CTMC Balance Equations

Definition

The balance equations for a CTMC equate the rate at which the process leaves a state with the rate at which it enters that state.

For each state $j\in S$:

${\nu }_j{\pi }_j={\sum }_{k\ne j}{\pi }_k{q}_{kj}$

where:

• ${\nu }_j={\sum }_{k\ne j}{q}_{jk}$ is the rate of leaving state $j$
• $q_{kj}$ is the transition rate from $k$ to $j$
• ${\pi }_j$ is the limit probability of state $j$

Interpretation

The left side (${\nu }_j{\pi }_j$) is the rate at which the process leaves state $j$. The right side ($\sum {\pi }_k{q}_{kj}$) is the rate at which it enters $j$. In steady state, these must be equal — otherwise the probability mass would accumulate or drain from state $j$.

Matrix Form

$\mathbf{\pi }Q=0,{\sum }_k{\pi }_k=1$

For Birth and Death Processes

State	Balance Equation
0	${\lambda }_0{\pi }_0={\mu }_1{\pi }_1$
1	$({\lambda }_1+{\mu }_1){\pi }_1={\mu }_2{\pi }_2+{\lambda }_0{\pi }_0$
2	$({\lambda }_2+{\mu }_2){\pi }_2={\mu }_3{\pi }_3+{\lambda }_1{\pi }_1$
$n\ge 1$	$({\lambda }_n+{\mu }_n){\pi }_n={\mu }_{n+1}{\pi }_{n+1}+{\lambda }_{n-1}{\pi }_{n-1}$

These simplify to: ${\lambda }_n{\pi }_n={\mu }_{n+1}{\pi }_{n+1}$ for all $n\ge 0$.

Derivation from Forward Equation

Starting from the Kolmogorov forward equation and taking $t\to \infty$:

${\lim }_{t\to \infty }\frac{d}{dt}{P}_{ij}(t)={\lim }_{t\to \infty }\left[-P_{ij}(t){\nu }_j+{\sum }_{k\ne j}{P}_{ik}(t)q_{kj}\right]$

At steady state, the derivative is zero, yielding: $0=-{\pi }_j{\nu }_j+{\sum }_{k\ne j}{\pi }_k{q}_{kj}$.

Related

• CTMC Limit Probabilities
• Existence of CTMC Limit Probabilities
• Solving BD Process Balance Equations
• Kolmogorov Differential Equations

Exercises

Back to Roadmap 📖 → 🃏 → ✏

Kuis 2 2025 No. 10. M/M/1/2: $\lambda =2$/menit, $\mu =3$/menit. Pelanggan yang datang saat sistem penuh (2 pelanggan) pergi (balking). Tuliskan dan selesaikan persamaan balance untuk ${\pi }_0,{\pi }_1,{\pi }_2$.

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

Kuis 2 2024 No. 3. Taksi tiba rate 1/menit, pelanggan tiba rate 2/menit. Pelanggan pergi jika tidak ada taksi. Modelkan state sebagai jumlah taksi menunggu (bisa 0, 1, 2, …). Tulis balance equation untuk state $n\ge 1$.

Jawaban: $\lambda =1$ (taksi), $\mu =2$ (pelanggan mengambil taksi). Balance: $\lambda {\pi }_n=\mu {\pi }_{n+1}\Rightarrow 1\cdot {\pi }_n=2\cdot {\pi }_{n+1}\Rightarrow {\pi }_{n+1}=\frac{1}{2}{\pi }_n$. ${\pi }_n=(\frac{1}{2}{)}^n{\pi }_0$.