Question

175-3. Consider the birth-and-death process with the following mean rates. The birth rates are Ao-2, A1 3,A.: 2. A 3 1, and A,s() for " > 3. The death rates are μ.-3,Pc-4. μ.-1,and = 2 for n > 4. (a) Construct the rate diagram for this birth-and-death process. (h) Develop the balance equations. (c) Solve these equations to find the steady-state probability dis- (di Use the general formulas for the birth-and-death process to cal- Also calculate L. L W.and

Answer 1

(a) Rate diagram for birth-and-death process

b) $2P_{0}=3P_{1}.....(1)$

$2P_{0}+4P_{2}=6P_{1}.....(2)$

$3P_{1}+1P_{3}=6P_{2}.....(3)$

$2P_{2}+2P_{4}=2P_{3}.....(4)$

$1P_{3}=2P_{4}.....(5)$

$P_{0}+P_{1}+P_{2}+P_{3}+P_{4}=1.....(6)$

(c) $\mathbf{{\color{DarkGreen} P_{1}=\frac{2}{3}P_{0}}}.....(from (1))$

  $2P_{0}+4P_{2}=6(\frac{2}{3}P_{0}).....(from (2))$

$2P_{0}+4P_{2}=4P_{0}$

  $4P_{2}=2P_{0}$

  ${\color{DarkGreen} \mathbf{P_{2}=\frac{1}{2}P_{0}}}$

$3P_{1}+P_{3}=6P_{2}.....(from (3))$

$3(\frac{2}{3}P_{0})+P_{3}=6(\frac{1}{2}P_{0})$

$2P_{0}+P_{3}=3P_{0}$

$\mathbf{{\color{DarkGreen} P_{3}=P_{0}}}$

$2P_{2}+2P_{4}=2P_{3}.....(from (4))$

$2(\frac{1}{2}P_{0})+2P_{4}=2P_{0}$

$P_{0}+2P_{4}=2P_{0}$

$\mathbf{{\color{DarkGreen} P_{4}=\frac{1}{2}P_{0}}}$

$P_{0}+P_{1}+P_{2}+P_{3}+P_{4}=1....(from(6))$

$\therefore P_{0}+\frac{2}{3}P_{0}+\frac{1}{2}P_{0}+P_{0}+\frac{1}{2}P_{0}=1$

$\therefore P_{0}(1+\frac{2}{3}+\frac{1}{2}+1+\frac{1}{2})=1$

$\therefore P_{0}(\frac{6+4+3+6+3}{6})=1$

$\therefore (\frac{22}{6})P_{0}=1$

$\therefore \mathbf{{\color{DarkBlue} P_{0}=\frac{6}{22}=\frac{3}{11}}}$

$\therefore \mathbf{{\color{DarkBlue} P_{1}=\frac{2}{3}(P_{0})=\frac{2}{3}\frac{3}{11}=\frac{2}{11}}}$

$\therefore \mathbf{{\color{DarkBlue} P_{2}=\frac{1}{2}(P_{0})=\frac{1}{2}\frac{3}{11}=\frac{3}{22}}}$

$\therefore \mathbf{{\color{DarkBlue} P_{3}=P_{0}=\frac{3}{11}}}$

$\therefore \mathbf{{\color{DarkBlue} P_{4}=\frac{1}{2}(P_{0})=\frac{1}{2}\frac{3}{11}=\frac{3}{22}}}$

d) $L=0P_{0}+1P_{1}+2P_{2}+3P_{3}+4P_{4}$

$=0(\frac{3}{11})+1(\frac{2}{11})+2(\frac{3}{22})+3(\frac{3}{11})+4(\frac{3}{22})$

$=0+\frac{2}{11}+\frac{3}{11}+\frac{9}{11}+\frac{6}{11}$

$\mathbf{{\color{DarkGreen} =\frac{20}{11}}}$

$L_{q}=0P_{1}+1P_{2}+2P_{3}+3P_{4}$

$=0(\frac{2}{11})+1(\frac{3}{22})+2(\frac{3}{11})+3(\frac{3}{22})$

$=0+\frac{3}{22}+\frac{6}{11}+\frac{9}{22}$

$\mathbf{{\color{DarkGreen} =\frac{24}{22}=\frac{12}{11}}}$

$\lambda ^{-}=\lambda_{0} P_{0}+\lambda_{1} P_{1}+\lambda_{2} P_{2}+\lambda_{3} P_{3}$

$=2(\frac{3}{11})+3(\frac{2}{11})+2(\frac{3}{22})+1(\frac{3}{11})$

$\mathbf{{\color{DarkGreen} =\frac{18}{11}}}$

$W=\frac{L}{\lambda^{-} }=\frac{20/11}{18/11}=\frac{10}{9}$

$W_{q}=\frac{L_{q}}{\lambda^{-} }=\frac{12/11}{18/11}=\frac{2}{3}$