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I was attempting this problem and I could not finish it.

I found the maximum frequency which was 67% as I divided 8 by 12(information bits/total bits), but I am having trouble finding the maximum utilization. The answer is 80%. I believe to find the utilization, you would divide the total packets per second over the total bits. To do that, I multiplied 300k with 0.67 and divided that number by 12. Can anyone explain to me where I went wrong?

Suppose a throughput of 10,000 16-bit samples is still required for the application, but the serial communication interface has the following characteristics:

  • a total packet size of 12 bits: 8 bits of information and 4 bits of overhead

  • does not require synchronization packets

  • a bandwidth of 300,000 bits per second

What is the maximum efficiency and maximum utilization?

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  • \$\begingroup\$ Sounds like homework \$\endgroup\$ – Passerby Oct 27 '17 at 4:05
  • \$\begingroup\$ quiz actually. No idea what went wrong. \$\endgroup\$ – AB408 Oct 27 '17 at 4:08
  • \$\begingroup\$ The link efficiency is simply total information bits / total link bits. The utilisation is different. That is the number of bits per second required / bits per second available. \$\endgroup\$ – Peter Smith Oct 27 '17 at 14:40
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Try this. For every 12 bits you transmit you must transmit 4 bits of overhead. So for every sample of 16 bits you must transmit 8 extra bits for a total of 24 bits per sample. You're required to transmit 10,000 samples (per sec?) so that's 240,000 bits per second required. You have a 300,000bps link so 240,000/300,000 = 80%.

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