Network Calculations

[thinboy]

Hi,

I need help with the following calculations i am having trouble with i dont want someone to do them just so me how to work them out they are as follows

• In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?
• In a given system, IP packets are used to convey CD quality music samples generated by various sources. The sampling rate at each source is 44 KHz (44000 samples are made per second) and each sample is encoded using 16 binary bits. Assuming the bandwidth to be 100 Mbps, how many sources can be accommodated?

• Calculate the delay associated with the transmission of a packet (15 Kbytes long) from source to destination, with a physical distance

[RevBooter]

alrite dude! hows she going? this is my first reply, losing my virginity, first for everything :P. in my humble opinion... 1 microsecond is one millionth of a second(10−6). if there is 13mbps or 13631488 bits per second, or 13.631488 bits per microsecond... 1703.936 bits transferred in 125 microseconds... or one source.

yeah?! lol... been a while!

is this a masters, or phd question! now way this is undergrad!

[RevBooter]

sorry man one more thing 1 us(microsecond) is the cycle time of 1MHz.

[Banfa]

1 microsecond is one millionth of a second(10−6). if there is 13mbps or 13631488 bits per second, or 13.631488 bits per microsecond... 1703.936 bits transferred in 125 microseconds... or one source.

Unfortunately all of this is correct except the last (highlighted step) which is wrong.

[Banfa]

• In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?
• In a given system, IP packets are used to convey CD quality music samples generated by various sources. The sampling rate at each source is 44 KHz (44000 samples are made per second) and each sample is encoded using 16 binary bits. Assuming the bandwidth to be 100 Mbps, how many sources can be accommodated?

The method I would use for this is calculate the bandwidth of the source in bps (or kbps or Mbps). Then having done that you can get the number of sources that can be accommodated by dividing the link bandwidth by the bandwidth per source.

• Calculate the delay associated with the transmission of a packet (15 Kbytes long) from source to destination, with a physical distance

Physical distance is not really relevant to electrical links (OK it is a bit but it is not normally thought of in those terms) it is normal to quantify the speed of link is bps which is not directly related to distanced travelled.

[RevBooter]

thats why you're the admin and im the newbie :P

[thinboy]

Can someone please do the calculations 4 me as they are
Thanks

[Cyberdyne]

Is this a homework question?

Can someone please do the calculations 4 me as they are
Thanks

[thinboy]

Hi,

Is not homework i need them for work

Thanks

[thinboy]

Hi,

Any chance someone could show me how to work each calculation out slowly in steps

Thanks

[Banfa]

It's really not hard, all you have to do is identify the amount of data that is passing in a given amount of time. The convert it to bits per second by multiplying by relevant factors.

Taking the first example

In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?

You are told that a source generates an 8 bit sample every 125 microseconds. You need to convert that to bits per second before you compare it with the available bandwidth.

So what factor do you need to use to convert 125 micro seconds to 1 second?

If you times your given bit rate (8 bits per 125 microseconds) by the same factor you will convert it to bits per second.

That is always how you do it, identify how much data you are getting how often, then multiply it by a factor to convert it to bits per second.

[thinboy]

is there any chance someone could do them for me as i really need help

[thinboy]

Hi,

Is there any chance someone could please do these for me as i really need help with them

• In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?
• In a given system, IP packets are used to convey CD quality music samples generated by various sources. The sampling rate at each source is 44 KHz (44000 samples are made per second) and each sample is encoded using 16 binary bits. Assuming the bandwidth to be 100 Mbps, how many sources can be accommodated?

Calculate the delay associated with the transmission of a packet (15 Kbytes long) from source to destination, with a physical distance between them of 106 m. The transmission rate is 10 Mbits/s. Assume the propagation speed of the signal in the transmission medium to be 2*10 8 m/s


Thanks

So Much

[thinboy]

Hi,

Is there any chance someone could please do these for me as i really need help with them

• In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?
• In a given system, IP packets are used to convey CD quality music samples generated by various sources. The sampling rate at each source is 44 KHz (44000 samples are made per second) and each sample is encoded using 16 binary bits. Assuming the bandwidth to be 100 Mbps, how many sources can be accommodated?

Calculate the delay associated with the transmission of a packet (15 Kbytes long) from source to destination, with a physical distance between them of 106 m. The transmission rate is 10 Mbits/s. Assume the propagation speed of the signal in the transmission medium to be 2*10 8 m/s


Thanks

So Much

[Banfa]

Please do not double post your questions.

Please do read this.

Please do attempt your problems yourself first and then post questions about the parts you do not understand.

Banfa
Administrator

[thinboy]

Hi, i cannot attempt the questions because i need to know how to work each one out

Thanks

[Banfa]

Hi, i cannot attempt the questions because i need to know how to work each one out

They are all basically the same. You use the data suuplied to get a bit rate in whatever units are provided, for instance bits per micro second.

You then multiply or divide that value my the factors required to convert it to the unit you require.

I have explained this already in this thread and someone else has in fact done most of the work for you.

[thinboy]

hi, what is a sources thn, could someone do one of the questions 4 me

Thanks

[MMcCarthy]

We are not here to do your work for you. We are a resource that's more than willing to help you do the work for yourself. Attempt one the the calculations yourself first and then post the results of your attempt. The experts will be more than happy to correct any errors you have made.

However, as I previously said no one is going to do it for you. You have already been given enough information to attempt the problem yourself. If you are not willing to do so then we can't help you.

ADMIN

[thinboy]

Hi i have had a go at what i think you do ive done this so far

8*125*13 = 13000 bits
13000 bits / (13*1000) = 100000

[thinboy]

Hi i need help with the last one

Thanks

[Banfa]

Hi i have had a go at what i think you do ive done this so far

8*125*13 = 13000 bits
13000 bits / (13*1000) = 100000

• In a given system, IP packets are used to convey speech samples generated by various sources. For each source a sample is generated every 125 microseconds, and the sample is encoded using 8 binary bits. For a bit rate of 13 Mbps, how many sources can be accommodated?

OK looks like you need to learn how to compare and adjust units, the values you are given are

8 bit sample
produced every 125 microseconds
with a maximum stream bandwidth of 13Mbps

The units on 2 sides of any equation are always the same (assuming the equation is correct) so when you do

8*125*13 = 13000 bits

the result is not bits because the equation is

8 bits * 125 microseconds * 13Mbps = 13000 bits microseconds Mbps

since micro is 10e-6 and M(mega from the Mbps) is 10e6 10e-6 * 10e6 = 1 so these cancel, you also have seconds and per second (ps from the Mbps) these also cancel leaving bits and bits (from the Mbps) or bits squared so

8*125*13 = 13000 bits squared

Did you want a result with a unit of bits squared?

Read this it will help explain how units and added, subtracted multiplied and divide, particularly the section at the bottom labeled "Calculations with Units".

You actually want the number of streams that can be accomodated, this is in fact a unitless number so you need to have canceled all the units out of the numbers you have been given by combining them in the correct manor.


Alternitively another way of performing the calculation is to divide the bit rate of the transport medium (give to you) by the bit rate of the media stream. The bit rate of the media stream is easy to calculate by dividing the number of bits that pass iby the time it takes them to pass.

[RedSon]

This thread is great, I have to subscribe!

[Banfa]

13000 bits / (13*1000) = 100000

And just in case you are wondering having corrected the maths to

13000 / (13*1000) = 1

the 13000 is in bits squared and the 13 in Mbps so this is actually

1 bits squared / Mbps k

The little k is for kilo and siginfies that for some reason an extra 1000 was included, this is equivilent to

1 micro bit second

I have no idea what it represents.

[RedSon]

I think its also important to note the difference between KB and KiB or anything with a little i in the middle. Since you are talking about digital systems all of your measurements should be done in *iB since it is base 2. The distinction between KB and KiB might not matter for this particular question but be aware that it is out there, and it has an important distinction.

For more info: http://en.wikipedia.org/wiki/Kibibyte

[thinboy]

Hi i have done this for the first one

1 source = 125 microseconds 1 sec = 1000000 microseconds/125microseconds = 8000 bits, 13Mbps*8000 bits/8 binary bits = 13000 Sources

How do you do the third one?

Thanks

[Banfa]

1 source = 125 microseconds 1 sec = 1000000 microseconds/125microseconds = 8000 bits,
13Mbps*8000 bits/8 binary bits = 13000 Sources

Yes but this is not the right answer. The first line is OK except that you have labelled the final value bits. How can the units be bits? Bits do not feature in either of the 2 values used in the calculation, they have units of microseconds per second and microseconds respectively. What is the unit of the value 8000 calculated? or put another way what does it represent?

The second line of this calculation is just wrong, again the final unit of the calculated value is wrong by your own labelling it would be bps. bps is clearly the wrong unit for number of sources. But you have the unit of the 8000 wrong anyway.

I strongly suggest you concentrate on working out the bit rate of the data stream first and leave attempting to work out the number of streams the transport machanism can carry till later (i.e. leave the 13 out of the equation for now).

How do you do the third one?

The third question does not included enough information to be able to calculate anything.