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?

  • \$\begingroup\$ Sounds like homework \$\endgroup\$
    – Passerby
    Commented Oct 27, 2017 at 4:05
  • \$\begingroup\$ quiz actually. No idea what went wrong. \$\endgroup\$
    – AB408
    Commented Oct 27, 2017 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\$ Commented Oct 27, 2017 at 14:40

1 Answer 1


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%.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.