Queueing System with Balking

Definition

A Queueing System with Balking is a variant of the 1 queue where an arriving customer who finds $n$ others already in the system will join the system with probability ${\alpha }_n$, and balk (leave without joining) with probability $1-{\alpha }_n$.

This is a birth and death process with:

${\lambda }_n=\lambda {\alpha }_n,n\ge 0$ ${\mu }_n=\mu ,n\ge 1$

Interpretation

Customers are deterred by long queues — the more people already waiting, the less likely a new arrival is to join. This models realistic behavior where customers have limited patience.

Special Case: Finite Capacity $N$

When the system has finite capacity $N$:

${\alpha }_n=\{\begin{array}{ll}1,& n<N\\ 0,& n\ge N\end{array}$

This is the M/M/1/$N$ queue, where customers finding $N$ in the system are lost (no waiting room).

Balking vs. Reneging

• Balking: Customer refuses to join upon arrival (depends on queue length at arrival time)
• Reneging: Customer joins but leaves before being served (depends on wait time)

Related

• 1 Queueing System
• Birth and Death Queueing Models
• Queueing System

Exercises

Back to Roadmap 📖 → 🃏 → ✏

Kuis 2 2025 No. 10. Loket tiket: 1 petugas, kapasitas 2 (1 dilayani + 1 menunggu). $\lambda =2$/menit, $\mu =3$/menit. Pelanggan datang saat penuh → pergi (balking). Tentukan ${\pi }_0,{\pi }_1,{\pi }_2$.

Jawaban: ${\lambda }_0={\lambda }_1=2$, ${\lambda }_2=0$ (balking). ${\mu }_1={\mu }_2=3$. ${\pi }_1=\frac{2}{3}{\pi }_0$, ${\pi }_2=\frac{2}{3}{\pi }_1=\frac{4}{9}{\pi }_0$. ${\pi }_0(1+\frac{2}{3}+\frac{4}{9})=1\Rightarrow {\pi }_0=\frac{9}{19}$, ${\pi }_1=\frac{6}{19}$, ${\pi }_2=\frac{4}{19}$.

Kuis 2 2024 No. 3. Taksi (rate 1/menit) dan pelanggan (rate 2/menit) di stasiun. Taksi selalu menunggu, pelanggan pergi (balking) jika tidak ada taksi. Tentukan rata-rata jumlah taksi menunggu.

Jawaban: State $n$ = jumlah taksi. ${\lambda }_n=1$ (taksi tiba), ${\mu }_n=2$ untuk $n\ge 1$ (pelanggan ambil taksi). ${\pi }_n=(\frac{1}{2}{)}^n{\pi }_0$, ${\pi }_0=1/2$. $L=\sum n{\pi }_n=\sum n(\frac{1}{2}{)}^{n+1}=\frac{1/2}{(1-1/2{)}^2}\cdot \frac{1}{2}=1$ taksi rata-rata.