Consider the transfer of a file containing one million 8-bit characters from one

Question

Consider the transfer of a file containing one million 8-bit characters from one station to another. What is the total elapsed time and effective throughput for the following cases: a) A circuit-switched, star-topology local network. Call setup time is negligible and the data rate on the medium is 64 kbps. b)A bus topology local network with two stations at distance D apart, a data rate of B bps and a frame size of P with 80 bits of overhead per frame. Each frame is acknowledged with an 88-bit frame before the next is sent. The propagation speed on the bus is 200m/us. Solve for: 1. D = 1 km, B = 1 Mbps, P = 256 bits 2. D = 1 km, B = 10 Mbps, P = 256 bits 3. D = 10 km, B = 1 Mbps, P = 256 bits 4. D = 1 km, B = 50 Mbps, P = 10 kbits

Explanation / Answer

Your first query...

Since,
1 character = 8 bits

Therefore,
1 million characters = 8 million bits

Given data rate is 64kbps or 64,000bps, therefore,

The total elapsed time,
T = 8,000,000bits / 64,000bps
= 125 seconds

The effective throughput = 64kpbs.


Your second query...

Since we are talking in terms of Transmission and Acknowledgements, therefore we have to deal with a term called Cycle as defined below.

One cycle is,
(data packet transmission time + propagation delay) + (Ack packet trans time + propagation delay)

Effective Data Rate = (actual frame size) / (time required for one cycle)

propagation delay = length of link / Velocity of propergation


So that one way propagation delay is,
1000m = 200m/microsec x t
t = 5 microsec

Round trip propagation delay is,
2t = 10 microseconds

Time to transmit a frame is,
number of bits = data rate x time
256 = 10,000,000bps x t
t = 25.6 microsec

Ack transmission time is,
number of bits = datarate x time
88 = 10,000,000bps x t
t = 8.8 microsec

- The transmitter seizes the channel: 10 microsec
- The transmitter transmits frame: 25.6 microsec
- The receiver seizes the channel: 10 microsec
- The receiver sends Ack frame: 8.8 microsec

Therefore, one cycle as mentioned above is,
(25.6 microsec + 10 microsec) + (8.8 microsec + 10 microsec)

The total time cycle for a packet of 256 bits = 54.4 microsec or (54.4 x 10^(-6))sec

But, we are sending 1 million characters or 8 million bits! (1 character = 8 bits)

So,
8,000,000 bits/256 bits = 31250 (keep in mind that we are calculating for the TOTAL elapsed time, so we have not subtracted the 80 bit overheard from the 256 bits packet! So, instead of dividing 8 million by 176 (256-80), we are calculating for the full packet size of 256)

It means that we are sending 256 bits exactly 31250 times.

Therefore,
We multiply total time cycle for a packet of 256 bits with 31250.
or,
54.4 microsec x 31250 = 1.7 seconds

Therefore, the total elapsed time for sending 1 million characters is 1.7seconds

But, since we are calculating for the EFFECTIVE throughput, therefore, we need to calculate for the actual frame size,

Actual frame size is 256 - 80 (overhead) = 176

So the effective throughput is,
176 / (54.4 x 10^(-6))sec = 3.235294 Mbps (approx.) or 3.24 Mbps..

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.