Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

3) Bandwidth-delay product – Suppose two hosts, A and B, are separated by 30,000

ID: 3871562 • Letter: 3

Question

3) Bandwidth-delay product – Suppose two hosts, A and B, are separated by 30,000 Km and are connected by a direct link of R=3 Mbps. Suppose the propagation speed over the link is 2.5*108 m/s.

a) Calculate the bandwidth-delay product R* dprop. (2 points)

b) Consider sending a file of 1,200,000 bits from host A to host B. Suppose the file is sent continuously as a large message. What is the maximum number of bits that will be in the link at any time? (2 points)

c) What is the width (in meters) of a bit in the link? (2 points)

d) Repeat (a), (b) and (c) but now with a link of 3 Gbps. (6 points)

Explanation / Answer

3) Bandwidth-delay product – Suppose two hosts, A and B, are separated by 30,000 Km and are connected by a direct link of R=3 Mbps. Suppose the propagation speed over the link is 2.5*10 ^8 m/s.

a) Calculate the bandwidth-delay product R* dprop.

solution:

1 Mbps = 10^6 bits/sec

1 Km = 10^3 m

R * dprop = Bandwidth Delay Product = bandwidth (bits per sec) * round trip time (in seconds)

=(3 * 10 ^6 bits /s ) * (30000 *10 ^3 m) / (2.5 * 10 ^8 m/s)

= 3.6 * 10 ^5 bits.

b) Consider sending a file of 1,200,000 bits from host A to host B. Suppose the file is sent continuously as a large message. What is the maximum number of bits that will be in the link at any time?

solution:

min(BDP, 1,200,000) = min(3.6 * 10^5 , 12* 10^5) bits

= BDP. 3.6 × 10^5 bits.

c) What is the width (in meters) of a bit in the link?

solution:

distance / BDP =( 30000 * 10^3) / (3.6 * 10^5) m

= 83.3 m

d) Repeat (a), (b) and (c) but now with a link of 3 Gbps.

(a) solution

1 Gbps = 10^9 bits/sec

R * dprop = Bandwidth Delay Product = bandwidth (bits per sec) * round trip time (in seconds)

=(3 * 10 ^9 bits /s ) * (30000 *10 ^3 m) / (2.5 * 10 ^8 m/s)

= 3.6 * 10 ^8 bits.

(b)solution:

min(BDP, 1,200,000) = min(3.6 * 10^8 , 12* 10^5) bits

= BDP. 12 × 10^5 bits.

(c)solution:

distance / BDP = (30000 * 10^3) / (3.6 * 10^8) m

= 83.3 *10^3 m

//any clarification ,please do comments

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote