The Profit Maximizing Cutoff in Observable Queues with State Dependent Pricing Christian Borgs (MSR-NE) Jennifer T. Chayes (MSR-NE) Sherwin Doroudi (CMU-Tepper) Mor Harchol-Balter (CMU-CS) Kuang Xu (MIT-LIDS) Support provided by Microsoft Computational Thinking grant 1. Background & Motivation 4. Problem Statement & Approach CHARGE ENTRY PRICE OBSERVABLE QUEUE Hmm… should I join? DELAY SENSITIVE CUSTOMERS w/ LIMITED BUDGET INPUTS: fixed 𝜆 and 𝑉 OBJECTIVE: Find cutoff 𝑘 ∗ to maximize the earning rate 𝑅(𝑘, 𝜆, 𝑉) TECHNIQUE: use a “discrete derivative” Define Δ𝑥 𝑓 𝑥 ≡ 𝑓 𝑥 + 1 − 𝑓 𝑥 Δ𝑥 is the forward difference operator Solve Δ𝑘 [𝑅 𝑘, 𝜆, 𝑉 ] = 0 for 𝑘 ∈ 𝐑 Obtain optimal cutoff 𝑘 ∗ = ⌈𝑘⌉ CLOUD COMPUTING SERVICE GOAL: Choose prices and admissions policy to maximize profits 2. The Queueing Model 𝑀/𝑀/1 queue 𝜆→ 𝜇=1 WLOG 5. The Analysis ln(𝜆) 𝐺 𝜆, 𝑉 − 𝑘 𝑒 ln All customers are identical with parameters: 𝑉 𝑣=1 Value for receiving service Waiting cost per unit time Solve: 𝑅 𝑘 + 1, 𝜆, 𝑉 − 𝑅 𝑘, 𝜆, 𝑉 = 0 1 Define: 𝑘 ≡ 𝑘 + 2 and 𝐺 𝜆, 𝑉 ≡ 1 − 𝜆 𝑉 + 1−𝜆 Question: How do we solve this for 𝑘? (𝑛 + 1) + 𝑝 ≤ 𝑉 Now solve: 𝑥𝑒 𝑥 = 𝐶 for 𝑥 Answer: Lambert 𝑊 (product log) function 𝑊 𝑥 𝑒 𝑛 : # of customers in (state of) system ln(𝜆)𝜆𝐺(𝑉,𝜆) = 1−𝜆 Call both of these 𝑥 𝑝: price customers must pay I wait only if: 𝜆 𝐺 𝜆,𝑉 −𝑘 3. How to Price? 𝑊(𝑥) =𝑥 Or equivalently 𝑊 𝑥𝑒 𝑥 = 𝑥 6. Conclusions Charge as much as customers are willing to pay in each state: 𝑝 𝑛 = 𝑉 − (𝑛 + 1) 𝝀 0 𝝀 𝝀 1 𝝀 = ceil 𝝀 1 1 ln 𝜆 ⋅ 𝜆 1−𝜆 𝑉+ − ⋅𝑊 1 − 𝜆 ln 𝜆 1−𝜆 𝑘 𝑘−1 $𝑉 − 1 𝟏 $𝑉 − 2 𝟏 1−𝜆 𝑉+ 𝑘∗ Cutoff at 𝑘 𝟏 $𝑉 − 𝑘 − 2 𝟏 $𝑉 − 𝑘 − 1𝟏 So what can the closed form do for us? But… we use an explicit early cutoff at state 𝑘. Why? Asymptotic Results For fixed 𝜆 < 1, as 𝑉 → ∞: 𝑘 ∗ ∼ 1 − 𝜆 𝑉 Answer: Cutoff shorter queue high prices more often When 𝜆 = 1, as 𝑉 → ∞: 𝑘 ∗ ∼ 2𝑉 So how much do we make? Probability: (𝑘) 𝜋0 0 Price charged: 𝝀 (𝑘) 𝜋1 𝝀 𝝀 1 (𝑘) 𝜋𝑘−1 $𝑉 − 1 𝟏 $𝑉 − 2 𝟏 For fixed 𝜆 > 1, as 𝑉 → ∞: 𝑘 ∗ ∼ log 𝜆 ( 𝑉) 𝝀 𝑘 𝑘−1 𝟏 $𝑉 − 𝑘 − 2 𝟏 Earning rate = (arrival rate) x (average price charged) Is the cutoff of practical use? When 𝜆 > 1: high cutoff huge losses! 𝑘−1 (𝑘) 𝜋𝑛 : Probability that a new arrival sees state 𝑛, given a cutoff of 𝑘 𝜆 1 𝑅(𝑘, 𝜆, 𝑉) = ⋅ 𝑘+1 1−𝜆 1−𝜆 𝑘∗ = 7 ⋅ 𝑝(𝑛)} 𝑝 𝑛 = 𝑉 − (𝑛 + 1) 1 − 𝜆𝑘 1 − 𝜆 𝑉 + 𝜆𝑘 1 + 𝑘 − 𝑘𝜆 − 1 Earning Rate 𝑅 𝑘, 𝜆, 𝑉 = 𝜆 ⋅ (𝑘) {𝜋𝑛 𝑛=0 Call this constant 𝐶 𝜆 = 1.2 𝑉 = 50 𝑘 1 1−𝜆 −2