Tema 2 - Teletráfico Soluciones de la hoja 1 de problemas
Tema 2 - Teletráfico Soluciones de la hoja 1 de problemas
Tema 2 - Teletráfico Soluciones de la hoja 1 de problemas
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E.T.S.I.I.T - Ingeniería <strong>de</strong> Telecomunicación<br />
Re<strong>de</strong>s Telefónicas - Curso 2011/2012<br />
<strong>Tema</strong> 2 - <strong>Teletráfico</strong><br />
<strong>Soluciones</strong> <strong>de</strong> <strong>la</strong> <strong>hoja</strong> 1 <strong>de</strong> <strong>problemas</strong><br />
Problema 1.<br />
(a) 2000 minutos.<br />
(b) 11.1 Er<strong>la</strong>ngs.<br />
(c) 11.1 l<strong>la</strong>madas.<br />
Problema 2.<br />
(a) 1 3 l<strong>la</strong>madas.<br />
(b) 2 segundos.<br />
(c) 2 3 circuitos.<br />
(d)<br />
Problema 3.<br />
(a) TO A = ρ = λ . µ<br />
TO B = ρ . 2<br />
1<br />
(b) PB A = .<br />
1+ 2 ρ + 2<br />
ρ 2<br />
PB B = 1<br />
1+ 2 ρ<br />
(c) T A = T B = 1 . 2µ<br />
(d) T A = T B = 1 . λ<br />
(e) A: p 1 =<br />
ρ<br />
1+ρ+ ρ2 2<br />
B: p 1 = ρ 2<br />
.<br />
1+ ρ 2<br />
1<br />
(f) A: p 0 =<br />
1+ρ+ ρ2 2<br />
B: p 0 = 1 .<br />
1+ ρ 2<br />
(g) N A = ρ+ρ2<br />
1+ρ+ ρ2 2<br />
B: N B = ρ 2<br />
.<br />
1+ ρ 2<br />
(h) A: λ C = λ(1−PB A ).<br />
B: λ C = λ(1−PB B ).<br />
p 2 = ρ2 2<br />
1+ρ+ ρ2 2<br />
Problema 4.<br />
(a) TO a = 10 Er<strong>la</strong>ngs.<br />
TO b = 20 Er<strong>la</strong>ngs.<br />
(b) PB a = PB b ≈ 0.245.<br />
(c) TC a ≈ 7.55 Er<strong>la</strong>ngs.<br />
TC b ≈ 15.1 Er<strong>la</strong>ngs.<br />
(d) TP a ≈ 2.45 Er<strong>la</strong>ngs.<br />
TP b ≈ 4.9 Er<strong>la</strong>ngs.<br />
(e) Pr a = 1 3<br />
Pr b = 2 3 .<br />
(f) Pr a = 1 3<br />
Pr b = 2 3 .<br />
Problema 5.<br />
(a) TC ≈ 7.85 Er<strong>la</strong>ngs.<br />
(b) TC ≈ 0.586 Er<strong>la</strong>ngs.<br />
(c) λ C ≈ 27.3 l<strong>la</strong>madas/hora.<br />
(d) λ C ≈ 17.6 l<strong>la</strong>madas/hora.<br />
(e) TC ≈ 7.85 Er<strong>la</strong>ngs.<br />
(f) TC ≈ 0.785 Er<strong>la</strong>ngs.<br />
(g) TC ≈ 0.785 Er<strong>la</strong>ngs.<br />
(h) λ C ≈ 23.6 l<strong>la</strong>madas/hora.<br />
Problema 6.<br />
(a) 0.4<br />
(b) 0.4<br />
(c) N = 1.2 circuitos.<br />
(d) 0.667<br />
(e) 0.2<br />
(f) 0.4<br />
(g) N = 1.2 circuitos.<br />
Problema 7.<br />
(a) 2)<br />
(b) 3)<br />
(c) 2)<br />
(d) 4)
Problema 8.<br />
(a)<br />
(b) p 00 ≈ 0.571 p 10 ≈ 0.286<br />
p 01 ≈ 0.057 p 10 ≈ 0.086<br />
(c) TO ≈ 0.55 Er<strong>la</strong>ngs.<br />
(d) N = 0.514 líneas.<br />
(e) PB ≈ 0.086.<br />
Problema 9.<br />
(a) TC ≈ 1.96 Er<strong>la</strong>ngs.<br />
(b) TC ≈ 1.04 Er<strong>la</strong>ngs.<br />
(c) 0.39.<br />
(d) 0.865.<br />
(e) 4.2 Er<strong>la</strong>ngs.<br />
Problema 10.<br />
(a) 0.28 segundos.<br />
(b) 0.23 segundos.<br />
Problema 11.<br />
(a) 2)<br />
(b) 2)<br />
(c) 3)<br />
(d) 3)<br />
Problema 12.<br />
(a)<br />
Problema 13.<br />
(a)<br />
(b) p i = A i p 0<br />
p i = 2 ( N A<br />
1<br />
p 0 =<br />
1−A N+1<br />
1−A<br />
(c) TO = A<br />
Problema 14.<br />
Problema 15.<br />
(a)<br />
Problema 16.<br />
Problema 17.<br />
( A<br />
) i−i<br />
p<br />
2 0 i > 0<br />
.<br />
2e A 2 −1<br />
( ) ]<br />
4<br />
e A 2 −1 −1 .<br />
A<br />
( ) )−4p 0 e A 2 −1 Problema 18.<br />
)<br />
N servidor = p 0<br />
(4e A 2 −4−A .<br />
[ ]<br />
Ae A 2<br />
−1 .<br />
4e A 2 −4−A<br />
W servidor = 1 . µ<br />
(b) p i = A i!<br />
p 0 = 1<br />
[<br />
(c) λ = p 0 λ<br />
(d) N = p 0 Ae A 2.<br />
(e) N co<strong>la</strong> = Ap 0<br />
(e A 2 +1<br />
(f) W co<strong>la</strong> = 1 µ<br />
i ≤ N<br />
i<br />
2)<br />
p 0 i > N<br />
.<br />
+AN+1 2−A<br />
(d) TC = Ap 0<br />
1−A N<br />
1−A .<br />
(e) PB = 1−p 0 .<br />
(f) p 0<br />
A N+1<br />
2−A .<br />
(a)<br />
(b) p 0 = 0.482 p 1 = 0.386 p 2 = 0.116<br />
p 3 = 0.015 p 4 = 0.01.<br />
(c) N = 0.667<br />
(d) 0.1 segundos.<br />
(b) p 0 = p 1 = p 2 = 4 15<br />
p 3 = 2<br />
15<br />
p 4 = 1<br />
15 .<br />
(c) p C bps = 8<br />
15<br />
p 2C bps = 1 5<br />
(d) PB = 1 15<br />
(e) W = 11<br />
7 segundos.<br />
(a) u = 320 paquetes/segundo.<br />
(b) λ 1 = 1600 pkt/s λ<br />
3 2 = 320 pkt/s.<br />
3<br />
(a) TO A = TC A = 2 Er<strong>la</strong>ngs<br />
TO B = TC B = 0.9 Er<strong>la</strong>ngs.<br />
(b) W A ≈ 26 segundos<br />
W B = 81 minutos.<br />
(c) 48 l<strong>la</strong>madas<br />
(a) PB 1 ≈ 0.21 PB 2 = 0.2.<br />
(b) 0.4.<br />
(c) N 1 = 30<br />
19<br />
N 2 = 0.8<br />
W 1 = 1 s N 2 = 0.5 s
Problema 19.<br />
(a) Original: p i = ( λL<br />
C<br />
p 0 = 2C−λL<br />
2C+λL<br />
Alternativo: p i = ( λL<br />
(b) Original: W co<strong>la</strong> =<br />
(c) α = 1+ λL C<br />
(d) Sí<br />
) i 1<br />
2 i−1 2C−λL<br />
2C+λL<br />
i > 0<br />
) i Cα−λL<br />
Cα Cα<br />
L(λL) 2<br />
C[(2C) 2 −(λL) 2 ]<br />
λL<br />
Alternativo: W co<strong>la</strong> = 2<br />
[<br />
1− λL ]<br />
4C<br />
Problema 20.<br />
(a) 2 5<br />
(b) N = 4 W = 1 5 2<br />
(c) 1 5<br />
(d) 17<br />
57<br />
(e) PB 1 = 17 PB<br />
57 2 = 25<br />
57<br />
Problema 21.<br />
(a) p i = p 0 A i i < S<br />
p i = p 0 A i q i−S i ≥ S<br />
1<br />
p 0 =<br />
1−A S+1<br />
1−A +AS+1 q<br />
1−Aq<br />
(b) q < C λL<br />
(c) (1−q)<br />
1<br />
1−A S<br />
1−A + AS<br />
1−Aq<br />
(d) p 0<br />
[<br />
A 2 (1−A S )<br />
(1−A) 2<br />
(e) 1 λ<br />
(f)<br />
A<br />
µ−λ<br />
A 2 (1−A S−1 )<br />
(1−A 2 )<br />
Problema 22.<br />
A S<br />
1−Aq<br />
− SAS+1<br />
1−A +<br />
Cα(Cα−λL)<br />
+ AS+2 q 2<br />
+ AS+1 Sq<br />
(1−Aq) 2 1−Aq<br />
− (S−1)AS<br />
1−A + AS Aq (S−1)<br />
(1−Aq) 2+AS 1−Aq<br />
1−A S q<br />
1−A +AS 1−Aq<br />
(a) PB ≈ 0.21 TC ≈ 1.58<br />
(b) W = 1 minuto<br />
Porcentaje ocupación: 52.63%<br />
(c) 4 operadores.<br />
(d) ≈ 0.21<br />
(e) 41.7%<br />
(f) PB premium ≈ 0.094<br />
PB no-premium ≈ 0.458<br />
(g) TC ≈ 1.37 Er<strong>la</strong>ngs<br />
W = 1 minuto.<br />
(h) ≈ 3 horas y 36 minutos.<br />
]<br />
Problema 23.<br />
(a) p i = 2 ( A i<br />
2)<br />
i > 0<br />
p 0 = 1−A 2<br />
1+ A 2<br />
W = 1 µ<br />
(b) N = A<br />
1− A2<br />
4<br />
(c) p i = A i (1−A)<br />
N = A<br />
1<br />
1− A2<br />
4<br />
W = 1 1−A µ 1−A<br />
A<br />
A<br />
(d) p 10 = p 0 p<br />
2 01 = p 0 2α<br />
A<br />
p i = p i<br />
0<br />
p 0 =<br />
(e) A = (1+α)<br />
1<br />
(1+α) i−2 2α<br />
2α(1+α−A)<br />
2α(1+α)+A(1+α 2 )<br />
Problema 24.<br />
(<br />
1−<br />
(a) 0.4<br />
(b) W espera = 1 2 minutos<br />
W total = 3 2 minutos.<br />
(c) ≈ 0.46<br />
(d) W espera = 0 minutos<br />
W total = 5 8 minutos.<br />
Problema 25.<br />
1<br />
√ )<br />
α(1+α)<br />
1+α 2<br />
(a) Pr espera = 1 3<br />
PB = 1 3<br />
(b) W espera = 10 minutos<br />
W total = 30 minutos.<br />
(c) Pr espera = 0.6 PB = 0.4<br />
(d) W espera = 20 minutos<br />
W total = 40 minutos.<br />
Problema 26.<br />
(a) Pr pérdida = 1 5<br />
(b) N = 0.8<br />
W aplicación = 3 segundos.<br />
(c) (Pr pérdida ) RRHH<br />
= 1 5<br />
(Pr pérdida ) No RRHH<br />
= 1 5<br />
(d) (Pr pérdida ) RRHH<br />
= 1 16<br />
(Pr pérdida ) No RRHH<br />
= 1 4<br />
(e) N procesos = 7 8<br />
N espera = 1 16<br />
(f) W espera = 3 segundos (≈ 0.214 s)<br />
14<br />
N espera = 45 segundos (≈ 3.214 s)<br />
14<br />
(g) Pr pérdida = 1 8
Problema 27.<br />
(a) Pr bloqueo = 2 7<br />
W espera = 1 min (12 s).<br />
5<br />
(b) Pr bloqueo = 3 13<br />
W espera = 1 min (10 s).<br />
6<br />
(c) Pr bloqueo = 1 6<br />
W espera = 1 min (6 s).<br />
10<br />
(d) α = 1 2<br />
(a) Pr bloqueo = 1 . 3<br />
(b) W espera = 5 min.<br />
W total = 15 min.<br />
(c) Pr bloqueo = 4 . 7<br />
(d) W simu<strong>la</strong>ción = 20 min.<br />
Problema 28.<br />
W espera = 40<br />
3<br />
W espera = 1 min (7.5 s). 8<br />
min (800 s).<br />
(e) Simu<strong>la</strong>ciones = 2.<br />
W simu<strong>la</strong>ción = 10 min.