patternMinor
Link utilization of sliding window protocol
Viewed 0 times
protocolslidingwindowutilizationlink
Problem
Before I ask my doubt I would like to state that problem which led me to my doubt. It can also serve as a good example scenario.
A $20\ Kbps$ satellite link has a propagation delay of $400\ ms$. The transmitter employs the "Sliding Window protocol" scheme with Window Size set to $10$. Assuming that each frame is $100$ bytes long. What is the link utilization of transmission medium.
From the question we can make out that Window Size $(WS)$ is $10$, transmission time $(T_t)$ of a frame is $40\ ms$ , and propogation time $(T_p)$ is $400\ ms$.
Now, the solution that book proposes simply states that Link Utilization $(LU)$ for sliding window protocol can be calculated as
$$LU = \frac{WST_t}{T_t + 2T_p}$$
That's where I'm facing the problem. From what I have learned about Link utilization
It is the fraction of total time the host was busy in transmission of data. In other words, it is the ratio of total transmission time over Total Time involved in transmission.
Now if I try to fit the formula proposed in book with what I have learned, the total time involved in transmission is $T_t+2*T_p$.
But that is only the total transmission time of one single packet over the medium. Why isn't it is $WST_t + 2T_p$, since we are sending $WS$ packets without wating for acknowledgement.
The only thing I could figure out after thinking a lot about my doubt is that I am probably going wrong because of my incomplete understanding of what Link Utilization means.
I will appriciate any kind of help.
A $20\ Kbps$ satellite link has a propagation delay of $400\ ms$. The transmitter employs the "Sliding Window protocol" scheme with Window Size set to $10$. Assuming that each frame is $100$ bytes long. What is the link utilization of transmission medium.
From the question we can make out that Window Size $(WS)$ is $10$, transmission time $(T_t)$ of a frame is $40\ ms$ , and propogation time $(T_p)$ is $400\ ms$.
Now, the solution that book proposes simply states that Link Utilization $(LU)$ for sliding window protocol can be calculated as
$$LU = \frac{WST_t}{T_t + 2T_p}$$
That's where I'm facing the problem. From what I have learned about Link utilization
It is the fraction of total time the host was busy in transmission of data. In other words, it is the ratio of total transmission time over Total Time involved in transmission.
Now if I try to fit the formula proposed in book with what I have learned, the total time involved in transmission is $T_t+2*T_p$.
But that is only the total transmission time of one single packet over the medium. Why isn't it is $WST_t + 2T_p$, since we are sending $WS$ packets without wating for acknowledgement.
The only thing I could figure out after thinking a lot about my doubt is that I am probably going wrong because of my incomplete understanding of what Link Utilization means.
I will appriciate any kind of help.
Solution
Link utilitization is about the steady state, i.e., what happens when you send a lot of data... not what happens at startup (at the very beginning of the connection).
Imagine you have a humongous file to send. It takes far more than 10 frames to send the entire file. Calculate the total time to send the entire file. Calculate the total transmission time for that file. Now divide.
Try doing that calculation again. I think you'll see that you get a different answer. For instance, try calculating what happens if you send a file that is 1000 frames long... or 10,000 frames long. How long does it take to send the entire file? What fraction of that was spent in transmission time, as opposed to waiting for acknowledgements?
Imagine you have a humongous file to send. It takes far more than 10 frames to send the entire file. Calculate the total time to send the entire file. Calculate the total transmission time for that file. Now divide.
Try doing that calculation again. I think you'll see that you get a different answer. For instance, try calculating what happens if you send a file that is 1000 frames long... or 10,000 frames long. How long does it take to send the entire file? What fraction of that was spent in transmission time, as opposed to waiting for acknowledgements?
Context
StackExchange Computer Science Q#62647, answer score: 3
Revisions (0)
No revisions yet.