Certainly! Let’s break down the problem step by step. **Given:** - Greg Davis operates a specialty sub-taco restaurant (SPS) near USC. - He wants to maximize productivity, especially during the busy lunch hour (11:30 a.m. – 12:30 p.m.). - Fulfilling a customer order involves **7 tasks** (details not provided, but we can proceed generically as is standard in assembly line balancing). - He wants to know: 1. If **one employee** does all 7 tasks, how many customers can be served per hour? 2. If **7 employees** (one per task), what is the **maximum output capacity**? And how many employees are *actually* needed to achieve this maximum? ## Assumptions Because the actual time for each of the 7 tasks is not given, we will use standard notation: - Let the time to complete each task be: \( t_1, t_2, t_3, t_4, t_5, t_6, t_7 \) (in minutes or seconds) ## **Step 1: All Tasks Done by One Employee** **Total time to serve one customer** = \( t_1 + t_2 + t_3 + t_4 + t_5 + t_6 + t_7 \) = \( T \) - Number of customers per hour \( = \frac{60}{T} \) (if times are in minutes; use 3600 if in seconds) - **So, answer to Q1:** > **Customers per hour = \( \boxed{\frac{60}{t_1 + t_2 + t_3 + t_4 + t_5 + t_6 + t_7} } \)** --- ## **Step 2: One Employee per Task (Assembly Line, 7 Employees)** This is a classic assembly line balancing scenario. - Each worker does just one task, and the order moves along the line. **Maximum output is determined by the slowest (longest) task.** - The time per customer is the **maximum** of the individual task times. - So, the bottleneck is \( t_{max} = \max(t_1, t_2, ..., t_7) \) - **Maximum output per hour = \( \frac{60}{t_{max}} \) customers per hour** > **Answer to Q2:** > - **Maximum output capacity = \( \boxed{\frac{60}{\max(t_1, t_2, ..., t_7)} } \)** **How many employees needed to achieve this maximum?** - One per task = 7 employees. --- ### **Summary Table** | Scenario | Output per hour formula | Number of employees | |------------------------------------|--------------------------------------------------------|--------------------| | All tasks by one employee | \( \frac{60}{t_1 + t_2 + ... + t_7} \) | 1 | | One employee per task (7 workers) | \( \frac{60}{\max(t_1, t_2, ..., t_7)} \) | 7 | --- ## **If Task Times Were Provided (Example)** Suppose task times in minutes are: | Task | Time (min) | |------|------------| | t1 | 2 | | t2 | 1 | | t3 | 1 | | t4 | 2 | | t5 | 0.5 | | t6 | 2 | | t7 | 1.5 | - **All tasks by one employee:** T = 2 + 1 + 1 + 2 + 0.5 + 2 + 1.5 = 10 min Customers/hour = 60 / 10 = **6 customers/hour** - **One employee per task:** Max task time = max(2,1,1,2,0.5,2,1.5) = 2 min Customers/hour = 60 / 2 = **30 customers/hour** (with 7 employees) --- ## **Conclusion** ### **Step-by-step answers:** **1. If all 7 tasks are handled by one employee, number of customers per hour:** - **Answer:** \( \boxed{\frac{60}{t_1 + t_2 + t_3 + t_4 + t_5 + t_6 + t_7}} \) **2. If each worker performs only one task (seven employees), what is the maximum output capacity?** - **Answer:** \( \boxed{\frac{60}{\max(t_1, t_2, ..., t_7)} } \) - **Number of employees needed:** 7 (one per task). --- **If you provide the actual times for each task, I can give you the numerical answers, but this is the general step-by-step solution you can apply to any given set of task times.**