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| Pavlov 2002-01-05, 4:04 pm |
| Can someone give me an easy way to remember/figure out how to solve these download time questions?
Downloading a page that's 7000k with a 56k modem will take how long?
What's the formula?
These will be the questions I give up on the exam  | |
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| (Theoretical, assuming no signal loss in your telephone line or noise/weather, etc.):
1) Internet D/U speed (Max by modem) = 56 Kbits/second;
2) Each Web Page size = 7,000 Kbits; (or was it in bytes?)
3) Thus, it will take (7000 Kbits)/(56 Kbits/sec) = 125 seconds = 125sec/(60sec/min) = 2.088888.. minutes!
(Practically, you wont get exactly 56kbits/ sec but around 45K+ depending on your neighborhood telephone lines/posts, etc..)
I assume this is CS college major question or basic web design homework type? | |
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| Thus, formula = Web Page size / Modem speed = Transfer Time Per Page theoretically.
Otherwise, you should take into account of TCP/IP packet/headers size, transimission errors/resend, actual data bits encapsulated in frames, handshaking, cookies size on pages, etc.. it is pain in the neck to figure out the exact actual bit rates in communications theory course!! | |
| wbafrank 2002-01-05, 5:02 pm |
|  Maths problem try this:
56000 bits per sec / 8 bits per byte = 7000 bytes per sec
Therefore 7000 bytes would take 1 second | |
| Pavlov 2002-01-05, 5:15 pm |
| makes sense wbafrank, but Exam Drill gives the answer as 1.25  "Oh, bother..." | |
| prometric 2002-01-06, 2:51 am |
| It depends on what the choices are in the real exam. On my test I didn't get 1.25, so I chose the answer that was exact because none of the others fitted the explanation in Exam Drill. Use wbafrank's formula. Remember that they will use the words "bits" & "bytes" in the exam to confuse you if you don't spot the difference. | |
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| quote:
Originally posted by Pavlov
Can someone give me an easy way to remember/figure out how to solve these download time questions?
Downloading a page that's 7000k with a 56k modem will take how long?
What's the formula?
These will be the questions I give up on the exam
Well, first off, are you sure its 7000k, not 7000 bytes? If it is, then thats quite a large web page. Well, assume it is.
Important thing to remember is that file size is measured in bytes and internet speeds are measured in bits 1 byte = 8 bits
First, take the file size and convert it to bits 7000000 * 8 = 56000000
Next, divide your connection speed by your file size in bits 56000/7000000 = 125 seconds
Finally, its usually a good idea to tack on a few seconds to make up for other variables (I dont know if this last part is correct for comptia, but me and exam drill agree on this).
If it was 7000 bytes instead of 7000K, then the 1.25 seconds answer would fit (tacked on an extra .25 seconds)
Remember that k tacks on 3 extra zeros, it can make a big difference if its there or not. | |
| neuralfx 2002-01-07, 3:27 am |
| ok .. i dont know where everyone is getting these numbers but this is how it works out mathematically.. first a KB is 1024bytes, not bits or 1000 bytes .. you CANT divide with different units, so all values in the equation have to be the same unit, in this case we will use bytes ( you could use bits, its just an extra step on each side)
a constant in these questions is the 56k
56,000 bits, which you convert to BYTES by
56,000 / 8 = 7000BYTES , of course 8bits is a byte
in this question Pavlov asked about 7000KB, so we convert this into bytes
7000 * 1024 = 7168000BYTES ( each KB is 1024BYTES, so multiply the KB by each BYTE)
so now we have 2 values that can be used in division, 7000BYTES and 7168000BYTES, now you DONT divide, the connection speed by file size, this will give an incorrect result..
You divide the Filesize by the "connection speed" , the speed is always in reference to seconds.. so we divide
7168000 / 7000 = 1024
now, the qoutient is in seconds.. Reason, file is 7000KB, and u are connected at 56kilobits per sec .. so each second you take a 56kilobit chunk out of the file..
so you then convert the 1024 seconds to minutes by of course
1024 / 60 = ~17 minutes
this is much simpler then it may look, basically you can remember that 56Kilobits is 7000bytes, and then convert the filesize the questions is asking to bytes, and divide accordingly ..
if the question had asked about 7MB, you would simply tack on another 1024, because 102KB is 1MB .. like this
7 * 1024 * 1024 = 7340032 ..
just play with it, you will see its very simple, and on your questions you should arrive at EXACT answers, because the questions dont care about tcp/ip overhead, or network traffic, its very straightforward .. the reason most come at unexact answers, is because they assume 1KB =1000 .. which after rounding excessivly throws the answer off by a lot .. well it was wordy but hope it was helpful to you ..
-neural
[edit] reread your question again, Pavlov the Examdrill or whatever you are using is completely wrong .. 7000KB will not transfer in 1.25, without using a forumla that is common knowledge .. and if it was using 7000BYTES, then it defies all mathematical logic, .. the forumla still works, of course in my example, if the choice was between
a)1.75minutes b) 70minutes c) 20min
of course the answer would be c), in any case, once you get the idea of how it works its very simple .. |
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