Answer: a) 0.77 and 0.39 b) 3.9 and 0.20
Explanation:
We have to find two things here i.e link parameter and channel utilization efficiency.
a)
We are given
Bitrate= 12 Mbps,
Distance= 6 miles
Propagation Velocity= 3 x 10⁸ m/s
Length of frame= 500 bits
We know that The link parameter is related to propagation time and frame time
Where Propagation time= Bitrate x Distance= 12 Mbps x 6 miles
and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 500
So,
Link parameter= [tex]\frac{12*10^6*6*1609.34}{3*10^8*500}[/tex]
Link parameter= 0.77
Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]
Channel utilization efficiency= 0.39
b)
We are given
Bitrate= 10 Mbps,
Distance= 15 km
Propagation Velocity= 3 x 10⁸ m/s
Length of frame= 256 bits
We know that The link parameter is related to propagation time and frame time, Here QPSK is used so bitrate is multiplied by 2,
Where Propagation time= Bitrate x Distance= 2 x 10 Mbps x 15 km
and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 256
So,
Link parameter= [tex]\frac{2*10*10^6*15*1000}{3*10^8*256}[/tex]
Link parameter= 3.9
Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]
Channel utilization efficiency= 0.20
