Consider a taxi station where taxis looking for passengers and passengers looking for taxis arrive according to Poisson processes, with mean rates per minute of 1 and 1.5

Consider a taxi station where taxis looking for passengers and passengers looking for taxis arrive according to Poisson processes, with mean rates per minute of 1 and 1.5. A taxi will wait no matter how many other taxis are in line, but an arriving passenger waits only if the number of passengers already waiting for taxis is two or less.

(a) Define the states, draw the transition diagram, and mark the rates. For clear understanding draw on a paper
(b) Write the balance equations and calculate the steady-state probabilities
(c) Find the mean number of taxis waiting for passengers.
(d) Find the mean number of passengers waiting for taxis.
(e) Find the mean number of passengers who in the course of an hour do not join the waiting line because at least three passengers were already waiting.

The Correct Answer and Explanation is:

Let’s work through this queueing system scenario step-by-step. The key idea is to model the number of people (taxis or passengers) waiting. This is a birth-death process, where the state of the system represents the net number of people waiting (positive for passengers, negative for taxis, zero for a perfect match).

(a) Define States and Transition Diagram

Let state n represent the number of people waiting:

If n = 0, no one is waiting — each arrival (passenger or taxi) instantly pairs with the other.
If n > 0, then n passengers are waiting for taxis.
If n < 0, then |n| taxis are waiting for passengers.

Allowed states:
n ∈ {…, -3, -2, -1, 0, 1, 2, 3}
(Note: Passengers are allowed to wait only if ≤ 2 are in queue, so maximum is 3 passengers — the arriving one plus two already waiting.)

Transitions:

Passenger arrival (rate λ = 1.5): causes state to move +1, unless already in state 3.
Taxi arrival (rate μ = 1): causes state to move –1.

Transition rates:

From state n to n+1: rate 1.5 if n < 3
From state n to n–1: rate 1 always (taxis always arrive and wait)

Diagram: States from –∞ to 3, with arrows between each pair of adjacent states and labeled rates.

(b) Balance Equations and Steady-State Probabilities

Let π_n be the steady-state probability of being in state n.

We can model this as a birth-death process, and define the balance equations recursively:

Let’s assume the system is stable and truncates at state 3 (since no more passengers can join when 3 are waiting). For states n ≤ 2:

$$
\pi_{n+1} = \left(\frac{\lambda}{\mu}\right) \pi_n = 1.5 \cdot \pi_n
$$

Starting from π_0, we get:

π₁ = 1.5 π₀
π₂ = (1.5)^2 π₀
π₃ = (1.5)^3 π₀

For taxis waiting (n < 0):

π₋₁ = (1 / 1.5) π₀ = (2/3) π₀
π₋₂ = (2/3)^2 π₀
π₋₃ = (2/3)^3 π₀
and so on…

Normalize: Total probability = 1

$$
\pi_0 \left[ \sum_{n=0}^{3} (1.5)^n + \sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n \right] = 1
$$

Compute the sums:

Forward (n=0 to 3): $\sum_{n=0}^{3} (1.5)^n = 1 + 1.5 + 2.25 + 3.375 = 8.125$
Backward (n=1 to ∞): geometric series $$
\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n = \frac{\frac{2}{3}}{1 – \frac{2}{3}} = 2
$$

So total = π₀ × (8.125 + 2) = π₀ × 10.125 → π₀ = 1 / 10.125 ≈ 0.09877

Now calculate other probabilities:

π₁ ≈ 1.5 × 0.09877 ≈ 0.14815
π₂ ≈ (1.5)^2 × 0.09877 ≈ 0.22222
π₃ ≈ (1.5)^3 × 0.09877 ≈ 0.33333
π₋₁ ≈ (2/3) × 0.09877 ≈ 0.06585
π₋₂ ≈ (2/3)^2 × 0.09877 ≈ 0.04390
π₋₃ ≈ (2/3)^3 × 0.09877 ≈ 0.02927
And so on…

(c) Mean Number of Taxis Waiting

Taxis wait in states: n = -1, -2, -3, ...

Expected number:

$$
E[T] = \sum_{n=1}^{\infty} n \cdot \pi_{-n} = \sum_{n=1}^{\infty} n \cdot \left(\frac{2}{3}\right)^n \cdot \pi_0
$$

Use formula for expected value of geometric series:

$$
\sum_{n=1}^{\infty} n r^n = \frac{r}{(1 – r)^2}, \text{ with } r = \frac{2}{3}
$$

$$
E[T] = \pi_0 \cdot \frac{\frac{2}{3}}{(1 – \frac{2}{3})^2} = \pi_0 \cdot \frac{\frac{2}{3}}{(\frac{1}{3})^2} = \pi_0 \cdot \frac{2}{3} \cdot 9 = 6 \cdot \pi_0 ≈ 6 × 0.09877 ≈ 0.5926
$$

(d) Mean Number of Passengers Waiting

Passengers wait in states n = 1, 2, 3.

$$
E[P] = \sum_{n=1}^{3} n \cdot \pi_n = 1 × π₁ + 2 × π₂ + 3 × π₃
= 0.14815 + 2 × 0.22222 + 3 × 0.33333
= 0.14815 + 0.44444 + 1.0 = 1.5926
$$

(e) Passengers Who Do Not Join Queue per Hour

Passengers are turned away when system is in state n = 3.

Arrival rate is 1.5/min, and fraction of time in state 3 is π₃ ≈ 0.33333. So:

$$
\text{Blocked rate} = 1.5 × π₃ = 1.5 × 0.33333 = 0.5 \text{ passengers/min}
$$

In 1 hour:

$$
0.5 × 60 = \boxed{30 \text{ passengers}}
$$

✅ Final Answers:

(c) Mean taxis waiting: ≈ 0.5926
(d) Mean passengers waiting: ≈ 1.5926
(e) Passengers turned away per hour: 30