VIPSolutionsTime-Varying Queue Surge: Max Queue and Clearance Time Queue: M/M/1 with time-varying arrival λ(t) and constant service μ. Parameters: μ=5/min; λ(t)= 2/min for 0 ≤ t < 30 min, 6/min for 30 ≤ t < 90 min, 2/min for t ≥ 90 min. Initial condition: empty at t=0. Rules: Fluid approximation for expected queue: dQ/dt = λ(t) − μ when λ≠μ, reflecting at 0; piecewise linear in expectation. Max queue occurs near end of overload interval; clearance time solves Q(t_clear)=0 with λ
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Time-Varying Queue Surge: Max Queue and Clearance Time Queue: M/M/1 with time-varying arrival λ(t) and constant service μ. Parameters: μ=5/min; λ(t)= 2/min for 0 ≤ t < 30 min, 6/min for 30 ≤ t < 90 min, 2/min for t ≥ 90 min. Initial condition: empty at t=0. Rules: Fluid approximation for expected queue: dQ/dt = λ(t) − μ when λ≠μ, reflecting at 0; piecewise linear in expectation. Max queue occurs near end of overload interval; clearance time solves Q(t_clear)=0 with λ
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Asked by: Vishal
Created at: 2025-09-24 15:12:09
Asked by: Vishal
Created at: 2025-09-24 15:12:09
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**Preparing results presentation**
I'm planning to present the results with details like "Expected maximum queue size: 60 customers; Clearance time after t=90: 20 minutes; Absolute clearance time: t=110 min." I think it's important to state the rule or formula before using it. So, I’ll include a subheading called "Rules" where I can outline the formula. Now, it's time to put everything together and produce the final output!
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# Time-Varying Queue Surge: Max Queue and Clearance Time
## Rule (Fluid Approximation)
- Differential equation: dQ/dt = λ(t) − μ when Q(t) > 0; reflecting at 0.
- Piecewise integrate with given λ(t).
## Piecewise Queue Dynamics
- 0 ≤ t < 30: λ=2, μ=5 ⇒ dQ/dt = −3; reflecting at 0 ⇒ Q(t)=0.
- 30 ≤ t < 90: λ=6, μ=5 ⇒ dQ/dt = +1 ⇒ Q(t) = (t − 30). So Q(90−) = 90−30 = 60.
- t ≥ 90: λ=2, μ=5 ⇒ dQ/dt = −3 ⇒ Q(t) = 60 − 3(t − 90), until it hits 0.
## Maximum Queue During Surge
- By rule, max occurs at end of overload interval (t=90).
- Q_max = Q(90) = 60 customers.
## Clearance Time After t=90
- Solve Q(t)=0 with λ<μ: 60 − 3(t − 90) = 0 ⇒ t − 90 = 20 min.
- Clearance occurs 20 minutes after t=90, i.e., at t=110 minutes.
## Final Answer
- Expected maximum queue size: 60 customers (at t=90).
- Time after t=90 to clear: 20 minutes (clears at t=110).
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