Difficult to do ?

Consider the transfer of a file containing ONE Million Characters from one station to another. What is the total Elapsed time effective throughtput for the following cases:

1) A circuit-switched, star topology local network. Call setup time is negligible and data rate on the medium is 64kbps.

2) A bus topology local network with 2 stations a distance D aparts, a data rate of B bps and a packet size P with 80 bits of overhead. Each packet is acknowledged with an 88-bit packet before the next is sent. The propagation speed on the bus is 200m/us. Solve for : D=1km, B=10Mbps, P=256bits

3) A ring topology with a total circular length of 2D, with the 2 stations a distance D apart. Acknowledgment is achieved by the allowing a packet to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time.
Solve for : D=1km, B=10Mbps, P=256bits, N=10

Your first query...

1) A circuit-switched, star topology local network. Call setup time is negligible and data rate on the medium is 64kbps.

The answer...

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...

2) A bus topology local network with 2 stations a distance D aparts, a data rate of B bps and a packet size P with 80 bits of overhead. Each packet is acknowledged with an 88-bit packet before the next is sent. The propagation speed on the bus is 200m/us. Solve for : D=1km, B=10Mbps, P=256bits

The answer...

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)

But first we must calculate the propagation delay,
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

So now you have both the Total Elapsed time = 1.7 seconds and the Effective throughput = 3.24 Mbps

I'm sure that now you will be able to do the calculations for the last query by yourself!!

3) A ring topology with a total circular length of 2D, with the 2 stations a distance D apart. Acknowledgment is achieved by the allowing a packet to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time.
Solve for : D=1km, B=10Mbps, P=256bits, N=10

Is this correct >

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

Total repeaters delay = 10 x (1/10 x 10^6) = 1microsec

The total time cycle for a packet of 256 bits = (25.6 microsec + 10 microsec) + (8.8 microsec + 10 microsec) + 1mircosec = 55.4 microsec

We multiply total time cycle for a packet of 256 bits with 31250.
or,
55.4 microsec x 31250 = 1.73 seconds

Therefore, the total elapsed time for sending 1 million characters is 1.73 seconds


Actual frame size is 256
There is no overhead

So the effective throughput is,
256/ (55.4 x 10^(-6))sec = 4.621 Mbps

time to go out and get some sunshine isnt it ?

Cheers,