Answer:
The bandwidth is 3200 Kbps
Explanation:
Given:
The length of the point to point link = 50 km = 50×10³ = 50000 m
Speed of transmission = 2 × 10⁸ m/s
Therefore the propagation delay is the ratio of the length of the link to speed of transmission
Therefore, [tex]t_{prop} = \frac{distance}{speed}[/tex]
[tex]t_{prop} = \frac{50000}{2*10^{8} }=2.5*10^{-4}[/tex]
Therefore the propagation delay in 25 ms
100 byte = (100 × 8) bits
Transmission delay = [tex]\frac{(100*8)bits }{(x)bits/sec}[/tex]
If propagation delay is equal to transmission delay;
[tex]\frac{(100*8)bits }{(x)bits/sec}= \frac{50000}{2*10^{8} }[/tex]
[tex]x=\frac{100*8*2*10^{8} }{50000}[/tex]
[tex]x=\frac{100*8}{2.5*10 ^-4}=3200000[/tex]
x = 3200 Kbps
Answer:
The complete question is here:
Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 x 108 m/s) equal transmission delay for
(a) 100-byte packets?
(b) 512-byte packets?
Explanation:
Given:
Length of the link = 50 km
Propagation delay = distance/speed = d/s
To find:
At what speed would propagation delay (at a speed 2 x 10^8 meters/second) equal transmit delay for 100 byte packets?
Solution:
Propagation Delay = t prop
= 50 * 10^3 / 2 * 10^8
= 50000 / 200000000
= 0.00025
= 25 ms
(a) When propagation delay is equal to transmission delay for 100 packets:
Transmission Delay = packet length / data rate = L / R
So when,
Transmission Delay = Propagation Delay
100 * 8 / bits/sec = 50 * 10^3 / 2 * 10^8
Let y denotes the data rate in bit / sec
speed = bandwidth = y bits/sec
100 * 8 / y bits/sec = 50 * 10^3 / 2 * 10^8
y bit/sec = 100 * 8 / 25 * 10^ -5
= 800 / 0.00025
y = 3200000 bit/sec
y = 3200 Kbps
(b) 512-byte packets
512 * 8 / y bits/sec = 50 * 10^3 / 2 * 10^8
y bit/sec = 512 * 8 / 25 * 10^ -5
= 4096 / 0.00025
y = 16384000 bit/sec
y = 16384 Kbps
