Chapter 5, Problem 5.9.1E

Explanation of Solution

Finding minimum of parity bit “p”:

Assume that “p” indicates the number of parity bits required and “d” indicates the data bit in “p+d” bit word.

Given that the word size “p+d” = 128 bits

The expression for single bit error correction will be as follows:

${\text{2}}^{\text{p}}\ge \text{p+d+1}$ bits

Therefore, $\text{p >= log}\left(\text{p+d+1}\right)$

Substitute the value of “p+d” as follows:

${\text{2}}^{\text{p}}\text{ >= p+128+1}$