Consider a point-to-point link 2 km in length. At what bandwidth would propagati

Question

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?

I have this is right?

Propagation Delay =  Distance     2000 / Speed 2*10^8

Transmition Delay  = Size 100*8/ Bandwith Mbps x*10^6

2,000/2 * 10^8= 100*8

2000/ (2*10^8) = (100 * 8) / (10^6)/100

X= 80 Mbps

Propagation Delay  = Distance     2000 / Speed 2*10^8

Transmition Delay  = Size 512*8/ Bandwith Mbps x*10^6

2000/ (2*10^8) = (512 * 8) / (10)

X= 409.6 Mbps

Explanation / Answer

let bandwidth in Mbps be X

so link transmission speed = X Mbps

here distance = 2Km = 2000 m , speed of signal = 2 * 10^8 m/s

Link propogation delay = distance / speed of signal

Link propaogation delay = 2000/2*10^8 = 10^-5 sec

and let number of bits to be transmisttedd is Y bits

then transmission delay is given by = No.of bits to transmit / Link Bandwidth = Y / X*10^6

so in problem we have to find X given Y different cases such that

Propogation delay link= Transmission delay of link

10^-5 =Y/X*10^6

X = Y/10

here two cases are :

(a) 100-byte packets?
(b) 512-byte packets?

so case a :

Y = 100 byte = 100 * 8 = 800 bits

from above equation :

X = 800/10 = 80

so link Bandwidth is = 80 Mbps

and in case b :

Y = 512 byte = 512 * 8 =   4096 bits

from above equation :

X = 4096 / 10 = 409.6

so link Bandwidth is = 409.6 Mbps

SO YOUR ANSWERSS MATCH WITH THESE . YOU CAN CHECK HERE FOR ANY NEW TYPE OF Y OR X INTERCHANGEBLY YOU CAN USE THIS FORMULAE .

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