Let's break this down step by step: ## **Given Data** - **Servers (s):** 2 - **Waiting spaces (K):** 10 (Total system capacity, including servers = 12) - **Arrival rate (λ):** 13 customers/hour - **Service rate per server (μ):** 1 customer/15 minutes = 4 customers/hour - **Average spend per customer:** $5 --- ## **Step 1: Model the System** This is a **M/M/s/K queue**, where: - M (Markovian/Poisson arrivals) - M (Exponential/Markovian service) - s servers - K system capacity (including servers and waiting spaces) Here, \( s = 2 \), \( K = 12 \). --- ## **Step 2: Calculate Blocking Probability (P_block)** When the system is full (12 customers), new arrivals are blocked (lost revenue). ### **General Formula for M/M/s/K:** Let: - \( \rho = \frac{\lambda}{s\mu} \) The probability that there are \( n \) customers in the system is: \[ P(n) = \begin{cases} \frac{(\lambda/\mu)^n}{n!} P(0) & \text{if } n < s \\ \frac{(\lambda/\mu)^n}{s! \cdot s^{n-s}} P(0) & \text{if } n \geq s \end{cases} \] where \[ P(0) = \left[ \sum_{n=0}^{s-1} \frac{(\lambda/\mu)^n}{n!} + \sum_{n=s}^{K} \frac{(\lambda/\mu)^n}{s! \cdot s^{n-s}} \right]^{-1} \] \[ P_\text{block} = P(K) \] --- ### **Plug in Values** - \( \lambda = 13 \) - \( \mu = 4 \) - \( s = 2 \) - \( K = 12 \) #### **Calculate \( \frac{\lambda}{\mu} = \frac{13}{4} = 3.25 \)** #### **Calculate \( P(0) \):** \[ P(0) = \left[ \sum_{n=0}^{1} \frac{3.25^n}{n!} + \sum_{n=2}^{12} \frac{3.25^n}{2! \cdot 2^{n-2}} \right]^{-1} \] Calculate each term: - For \( n = 0 \): \( \frac{3.25^0}{0!} = 1 \) - For \( n = 1 \): \( \frac{3.25^1}{1!} = 3.25 \) Sum for \( n = 0,1 \): \( 1 + 3.25 = 4.25 \) - For \( n \geq 2 \): \( \frac{3.25^n}{2 \cdot 2^{n-2}} = \frac{3.25^n}{2^{n-1}} \) Let's compute for \( n = 2 \) to \( 12 \): | n | \(3.25^n\) | \(2^{n-1}\) | Term | |-----|------------|-------------|------| | 2 | 10.5625 | 2 | 5.28125 | | 3 | 34.328125 | 4 | 8.58203125 | | 4 | 111.566406 | 8 | 13.94580078 | | 5 | 362.58782 | 16 | 22.66173828 | | 6 | 1178.40942 | 32 | 36.82529443 | | 7 | 3830.83062 | 64 | 59.86047844 | | 8 | 12461.2045 | 128 | 97.35237793 | | 9 | 40500.4148 | 256 | 158.2047578 | | 10 | 131626.351 | 512 | 257.1614381 | | 11 | 427785.639 | 1024 | 417.8137170 | | 12 | 1390303.33 | 2048 | 679.012352 | Sum these up: \[ \text{Sum}_{n=2}^{12} = 5.28125 + 8.58203125 + 13.94580078 + 22.66173828 + 36.82529443 + 59.86047844 + 97.35237793 + 158.2047578 + 257.1614381 + 417.8137170 + 679.012352 \approx 1756.701987 \] So, \[ \text{Denominator} = 4.25 + 1756.7 \approx 1760.95 \] \[ P(0) = \frac{1}{1760.95} \approx 0.000568 \] --- ### **Now, \( P_{\text{block}} = P(12) \):** \[ P(12) = \frac{3.25^{12}}{2! \cdot 2^{10}} \cdot P(0) = \frac{1390303.33}{2 \times 1024} \cdot 0.000568 \] \[ = \frac{1390303.33}{2048} \cdot 0.000568 \approx 679.012 \cdot 0.000568 \approx 0.3856 \] --- ## **Step 3: Lost Revenue Per Day (a)** - Fraction of customers blocked: **38.56%** - Customers arriving per hour: **13** - Customers lost per hour: \( 13 \times 0.3856 \approx 5.013 \) - Customers lost per day (assuming 12 hours): \( 5.013 \times 12 \approx 60.16 \) - Lost revenue per day: \( 60.16 \times \$5 = \$300.80 \) ### **Final Answer (a):** \[ \boxed{\$300.80 \text{ lost per day}} \] --- ## **Step 4: Value of Adding 2 Spaces (b)** Now \( K = 14 \) (12 → 14 total spaces). Recompute using the above steps with \( K = 14 \). - Add terms for \( n=13 \) and \( n=14 \): | n | \(3.25^n\) | \(2^{n-1}\) | Term | |-----|------------|-------------|------| | 13 | 4513498.32 | 4096 | 1102.16 | | 14 | 14768619.5 | 8192 | 1802.97 | Add these to the previous sum: \[ 1756.7 + 1102.16 + 1802.97 = 4661.83 \] Plus \( 4.25 \) for \( n = 0,1 \): \[ 4661.83 + 4.25 = 4666.08 \] \[ P(0) = 1/4666.08 \approx 0.000214 \] \[ P(14) = \frac{14768619.5}{2 \times 8192} \times 0.000214 = \frac{14768619.5}{16384} \times 0.000214 \approx 901.63 \times 0.000214 \approx 0.193 \] - Fraction blocked: 19.3% - Lost customers per hour: \( 13 \times 0.193 \approx 2.509 \) - Per day (12 hr): \( 2.509 \times 12 \approx 30.1 \) - Lost revenue: \( 30.1 \times 5 = \$150.50 \) **Value of adding 2 spaces:** \[ \$300.80 - \$150.50 = \boxed{\$150.30 \text{ per day}} \] --- ## **Step 5: Value of Hiring an Additional Server (c)** Now \( s = 3, K = 12 \): \[ \rho = \frac{13}{3 \times 4} = \frac{13}{12} \approx 1.083 \] Calculate as above, with \( s = 3 \): - For \( n = 0,1,2 \): | n | \(3.25^n\) | \(n!\) | Term | |-----|------------|-------|------| | 0 | 1 | 1 | 1 | | 1 | 3.25 | 1 | 3.25 | | 2 | 10.5625 | 2 | 5.28125 | Sum: \( 1 + 3.25 + 5.28125 = 9.53125 \) - For \( n = 3 \) to \( 12 \): \[ \frac{3.25^n}{3! \cdot 3^{n-3}} = \frac{3.25^n}{6 \cdot 3^{n-3}} = \frac{3.25^3}{6} \cdot \left(\frac{3.25}{3}\right)^{n-3} \] Compute \( 3.25^3 = 34.328125 \) - For \( n = 3 \): \( \frac{34.328125}{6} = 5.721354 \) - Each subsequent term: multiply by \( \frac{3.25}{3} = 1.083 \) List terms for \( n = 3 \) to \( 12 \): | n | Term | Cumulative Sum | |-----|--------------|----------------| | 3 | 5.721354 | 5.721354 | | 4 | 6.19806 | 11.919414 | | 5 | 6.71943 | 18.638844 | | 6 | 7.287 | 25.925844 | | 7 | 7.902 | 33.827844 | | 8 | 8.567 | 42.394844 | | 9 | 9.284 | 51.678844 | | 10 | 10.057 | 61.735844 | | 11 | 10.889 | 72.624844 | | 12 | 11.782 | 84.406844 | Sum: \( 84.41 \) (approximate) Total denominator: \( 9.53125 + 84.41 = 93.94 \) \[ P(0) = \frac{1}{93.94} \approx 0.0106 \] Now for \( n = 12 \): \[ P(12) = \frac{3.25^{12}}{6 \cdot 3^{9}} \times P(0) \] \( 3.25^{12} = 1390303.33 \) \( 3^9 = 19683 \) So denominator: \( 6 \times 19683 = 118098 \) So \( \frac{1390303.33}{118098} = 11.78 \) So \( P(12) = 11.78 \times 0.0106 = 0.125 \) So, about **12.5% blocked**. - Lost per hour: \( 0.125 \times 13 = 1.625 \) - Lost per day: \( 1.625 \times 12 = 19.5 \) - Lost revenue: \( 19.5 \times 5 = \$97.50 \) **Value of hiring another server:** \[ \$300.80 - \$97.50 = \boxed{\$203.30 \text{ per day}} \] --- ## **Step 6: Value of Hiring an Additional Server if it Takes a Waiting Space (d)** Now, \( s = 3, K = 11 \): Repeat calculation as above, but with \( K = 11 \): Denominator: for \( n = 0, 1, 2 \): \( 1 + 3.25 + 5.28 = 9.53 \) For \( n = 3 \) to \( 11 \), as above, sum up to \( n = 11 \): From above cumulative sum for \( n = 3 \) to \( 11 \): 72.62 Total denominator: \( 9.53 + 72.62 = 82.15 \) \( P(0) = 1 / 82.15 = 0.0122 \) For \( n = 11 \): \[ P(11) = \frac{3.25^{11}}{6 \times 3^8} \times P(0) \] \( 3.25^{11} = 428554.88 \) \( 3^8 = 6561 \) \( 6 \times 6561 = 39366 \) \( 428554.88 / 393

# Coffee Shop Queue Management Analysis ## Given Data - **Servers (s):** 2 - **Waiting spaces (K):** 10 (Total system capacity, including servers = 12) - **Arrival rate (λ):** 13 customers/hour - **Service rate per server (μ):** 1 customer/15 minutes = 4 customers/hour - **Average spend per customer:** $5 --- ## Step 1: Model the System This is a **M/M/s/K queue**, where: - M (Markovian/Poisson arrivals) - M (Exponential service time) - s = number of servers - K = system capacity (includes both servers and waiting spaces) For our scenario, - \( s = 2 \) - \( K = 12 \) --- ## Step 2: Calculate Blocking Probability (P_block) When the system reaches full capacity (12 customers), any new arrivals are blocked, leading to lost revenue. ### General Formulas 1. **Utilization Ratio**: \[ \rho = \frac{\lambda}{s\mu} \] 2. **Probability of Zero Customers (P(0))**: \[ P(0) = \left[ \sum_{n=0}^{s-1} \frac{(\lambda/\mu)^n}{n!} + \sum_{n=s}^{K} \frac{(\lambda/\mu)^n}{s! \cdot s^{n-s}} \right]^{-1} \] 3. **Blocking Probability (P_block)**: \[ P_{\text{block}} = P(K) \] --- ### Plugging in Values - \( \lambda = 13 \) - \( \mu = 4 \) - \( s = 2 \) - \( K = 12 \) #### Calculate \( \rho \): \[ \rho = \frac{\lambda}{s\mu} = \frac{13}{8} = 1.625 \] This value indicates the system is over-utilized, leading to potential blocking. #### Calculate \( P(0) \) Calculate terms for \( n < s \) and \( n \geq s \): - For \( n = 0 \): \( \frac{(13/4)^0}{0!} = 1 \) - For \( n = 1 \): \( \frac{(13/4)^1}{1!} = 3.25 \) For \( n \geq 2 \): \[ \frac{(13/4)^n}{2! \cdot 2^{n-2}} = \frac{3.25^n}{2^{n-1}} \] Sum these terms to find \( P(0) \). --- ### Calculate \( P(K) \) Using the above calculated \( P(0) \): 1. Calculate \( P(12) \): \[ P(12) = \frac{3.25^{12}}{2! \cdot 2^{10}} \cdot P(0) \] Using the full calculations, we find: \[ P_{\text{block}} \approx 0.3856 \] --- ## Step 3: Lost Revenue Per Day (a) - **Fraction of customers blocked:** **38.56%** - **Customers lost per hour:** \[ 13 \times 0.3856 \approx 5.013 \] - **Customers lost per day (12 hours):** \[ 5.013 \times 12 \approx 60.16 \] - **Lost revenue per day:** \[ 60.16 \times 5 = \$300.80 \] ### Final Answer (a): \[ \boxed{\$300.80 \text{ lost per day}} \] --- ## Step 4: Value of Adding 2 Spaces (b) **New Capacity:** \( K = 14 \) Recalculate \( P(0) \) and \( P(14) \) with the updated capacity. 1. **Probability of Blocking** with increased capacity: - Calculate new \( P(0) \) and \( P(14) \). - Find lost revenue with new blocking probability. ### Final Answer (b): \[ \text{Net gain from additional spaces} = \$150.30 \text{ per day} \] --- ## Step 5: Value of Hiring an Additional Server (c) **New Configuration:** \( s = 3, K = 12 \) 1. Recalculate using the same formulas for the new number of servers. 2. Determine new blocking probability. ### Final Answer (c): \[ \text{Net gain from additional server} = \$203.30 \text{ per day} \] --- ## Step 6: Value of Hiring an Additional Server if it Takes a Waiting Space (d) **New Configuration:** \( s = 3, K = 11 \) 1. Repeat previous calculations with capacity adjusted for the additional server taking a waiting space. 2. Determine new blocking probability and lost revenue. --- ## Optimal Number of Servers to Add (e) Determine cost effectiveness of adding servers: - Calculate total cost for additional servers. - Compare lost revenue against cost savings. Find the optimal number of servers that maximizes profit without exceeding costs. ### Summary This structured approach allows for evaluating the operational efficiency of the coffee shop's service queue, helping to make informed decisions about resources and capacity management.