λ
5 7
µ
1 1
ρ =
λ µ
=
5 7
= 0,7142
1
1x1 ρ = ( 0,714 1! x( 1x1 0,714 ) + −
0,7145 −1 5! )
ρ = ( 52 + 0, 00015)1 = 0, 0 , 399 1
Lq = = 0, 399 399x x 00!,714 (1 ∗
W q = =
Lq λ
0,714∗1 0,714) 2
∗
−
= 2,49
= 3, 3 , 48 48horas horas
P q = =
1 S !
∗
ρ ∗ ( s su u λ
P w =
1 1!
∗
0,7141 ( 1
∗
∗
−
∗
ρ)
1∗1 1−0,714
∗
∗
0, 399) = 99, 99,93 %
λ
1 3
µ
1 2
Ls = λ ∗ W s W s = Ls =
W s =
1 3
1 µ−λ
∗
1 µ−λ
=
1 1/2−1/3
6 =2
=
1 1/2−1/3
= 6min
Pn > k = ( µλ )k+1 P n > 2 = ( 23 )3 = 0, 296
λ = 3aver´ıa/hora
= 6min
Wa = W c =
1 µc−λc
=
1 8−3
1 µa−λa
=
1 5−3
=
=
λ µ λ
W q = ρ =
λ µ
=
3 4
W s =
λ µ∗(µ−λ)
=
1 µ−λ
3 4(4−3)
=
1 4−3
= 1minutoLs = 3 ∗ 1 = 3clientes
= 0, 75
= 0, 75 ρ ∗ 100
W q = Pn > k = ( µλ )k+1 P n > 2 = ( 34 )3 = 0, 4218 P n > 2 ∗ 100
Wb =
1 µb−λb
=
1 7−3
=
λ µ
T =
1 µ−λ
=
1 10−8
Alquilar = 7488 2000 = 3, 74
µ LLR = 250[LLR] A = V = 2500(llamadas )∗12(segundos ) 120(1segundos )
8, 33[Er]x
30 LLR HC 1[Er ] =
250LLRHC
A = 200LLRHCx 301Er = 6, 67[Er] LLR HC
V T
x12(segundos) = 2500 = 3600 segundos
P B = 1 A = 750LLRHC LLR A = 750 LLR HC x1[Er]30 HC = 25[Er].
ρ =
A C
=
25 36
= 0, 69
ρ =
A C
=
5 11
= 0, 45
1 µ
1 µ
min = 3sx 160 s =
1 20 min
min = 1sx 160 s =
µ = 20 peticiones/min λ = 50 peticiones/min
1 60 min
µ = 60 peticiones/min
λ µ
λ µ
λ µ
τ = M s = M
C = 1 ρ = µλ λ = ρ.µ N = λ.T T = W +
λ µ
N = λ × (W + µ1 ) λ = ρ.µ N = µρ × (W + µ1 ) N = ρ × (µW + 1) N q = λ.W = µρW
N − N q = ρ(µW + 1)
−
µρW N − N q =
ρ(µW + 1) − µW N − Nq = ρ
T =
N λ
=
ρ(µW +1) µρ
= W +
1 µ
∞
λ = 10 usuarios hora
W =
λ µ(µ−λ)
1 µ
= 5min ×
= 0, 4166[horas] T =
1hora 60min
1 µ−λ
=
1 12
µ = 12 usuarios ρ= hora
= 0, 5[horas]
λ µ
=
10 12
= 0, 83
P RX 1 = P T X 1 + GRX 1 + GT X 1 − Lc − Lb Lb1 = 32,4 + 20 ∗ log(f 1(MHz)) + 20 ∗ log(d1(Km) P RX 1 = M D + S
a1 P T X 1 + GRX 1 + GT X 1 − Lc
a1 = P RX 1 + Lb1 a2 = P RX 2 + Lb2 P RX 2 + Lb2 = P RX 1 + Lb1 S = P RX 1 − M D P RX 2 − M D = P RX 1 − M D M D = 0 Lb2 = Lb1 32,4 + 20 ∗ log(f 2) + 20 ∗ log(d2) = 32,4 + 20 ∗ log(f 1) + 20 ∗ log(d1) log(f 2) + log(d2) = log(f 1) + log(d1) 2 d1 log( f f 1 ) = log( d2 ) d1 log( 1800 900 ) = log( d2 ) d1 d2
=2
d1 = 2 ∗ d2
Lb2 = Lb1 40 ∗ log(d1) − 20 ∗ log(ht ∗ hr) = 40 ∗ log(d2) − 20 ∗ log(ht ∗ hr) 40 ∗ log(d1) = 40 ∗ log(d2) log( dd12 ) = 1 d1 d2
= 10
d1 = 10 ∗ d2
46,3 + 33,9 ∗ log(f 2[MHz]) − 13,82 ∗ log(ht) − a(hr) + (44,9 − 6,55 ∗ log(ht)) ∗ log(d2) = 46,3 + 33,9 ∗ log(f 1[M Hz ]) − 13,82 ∗ log(ht) − a(hr) + (44,9 − 6,55 ∗ log(ht)) ∗ log(d1) 156,65 − a(hr) + 38,5 ∗ log(d2) = 146,44 − a(hr) + 38,5 ∗ log(d1) 38,5 ∗ log( dd12 ) = 10,21 log( dd12 ) = 0,26 d1 = 1,84 ∗ d2
Loku = A + B ∗ log(d) + C A = 69,55 + 26,16 ∗ log(f [MHz]) − 13,82 ∗ log(hb) − a(hm) B = 44,9 − 6,55 ∗ log(hb) P RX 1 = P T X 1 + GRX 1 + GT X 1 − Lc − Lb
B = 44,9 − 6,55 ∗ log(15) = 37,2
89,52 = 30 + 12 − 1,5 − Lb
−
Lb = 130,02 104 = 30 + 12 − 1,5 − Lb
−
Lb = 130,02 Lb = L oku A = 69,55 + 26,16 ∗ log(f ) − 13,82 ∗ log(15) − a(hm) a(hm) = 3,29(log(11,75 ∗ hm)2 ) − 4,97 a(hm) = 3,29(log(11,75 ∗ 1,5)) − 4,97 a(hm) = −0,87 A = 69,55 + 26,16 ∗ log(900) − 13,82 ∗ log(15) + 0,87 A = 131,45 Loku = A + B ∗ log(d) + C 130,02 = 131,45 + 37,2 ∗ log(d) 0,038 = log(d)
−
d = 0,91km 144,5 = 131,45 + 37,2 ∗ log(d) dmax = 2,24km
