EE465-USC-Fall08-Assignment7_Solution

EE465-USC-Fall08-Assignment7_Solution - EE 465 : Homework 7...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 465 : Homework 7 Solutions 1. (a) The desired expected time is E [ T 0 ] + E [ T 1 ] + E [ T 2 ] + E [ T 3 ] where E [ T i ] is the expected time to go from state i to state i + 1. E [ T 0 ] = 1 λ 0 = 1 λ . E [ T 1 ] = 1 λ 1 + μ 1 λ 1 E [ T 0 ] = 1 2 λ + μ 2 λ 2 E [ T 2 ] = 1 λ 2 + μ 2 λ 2 E [ T 1 ] = 1 3 λ + 2 μ 3 λ 1 2 λ + μ 2 λ 2 E [ T 3 ] = 1 λ 3 + μ 3 λ 3 E [ T 3 ] = 1 4 λ + 3 μ 4 λ 1 3 λ + 2 μ 3 λ ± 1 2 λ + μ 2 λ 2 ¶‚ Add E [ T 0 ] ...E [ T 3 ] to get the desired result. (b) E [ T 2 ]+ E [ T 3 ]+ E [ T 4 ], where E [ T 2 ] and E [ T 3 ] is as given in part (a) and E [ T 4 ] = 1 5 λ + 4 μ 5 λ E [ T 3 ]. 2. (a) We have the system shown in Figure 1. In other words, 0 1 2 n-1 n n+1 . . . . . . . . . N N+1 N+2 . . . Θ μ λ + Θ 2 μ (n-1) λ + Θ n μ n λ + Θ (n+1) μ N λ (N+1) λ (N+1) μ (N+2) μ Figure 1: The transition graph. λ n = + θ, n < N λ n = nλ, n N μ n = (b) Immigration is restricted if n 3. Therefore we need to find n =3 P n = 1 - [ P 0 + P 1 + P 2 ]. P 1 = θ μ P 0 P 2 = λ + θ 2 μ P 1 = θ ( λ + θ ) 2 μ 2 P 0 P 3 = 2 λ + θ 3 μ P 2 = θ ( λ + θ )(2 λ + θ ) 6 μ 3 P 0
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
For n 4, we get P n = ( n - 1) λ P n - 1 = ( n - 1)( n - 2) ... (3) n ( n - 1) ... (4) ± λ μ n - 3 P 3 = 3 n ± λ μ n - 3 P 3 Therefore, n =3 P n = 3 ( μ λ ) 3 P 3 n =3 1 n λ μ · n . But if λ μ < 1, n =1 1 n λ μ · n = log ± 1 1 - λ μ = log μ μ - λ · . So, n =3 P n = 3 ( μ λ ) 3 P 3 log μ μ - λ · - λ μ - 1 2 λ μ · 2 = 3 ( μ λ ) 3 log μ μ - λ · - λ μ - 1 2 λ μ · 2 θ ( λ + θ )(2 λ + θ ) 6 μ 3 P 0 . Now, i =0 P i = 1 implies P 0 = " 1 + θ μ + θ ( λ + θ ) 2 μ 2 + " log ± μ μ - λ - λ μ - 1 2 ± λ μ 2 # θ ( λ + θ )(2 λ + θ ) 2 λ 3 # - 1 And finally, X n =3 P n = log μ μ - λ · - λ μ - 1 2 λ μ · 2 θ ( λ + θ )(2 λ + θ ) 2 λ 3 1 + θ μ + θ ( λ + θ ) 2 μ 2 + log μ μ - λ · - λ μ - 1 2 λ μ · 2 θ ( λ + θ )(2 λ + θ ) 2 λ 3 3. The corresponding birth and death process is shown in Figure 2. The birth rate
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/05/2009 for the course EE 465 taught by Professor Chow during the Fall '04 term at USC.

Page1 / 8

EE465-USC-Fall08-Assignment7_Solution - EE 465 : Homework 7...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online