Answer to Question #171378 in Computer Networks for samuel

Question

• Consider a bit streaming scenario for a video where the following values apply:

– The buffer size is 1 MiB

– The low watermark is set at 100KiB

– The high watermark is set at 900KiB

– The incoming data rate is 1Mbps

– The video display rate is 300 Kbps

Assume that the video is playing and that the buffer content has dropped to the low-water mark. The media player sets the controls for data input to begin again.

– Calculate the amount of data that will be input in two seconds to the buffer and the amount of data that will be removed from the buffer in the same period of time.

– Repeat the calculation for 4,6,8,10,12 seconds.

– From this data estimate when the buffer will have filled up to the high water mark.

– Assuming that the incoming transmission is halted at this time, calculate how long it will be before the buffer content has again fallen to the low-water mark level.

Answer 1

Given,

Low watermark is set at $=100kb$

High watermark is set at $=900kb$

Incoming data rate KB/sec $=1Mbps$

$=1000 kbps=125kb/s$

Video display rate $=300 Kb/s$

$=300/8 KB/s =37.5KB/s$

Input buffer coming in 2 sec =incoming Data Rate * time = 125*2 = 250 KB

2 sec, the data in the buffer = 250KB

As the low watermark is 100KB so only 150 KB of data can be displayed on the screen.

Hence, the amount of data which need to remove in the same period of time $= 37.5\times 2 =75KB$

After 2 sec, data remaining in the buffer = 100-75 =25KB

similarly,

t=4sec

Data coming to buffer = incoming data rate * t $=125\times 4= 500KB$

Data empty in 4 sec = video display rate *t = 150KB

Data Buffer = 500-150KB = 350KB

as highest limit is 900KB, Hence the data is not filled up to high watermark.

For t = 6

Data coming to buffer = incoming data rate * t $=125\times 6= 750KB$

Data empty in 4 sec = video display rate *t = $37.5\times 6 =225KB$

Data Buffer = 750-225 = 525KB

As highest limit is 900KB, Hence the data is not filled up to high watermark.

Now, for t=8sec

Data coming to buffer = incoming data rate * t $=125\times 8= 1000KB$

Data empty in 4 sec = video display rate *t = $37.5\times 8 =300KB$

Data Buffer = 1000-300 = 700KB

As highest limit is 900KB, Hence the data is not filled up to high watermark.

For t=10 sec

Data coming to buffer = incoming data rate * t $=125\times 10= 1250KB$

Data empty in 4 sec = video display rate *t = $37.5\times 10 =375KB$

Data Buffer = 1250-375 = 875KB

As highest limit is 900KB, Hence the data is not filled up to high watermark.

For t= 12sec

Data coming to buffer = incoming data rate * t $=125\times 12= 1500KB$

Data empty in 4 sec = video display rate *t = $37.5\times 12 =450KB$

Data Buffer = 1500-450 =1050KB

As here the maximum buffer limit is 900KB but here actual buffer size is 1050KB

Data in buffer will be the upper limit 900KB.

Hence high watermark will be between 10 to 12 sec