Question

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 10^8 meters/sec.a. Calculate the bandwidth-delay product, R _ dprop.b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time?c. Provide an interpretation of the bandwidth-delay product.d. What is the width (in meters) of a bit in the link? Is it longer than a football field?e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m.

Answer 1

a ) The bandwidth-delay is 160,000 bits.

b)The maximum number of bits at a given time will be 160,000 bits.

c ) The maximum number of bits on the transmission line is equal to the product of he bandwidth and the delay

d) The length is 125 meters which is longer than a football field

e) The general expression is given as
`(S)/(R)`

Step-by-step explanation:

Step-by-Step Explanation

(a) We are old from he question that A and B are connected with a direct link.

The length between A and B is given as ( d )
`= 20,000 Km = 2×10⁷ meters` given that (1 Km = 10³ )

The rate of transmission on the direct connection between A and B

Denoted as (R) = 2 Mbps = 2 × 10⁶bps given that (1Mbps = 10⁶bps)

Speed of propagation of the link between A and B

Denoted as (S) = 2.5 × 10⁸ meters/sec

Formula to calculate propagation delay denoted as dₐ =
`(D)/(S)`

dₐ =
`(D)/(S)`

=
`(2 * 10 ^7)/(2.5*10^8)`

= 0.08 sec

We multiply the rate of transmission with the propagation delay to get the bandwidth delay

i.e. R × dₐ = (2 × 10⁶) × 0.08

= 16 × 10⁴ bits

Hence the bandwidth delay is given as 160,000 bits.

Step 2 :

b) From the question we are given that file is been transmitted from host A to host B consistently as one large message.

Given that the size of the file o be transmitted is =800,000 bits

= 8 × 10⁵bits

The rate of transmission of the direct connection between A and B (R)

= 2 Mbps

= 2 × 10⁶bps (Given that 1Mbps = 10⁶ bps)

Note : That the maximum number of bits the connection can carry is not dependent on the file size

but on the bandwidth -delay i.e (R * da). It is also equal in value hence

At a given time the maximum number of bits is equal to the bandwidth product delay = 160,000 bits

Step 3

c) The highest number of bits on the link between A and B is gotten from the product of bandwidth and delay

Step 4:

Given from the question that the propagation speed of the connect A and B (S) = 2.5 × 10⁸ meters/sec

The rate of transmission through the direct connection between A and B (R) = 2 Mbps

= 2 × 10⁶ bps (Given that 1Mbps = 10⁶ bps)

We then have that the length of 1 bit on the link can be calculated with the formula
`(S)/(R)`

Length of 1 bit =
`(S)/(R)`

=
`(2.5 * 10^8)/(2*10^6)`

= 125 m/bit

We can see from the answer that the length of 1 bit is longer than a football field

Step 5:

We can deduce the width of a bit by looking at the unites of the given parameters

Unit for the length of the link = m meters

Unit of propagation speed = S m/s

the transmission rate is = R bps

Note :Width of a bit is not dependent on the length but on the propagation speed and the rate of transmission and it is the ratio these parameter

Hence the width of a bit is given as =
`(S)/(R)`