consider (A) point to point link 4 kilometers in length with a bandwidth of 100

Question

consider (A) point to point link 4 kilometers in length with a bandwidth of 100 megabits per second (Mbps). Consider another link (B) 800 meters in length with bandwidth 1000 kilobits per second ( Kbps ). Assume you are sending a packet that is 125 kilo bytes (KB) in size. Which link would be faster? Show your calculations. 3. Consider a point-to-point link (A) 4 kilometers in length with a bandwidth of 100 Megabits per second (Mbps. Consider another link (B) 800 meters in length with a bandwidth of 1000 Kilobits per second (kbps). Assume you are sending a packet that is 125 Kilo Bytes (KB) in size. Which link would be faster? Show your calculations. (3 points).

Explanation / Answer

given

link A: with

bandwidth 100 mbps

length 4km

link B: with

bandwidth 1000 kbps

length 800 m

and

packet size 125 kiloBytes

now..

the progation speeds for the both links is not given...

lets consider the propation speed for the both the links is same and that is 1mbps

so, now lets calculate the time taken for transefering the packet through link A

that is = transmission delay + propagation delay

transmission delay = (packet length/bandwidth)

propagation delay = (link length/propagation speed)

packet length = 125kilobytes

1 Byte = 8bits

converting packet length Byte to bits

packet length = 125*8 kilobits = 1000 kilobits

bandwidth(A) = 100 mbps

transmission delay = 1000 kilo bits/100 mbps = 10 micro secs = 0,01secs

(link length/propagation speed) = 4 km/1mps = 0.004 secs

total = 0.01+0.004=0.014 secs

now for link B

transmission delay = (1000 kilo bits)/1000kbps =1sec

propagation delay = 800 m/1 mps = 0.8 micro secs

total = 1.0008 secs

The total time taken to transmit the packet is lesser in linkA

hence the Link A is efficient

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