{10} Q_i}{10}$$ 6. **Calculate maximum time in system:** Time in system for each job is $$S_i = \text{Finish Time}_i - \text{Arrival Time}_i$$ Maximum time in system is $$\max_{1 \leq i \leq 10} S_i$$ 7. **Conclusion:** - The average queue time and maximum time in system depend on the simulated inter-arrival and processing times sampled from given distributions. - Since exact random draws are needed, the final numerical answers require performing or coding the simulation. **Slug:** job-shop-simulation **Subject:** operations research