Calculating Utilization in a Stop-And-Wait Protocol

Question

I have a problem in my book about calculating the utilisation, but I'm not being able to find any substantial information regarding this topic in order to solve it.

Anyway, here's the question:

The distance from earth to a distant planet is approximately 9 × 10^10 m. What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64 Mbps point-to-point link? Assume that the frame size is 32 KB and the speed of light is 3 × 10^8 m/s.

Suppose a sliding window protocol is used instead. For what send window size will the link utilization be 100%? You may ignore the protocol processing times at the sender and the receiver.

Voluptuous

"assuming base 10 for kilo and mega" You are wrong man... KB=1024 BYTES=1024x8 bits (base 2) Mbps=1000 bits per second (base 10) – Ramtil

en.wikipedia.org/wiki/Data_rate_units#Kilobit_per_second – Happening

Answer 1

This is a fairly simple question. Utilization is the part of time that the medium is being used with "good" byte, that is bytes of payload (no headers or ack/nack frames).

In your question, there is no header specified and ACK size is not given, thus I will assume they are both of size 0.

For the S&W case:

Each period is Tx + propagation + ACK propagation = Tx + RTT

Tx = 32KB/64Mbps = 0.004 seconds (assuming base 10 for kilo and mega)

RTT = 2 * (9*10^10m) / (3*10^8m/s) = 600 seconds

Utilization = 0.004 / (600 + 0.004) = 6.667x10^-6 = 6.667x10^-4 %

This is very bad utilization, since the medium is very long and there is a lot of time wasted for waiting for the ACK.

For sliding window:

Since there is no error probability, I assume it is 0. To get to 100% utilization, you need to keep transmitting packets while you wait for the ACKs, that means for the whole period.

period = 600.004 seconds
1 Tx = 0.004 seconds

For non stop Tx you need to transmit 600.004/0.004 packets each period, therefore 150001 should be your window size.

Happening