Case Study: Utilization on a 56k Modem

On a 56k modem, the theoretical bit rate is 53000 bits per second. It is not 56000 bps due to FCC regulations. However, the typical data transfer rate is somewhat slower. This disparity is due to several factors:

  • Bandwidth limitations
  • Noise on the line
  • Digital to Analog conversion
  • Analog to Digital conversion
  • Framing overhead

    Bandwidth
    The bandwidth of telephone lines is 3000 Hz. This value directly affects the capacity of the line. Using the Shannon-Hartley law...

  • bit rate = bandwidth*(log2(1 + S/N))
  • 53000 bps = (3000 Hz)*(log2(1 + S/N))
  • so the S/N must be at least 208545. SNR = 10 log(S/N) = 53.2 dB

    This high a signal-to-noise ratio is only available on a digital line. The typical SNR on the telephone system is only 35 dB.

    As noted in the previous discussion on the V.90 standard, only downloading can be done in this purely digital state. Uploading still goes through the digital-analog-digital conversion, so the actual maximum bit rate in that direction...

  • SNR = 35 dB = 10 log(S/N) S/N = 3162.3
  • bit rate = (3000 Hz)*(log2(1 + 3162.3) = 34881 bps

    Digital-Analog-Digital Conversion
    This conversion pretty much only concerns the upload condition, but it deserves treatment here anyway. Say, for argument's sake that the only noise in the channel is due to this conversion between analog and digital. That value is called Quantization noise.

  • quantization noise = (1/2^q)^2
  • if signal = 1, then S/N = 2^2q
  • 2^2q = 3162.3
  • q = 5.81 bits

    This means that there are 5.81 -> 6 bits needed for this conversion.

    Framing
    Point-to-Point Protocol (PPP) uses the HDLC frame format. PPP is the framing format used by modem users who want to make more than just a terminal connection to the internet. The HDLC frame format is below:

    Flag Address Control Data FCS Flag
    # of bits 8 8 or 16 8 or 16 variable 16 or 32 8

    According to this, there are a minimum of 48 bits that are not part of the data. These extra bits lower the effective data rate and utilization of the channel.

    The optimum frame size in a selective repeat system with a burst error rate of 10^-5 (which are the stats for the phone system) is 277 bytes. That is 2216 bits (F). With the information that there are 48 bits per frame that are not part of the data, that means there are 2168 bits left for data (N). The transmission rate is assumed to be 53000 bps (R) because of the FCC limitation on download data rates. For argument's sake, it takes 1 microsecond to create the frame (T).

  • effective data rate = N/(T + (F/R)) = 2168 bits/(.000001 s + (2216 bits/53000 bps)) = 51850 bps
  • so using that as a measure of channel utilization = effective data rate/theoretical data rate = 51850 bps/53000 bps = 97.8 %

    These, of course, only apply to download rates. In the upload case...

  • effective data rate = 2168 bits/(.000001 s + (2216 bits + 6 bits for DAD conversion/34881 bps)) = 34033 bps
  • channel utilization = 34033 bps/34881 bps = 97.6 %
  • but if we assume the channel capacity is 53000 bps, then utilization = 34033 bps/53000 bps = 64.2 %

    So 53000 bps on a 56k modem is only a theoretical capacity. There are several contributing factors that slow the transmission rates down. For instance, high network load will slow down one's individual transmission rates. Excessive noise on the channel can be a factor as well.

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