CSSA Qu. 16 (9600bps etc.) (1 Viewer)

[xlr9]

Hi was having a little trouble with CSSA 2003 IPT Question 16. Does anyone have a sec to post a worked solution? I keep getting 3.413 for the answer :-s and the way that i figured it out really did call for the use of a calculator (no calculators allowed)

Here is the question...

Max would like to send his friend a photograph of himself on the computer, the photo is 4kb in size and his modem's speed is 9600bps. How long would it take him in seconds (approx) for Max's friend to receive the picture?

Possible responses are:
a. 3 secs
b. 0.4 secs
c. 6 secs
d. 4.3 secs

(no 3.4 answer in the list )

any help / gidence would be much appreciated.
thanks

simon

 

[skip to my lou]

The answer is posted in some other thread!!

but you got the idea! There wasnt an "exact" answer...i think you were suppose to estimate between 3 and 4.3
so yeah

 

[Huy]

9600 bps / 8 = 1200 Kb/s
1200 / 1024 = 1.171875 KB/s
4 KB / 1.171875 = 3.4133333333333333333333333333333

or

4KB = 4*1024 Bytes = 4096 Bytes
4096 Bytes = 4096 * 8 Bits = 32768 Bits

Since modem is 9600 bits per second

32768 / 9600 = approx 3.4 seconds

or

9600 bits per sec / 8 bits per byte = 1200 kilobytes per sec
1200 / 1024 bytes = 1.171875 kilobytes per sec
4 KB / 1.171875 kilobytes per sec
= 3.4133333333333333333333333333333

that's how i usually work things out, anyway,
for example:
56K = 56,700 bps
56,700 bps / 8 = 7087.5 Kbps
7087.5 Kbps / 1024 = 6.92138671875 KBps (Kbytes)

Theoretical bandwidth for 56K is about 7KB/s.
I should have started off by knowing this, (slipped my mind),
then dividing 56K by 9.6K
(about 6)

Dialup downloads at 7KB/s (tops)
meaning for a 9.6K modem, it'll be about 1/6th of 56K,
7/6 is roughly 1.1KB/s

4KB file, going at a max of 1.1KB/s
4 / 1.1 = ~3.6
Which is close enough to 3.4 secs, moreso than 4.3 seconds

But the preferred method:
4KB image/file
1000 bytes for every kilobyte (1KB)

9600 bits per second
Convert 4KB to 4000 bytes
Multiply 4000 by 8 as there are 8 bits per byte, giving 32000 bits for the image file.

Transfer rate of 9600 bps for a 32000 bit file.

32000 / 9600
= 3.33*

The answer was (A) 3 seconds

In practice, however, I tend to think that it would be (D) 4.3 seconds as there are latencies/delays in transmission, and with a 9600 modem, you won't be achieving the full speed for extended periods of time, maybe in short bursts.

But theoretically, (A) is the answer.
(Think about dialup, you wont download at 5, 6 or even 7KB/s constantly, even though the maximum throughput/bandwidth is 56K (or 56,700 bps)).

You were right anyway xlr9

 

[fatmuscle]

lol

IPT doesn't care about latencies or delays...
I wish the world was like the IPT world. no latencies or delays. 56k was 56k, 512/256 was what it is. How nice that would be =)

 

[Huy]

"Wouldn't it be nice if the world was ...IPT"

*Cadbury chocolate tune*

 

[stargater74]

I think the answer is 4.3 seconds, as the image file will not be fully transmitted after 3 seconds. The question asks how long it will take for the other person to receive the photo, and after 3 seconds, the photo will not be displayed, as it doesn't have all the data to display it, so it will not be completely downloaded until at least 4 seconds. So I think that 4.3 is more likely to be correct than 3 seconds, even though that is not the answer

 

[Huy]

Originally posted by stargater74
I think the answer is 4.3 seconds, as the image file will not be fully transmitted after 3 seconds. The question asks how long it will take for the other person to receive the photo, and after 3 seconds, the photo will not be displayed, as it doesn't have all the data to display it, so it will not be completely downloaded until at least 4 seconds. So I think that 4.3 is more likely to be correct than 3 seconds, even though that is not the answer

True, but as fatmuscle and I have pointed out in other threads (regarding the same question and discussion about transmission rates, speeds, lag/delays and other things which could have an affect on the transmission), IPT doesn't care about it.

The answer, in practice, would be 4.3 seconds, (which was what I had in my CSSA paper), but from an IPT point of view, disregarding real-world situations and scenarios, all the examiners are looking for is an application of theory, proof that you know the formulae and can manipulate things like bits/bytes, multiplication and division, and so on to achieve an answer, as would be expected from the numbers -- not from everyday usage and real-world situations

The other friend wants to receive it, he/she may not wish to view it until later (But that wasn't in the question, so it could be stipulated that they only want to receive the picture for later viewing). You have a valid point, and it was my understanding at the time, during the CSSA trial, but from theory to it's application, 3.41 seconds (or rather, 3 seconds, answer A), is the most correct from the data given, (and it's expected usage and manipulation of the data/numbers, as an IPT student).