logo elektroda
logo elektroda
X
logo elektroda
Advanced Search
logo elektroda
UV printer on your desk?
UV printer on your desk? Discover eufyMake E1 »
ADVERTISEMENT
Dostępna jest polska wersja

Czy wolisz polską wersję strony elektroda?

Nie, dziękuję Przekieruj mnie tam

Amount of Real Data Transmitted at 512 Kbps: Synchronous Link & 64kB/sec without Compression

kalkowska 15543 12
Best answers

How much actual data can be transmitted in one second over a 512 kbps synchronous link without compression, and why is it not simply 64 kB?

At 512 kbps, the theoretical one-way rate is 512,000 bits/s ÷ 8 = 64,000 bytes/s, or 62.5 KiB/s when bytes are expressed with 1024-based units [#14311778][#14312437] You should not divide by 2 just because the link is synchronous; synchronous is not the same as symmetrical, so the 512 kbps figure applies to the link rate itself, not half of it [#14313813] If you mean usable payload rather than raw line rate, checksums, control packets, and other protocol overhead reduce it further, which explains lower figures such as “over 55 kB” [#14311783]
Generated by the language model.
ADVERTISEMENT
Treść została przetłumaczona polish » english Zobacz oryginalną wersję tematu
Report a violation of the law
Reply | New topic
  • #1 14311535
    kalkowska
    Level 2  
    Posts: 2
    Rate: 1
    How much real data can be sent in 1 s over a synchronous link with a bandwidth of 512 kbps, without hardware and software compression?
    I've seen a question on the forum before, the answer is over 55 kB.
    Why? Please give me a precise explanation.
    Since 1B=8b, then 512kb/8=64kB
  • ADVERTISEMENT
  • #2 14311625
    McMarycha
    Level 31  
    Posts: 1526
    Help: 114
    Rate: 198
    512kbps : 2 = 256kbps because it's synchronous
    256kbps : 8 = 32kBps it can send in 1 second
    I hope its good.
  • ADVERTISEMENT
  • #4 14311778
    Jones_
    Level 10  
    Posts: 14
    Rate: 4
    I think I will surprise you, I get 62.5KB of data :D
  • ADVERTISEMENT
  • #5 14311783
    excray
    Level 41  
    Posts: 5500
    Help: 739
    Rate: 656
    There are also checksums, control packages and other things that make life difficult for programmers, so it will be even less.
  • #6 14311805
    McMarycha
    Level 31  
    Posts: 1526
    Help: 114
    Rate: 198
    Jones how did it go?
  • #7 14311824
    Jones_
    Level 10  
    Posts: 14
    Rate: 4
    McMarycha wrote:
    Jones how did it go?

    apart from other matters, only theoretically the bandwidth of the connection
    512/8/1.024=62.5
    throughput is wka data is in K
  • ADVERTISEMENT
  • #8 14312379
    McMarycha
    Level 31  
    Posts: 1526
    Help: 114
    Rate: 198
    I thought the author meant how much he could send. But unless I understood.
  • #9 14312437
    Jones_
    Level 10  
    Posts: 14
    Rate: 4
    In my opinion, that's how a kilogram is not equal to a kilogram
    speed small k = kilo, decimal 10 to 3 = 1000B
    big data K = Kilo, binary 2 to 10 = 1024B

    I don't understand why I have to divide it by 2
    After all, the type of connection does not matter, only what speed we have assigned
    it doesn't even matter if we receive or transmit in a given direction, we have 512 kilos
  • #10 14312867
    McMarycha
    Level 31  
    Posts: 1526
    Help: 114
    Rate: 198
    I meant that if the link is 512kbps and it is synchronous, the upload and download are equal. So we divide by 2
  • #11 14313813
    rwisniewski1
    Level 23  
    Posts: 502
    Help: 40
    Rate: 39
    McMarycha wrote:
    I meant that if the link is 512kbps and it is synchronous, the upload and download are equal. So we divide by 2


    Does your colleague distinguish between the terms: synchronous - asynchronous link from symmetrical - unsymmetrical?
    With a 512kbps synchronous link, division by 2 makes no sense.
  • #12 14314174
    McMarycha
    Level 31  
    Posts: 1526
    Help: 114
    Rate: 198
    And I actually got those ideas wrong.
Reply | New topic
Report a violation of the law

Topic summary

✨ The discussion centers on the amount of real data that can be transmitted over a synchronous link with a bandwidth of 512 kbps without compression. Various participants provide calculations and insights, with some suggesting that the effective throughput is around 62.5 kB/s, while others argue that factors like checksums and control packets reduce this figure further. The debate includes clarifications on the distinction between synchronous and asynchronous links, with some participants mistakenly dividing the bandwidth by two, assuming equal upload and download speeds. Ultimately, the consensus leans towards a theoretical maximum of 64 kB/s, but practical considerations suggest lower actual throughput.
Generated by the language model.
ADVERTISEMENT