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WHAT'S YOUR ENGINEERING QUOTIENT?

Problem 8—How many check bytes would be required in a block of 200 in order to reduce the probability of an uncorrectable block to approximately 0.001, given a BER of 10^-3?

Continuing the table from the previous question gives us:

k	Probability of k errors C(n, k) × pk × (1–p)n–k	Probability of k or fewer errors (cumulative sum)	Probability of more than k errors (1 – sum)
0	0.20173	0.20173	0.79827
1	0.32423	0.52597	0.47403
2	0.25925	0.78522	0.21478
3	0.13750	0.92273	0.07727
4	0.05442	0.97715	0.02285
5	0.01714	0.99429	0.00571
6	0.00448	0.99877	0.00123
7	0.00100	0.99977	0.00023

It requires the ability to correct up to six errors per block in order to get down to a probability of 0.00123 of finding an uncorrectable block. Correcting 6 errors requires 12 check bytes, so now the overhead is 12/200, or 6%.

Contributor: Dave Tweed

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