Peterson Chap 1 Q5

Question

 5. Consider a point-to-point link 4 km in length. At what bandwidth
would propagation delay (at a speed of 2 × 10⁸m/s) equal transmit delay for 100-byte packets? What about 512-byte packets?
 

Comments

for 100 byte is this 20*10^6bps.and 512 byte is 102.4*10^6 bps?

Answer 1

Propagation delay(pd) = d/v

Now it(pd) should be equal to L/B

so L/B = d/v

B = L*v/d = 100*8*2*10^8/4*1000 bits/sec = 4*10^7 bits/sec = 40 Mbps

For 512 byte packet size B = 40*512/100 Mbps = 204.8 Mbps

Answer 2

point to point link length=4km

speed of propagation delay=2*10⁸ m/sec

packet size 100 B

L/B = 2*d/v

or,B=L*v/(2*d)= 100*8 *2*10⁸ /(2*4*1000) =2*10⁷ bits/sec

                                                                   =25 * 10⁵ B/sec

For 512 B packet

 512*8*2*10⁸ /(8*1000) =1024*10⁸ b/sec

                                      =128*10⁸ B/sec

Comments

why have u taken tran. time = 2* PD . they want tran. time = PD. So answer would be 40 Mbps