Columbia Pizza Has One Oven That Can Make A Whole Pie ✓ Solved

Columbia Pizza Has One Oven That Can Make A Whole Pie In About 10 Minu

Columbia Pizza Has One Oven That Can Make A Whole Pie In About 10 Minu

Columbia Pizza has one oven that can make a whole pie in about 10 minutes. Their late-night business among hungry college students is booming, and from 9 pm – 11 pm, they average 5 pie orders per hour. However, from 5 pm – 9 pm, they average only 3 pie orders per hour. Both order rates can be assumed to arrive according to a Poisson distribution. The owner of Columbia Pizza has discovered that if people have to wait more than 20 minutes from when they get in the door to when they receive their pie, they are likely to simply go next door to University Falafel, meaning that Columbia Pizza would lose out on their $20 sale.

The owner is thinking of buying a second pizza oven and register (e.g., a second channel) that he thinks would cost $75 per night to operate (for the sake of this example, that is both the amortized cost of the oven and the pay of the person who is taking orders for the six hours that the pizza shop is open). Details Answer each of the questions below. You must show all work. Be sure to highlight your final answer and when asked, please be sure to justify your response.

Questions and calculations:

1. Average number of orders in the system and waiting time for 5-9 shift

Demand rate (λ) during 5-9 shift: 3 orders/hour.

Service time per pizza: 10 minutes = 1/6 hour.

Service rate (μ): 1 pizza / 10 minutes = 6 pizzas/hour.

Since ρ = λ / (number of servers * μ), with 1 server (oven), ρ = 3 / 6 = 0.5.

Using M/M/1 queue formulas:

  • Average number of orders in the system (L):
  • L = ρ / (1 - ρ) = 0.5 / (1 - 0.5) = 0.5 / 0.5 = 1.
  • Average time an order spends in the system (W):
  • W = 1 / (μ - λ) = 1 / (6 - 3) = 1 / 3 hour = 20 minutes.
  • Average number of orders in the queue (Lq):
  • Lq = ρ^2 / (1 - ρ) = (0.5)^2 / (1 - 0.5) = 0.25 / 0.5 = 0.5.
  • Average waiting time in the queue (Wq):
  • Wq = Lq / λ = 0.5 / 3 ≈ 0.1667 hour ≈ 10 minutes.

2. Average number of orders in the system and waiting time for 9-11 shift

Demand rate (λ): 5 orders/hour.

Service rate (μ): 6 pizzas/hour (same as above).

> ρ = 5 / 6 ≈ 0.8333.

Applying M/M/1 formulas:

  • L = ρ / (1 - ρ) = 0.8333 / (1 - 0.8333) ≈ 0.8333 / 0.1667 ≈ 5.
  • W = 1 / (μ - λ) = 1 / (6 - 5) = 1 hour = 60 minutes.
  • Lq = ρ^2 / (1 - ρ) ≈ (0.8333)^2 / 0.1667 ≈ 0.6944 / 0.1667 ≈ 4.17.
  • Wq = Lq / λ ≈ 4.17 / 5 ≈ 0.834 hour ≈ 50 minutes.

3. Total maximum revenue opportunity assuming all customers are served

For 5-9 shift:

Average number of arrivals in 4 hours: 3 orders/hour * 4 hours = 12 orders.

All served, total revenue: 12 orders * $20 = $240.

For 9-11 shift:

Average arrivals: 5 orders/hour * 2 hours = 10 orders.

Total revenue: 10 * $20 = $200.

4. Revenue if a 20-minute total wait time is enforced with one oven

Since W is 20 minutes (from earlier calculations) at 5-9 shift and 60 minutes at 9-11, only the 5-9 shift can meet the target without additional measures. For the 9-11 shift, we need to reduce system load or add resources to meet the 20-minute total time.

Maximum served customers in 5-9 shift: limited to those who can be served within 20 mins. Since the average W (20 mins) matches the acceptable wait, all customers can be served, with revenue of $240.

In 9-11 shift, with 60-minute wait, the maximum customers served within 20 minutes is less, so the effective revenue is less than $200. Specifically, the number of customers served within 20 mins is equivalent to those in the queue with waiting time ≤20 mins.

5. Daily loss of sales to University Falafel

During 5-9 shift: full capacity served, no loss.

During 9-11 shift: only customers with W ≤ 20 mins are served. Since W ≈ 50 minutes, customers exceeding that leave, causing potential loss.

Number of customers not served within 20 minutes in 9-11 shift :

  • Expected arrivals: 10 orders.
  • Since average W = 50 mins, some customers wait longer than 20 mins. Specific loss depends on queue distribution, but approximating with Lq and Wq, the unserved demand is significant, leading to estimated loss of approximately 4-5 customers worth $20 each, totaling around $80-$100 per day.

6. Effects of buying a second oven (assuming operational cost of $75/night)

For 5-9 shift with two ovens:

  • Demand rate: 3 orders/hour, service rate per oven remains 6/hour.
  • Utilization per oven with two servers: ρ = λ / (number of ovens μ) = 3 / (2 6) = 0.25.
  • Average number of orders in the system (L):
  • L = ρ / (1 - ρ) / number of servers = 0.25 / (1 - 0.25) = 0.25 / 0.75 ≈ 0.333.
  • Average time in the system (W):
  • W = L / λ = 0.333 / 3 ≈ 0.111 hour ≈ 6.7 minutes.
  • Average number of orders in queue (Lq):
  • Lq = ρ^2 / (1 - ρ) ≈ (0.25)^2 / 0.75 ≈ 0.0625 / 0.75 ≈ 0.0833.
  • Average waiting time in queue (Wq):
  • Wq = Lq / λ ≈ 0.0833 / 3 ≈ 0.028 hour ≈ 1.67 minutes.

For 9-11 shift with two ovens:

  • Demand rate: 5 orders/hour.
  • Utilization ρ = 5 / (2 * 6) ≈ 0.4167.
  • L ≈ 0.4167 / (1 - 0.4167) / 2 ≈ 0.4167 / 0.5833 / 2 ≈ 0.357 / 2 ≈ 0.1785.
  • W = L / λ = 0.1785 / 5 ≈ 0.0357 hours ≈ 2.14 minutes.
  • Lq = ρ^2 / (1 - ρ) / 2 ≈ (0.4167)^2 / 0.5833 / 2 ≈ 0.1736 / 0.5833 / 2 ≈ 0.2976 / 2 ≈ 0.149.
  • Wq = Lq / λ ≈ 0.149 / 5 ≈ 0.0298 hours ≈ 1.79 minutes.

Maximum revenue with two ovens (assuming all customers served):

  • 5-9 shift: 3 orders/hour * 4 hours = 12 orders, revenue = $240. Since capacity increases, actual served probably remains close to demand, so revenue remains similar or slightly higher due to capacity.
  • 9-11 shift: 10 orders, revenue = $200.

7. Revenue opportunity with two ovens to meet 20-min wait time

With two ovens, system W drops to roughly 6-7 minutes in 5-9 shift and about 2-3 minutes in 9-11 shift, improved queue management allows all customers to be served within 20 minutes.

Maximize served customers to the demand level -> same as above, but possibly more customers served if demand exceeds existing capacity. Thus, revenue potential increases accordingly.

8. Total daily loss of sales to University Falafel with two ovens

Losses reduce significantly because more customers can be served within the 20-minute threshold, potentially approaching full demand capture, minimizing lost sales.

9. Should Columbia Pizza buy the second oven? Why? What is the net financial impact?

Based on the above, the second oven reduces waiting times, increases service capacity, and captures more sales, especially during peak hours. Cost of $75 per night is offset by increased revenue (~$40 more during 9-11 shift and improved service in 5-9 shift). The net financial impact is positive if additional demand exists beyond current capacity.

10. Effect of reducing pizza-making time to 8 minutes on recommendation

With a shorter preparation time, service rate increases (μ becomes 1 pizza/8 minutes ≈ 7.5 pizzas/hour), further decreasing waiting times and queues with either one or two ovens. This improvement strengthens the case for investing in additional capacity as demand grows, or possibly negating the need for a second oven altogether if demand remains the same. Operational efficiency gains support a strategic decision to expand capacity or improve margins.

Summary and conclusion:

By applying queuing theory formulas to the demand and service rates during different shifts, we see that current capacity handles off-peak well but faces issues during peak hours leading to potential lost sales. Adding a second oven significantly improves queuing metrics and revenue potential. Lowering pizza preparation time further enhances capacity and reduces wait times, favoring investment in capacity expansion if demand forecasts justify it. Financially, investing in a second oven appears justified provided demand levels are sufficient to offset operational costs, especially considering the competitive advantage gained by reducing customer wait times and preventing lost sales to competitors like University Falafel.

References

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