Chapter 2, Problem 67E

a.

Explanation of Solution

Number of parity bits:

The length of memory word is 12

The number of parity bits for memory word can be calculated using the following formula

$\text{m}\text{+}\text{r}\text{+}\text{1}\le {\text{2}}^{\text{r}}$

Substitute, “12” for “m” in the above formula

$\begin{array}{l}\text{12}\text{+}\text{r}\text{+}\text{1}\le {\text{2}}^{\text{r}}\\ \text{13}\text{+}\text{r}\le {\text{2}}^{\text{r}}\end{array}$

Hence, r must be greater than or equal to 5.

Therefore, the number of parity bits required for 12 length memory word is “5”.

b.

Explanation of Solution

Code word representation using hamming algorithm:

To create the code for memory word with 12-bit length, 5 parity bits are added compulsorily.

The number of bits present in the code word can be calculated using the following formula,

$\text{n}\text{=}\text{m}\text{+}\text{r}$

Substitute, “12” for “m” and “5” for “r” in the above formula,

$\begin{array}{c}\text{n}\text{=}\text{12}\text{+}\text{5}\\ \text{=}\text{17}\end{array}$

Thus, the number of bits present in the code word is 17.

The number bits starts from right to left with 1.

The number equal to the power of 2 is known as parity bit.

For example, $2^0=1$, $2^1=2$, $2^2=4$, $2^3=8$ and $2^4=16$

The positions of the parity bits are 1, 2, 4, 8 and 16.

Now, write the sum of the numbers with powers of 2.

$\begin{array}{l}1=1\\ 2=2\\ 3=1+2\\ 4=4\\ 5=1+4\\ 6=2+4\\ 7=1+2+4\\ 8=8\\ 9=1+8\\ 10=2+8\\ 11=1+2+8\\ 12=4+8\\ 13=1+4+8\\ 14=2+4+8\\ 15=1+2+4+8\\ 16=8+8\\ 17=1+8+8\end{array}$

The data bits are filled with the given code word by leaving the positions related to the parity bits.

$\begin{array}{l}100100011010\\ 1716151413121110987654321\end{array}$

Parity bit 1 checks parity over 3, 5, 7, 9, 11, 13, 15 and 17.

Now perform the modulo 2 sum on these bits.

$0+1+1+1+0+1+0+1$

Thus, the parity bit of 1 is 1.

Parity bit 2 checks parity over 2, 3, 6, 7, 10, 11, 14 and 15

Now perform the modulo 2 sum on these bits.

$0+0+1+0+0+0+0+0$

Thus, the parity bit of 2 is 1.

Parity bit 3 checks parity over 5, 6, 7, 12, 13, 14 and 15.

Now perform the modulo 2 sum on these bits.

$1+0+1+0+1+0+0$

Thus, the parity bit of 4 is 1.

Parity bit 4 checks parity over 9, 10, 11, 12, 13, 14 and 15.

Now perform the modulo 2 sum on these bits.

$1+0+0+0+0+1+0+0$

Thus, the parity bit of 8 is 0.

Parity bit 5 checks parity over 16 and 17

Thus, the parity bit of 16 is 1.

The data bits are filled as follows,

$\begin{array}{l}11001000101011011\\ 1716151413121110987654321\end{array}$

Therefore, the code word for 100100011010 is “11001000101011011”.