Revise in a moth question! (1 Viewer)

[mattstonestreet]

Q-- An office has a small ethernet bus network which consists of six worstations a laster printer, a colour printer, an e-mail server connected to the internet using a 56k modem, and a file server.

iii. Assuming the modem was functioning at its maximium capacity, approximatly how long would it take to download a 900k file?

Here is the worker solution 900 x 8/56 = 128 secs = approx. 2mins...

I Don't understand this solution why isn't it just 900/56 = 16seconds?

 

[orange_blob]

I don't do IPT but:

it's not just 900/56 because the 56k on a modem means it goes at 56 kilobits per second, and the file being 900k means it's 900 kilobytes. Now since there are 8 bits in every byte, the modem will download at 7 kilobytes per second (ie 56/8).

So you then do 900/7 which gets you to approximatly 2 minutes.

Don't forget to actually think about whether the answer you get is reasonable - 900k is nearly 1MB, I don't know if you've ever used dial-up, but it definatly takes much longer than 16 seconds to download 1MB.

 

[maskd]

Like you may have already known, you can do this answer 2 ways..

The first way is to convert 56k into 7K, then do 900/7. Meaning that you are always working in KiloBYTES, I personally prefer this method.

The other way is to covert 900K into 7200k, then do 7200/56. With this method you are working with Kilobits.

Just remember between bits and bytes, it comes up a fair bit in communication systems and knowing the differences will most likely allow you gain more marks.

 

[seremify007]

Yep it's the difference between bits and bytes.

I personally prefer to work in kilobits because it seems more natural to me (ie. multiply by 8 and then divide by 56) but really it comes down to personal preference. maskd's method would probably be preferable if you didn't have a calculator on you