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!!