Chapter 2, Problem 20CRP

a.

Explanation of Solution

Determine the bit pattern in register $0\times 0,0\times 1\text{and}0\times 2$ when the machine halts:

The address content of the program is shown as follows.

$\begin{array}{l}\text{20}\text{12}\\ \text{21}\text{20}\\ \text{22}\text{32}\\ \text{23}\text{30}\end{array}$

$\begin{array}{l}\text{24}\text{B0}\\ \text{25}\text{21}\\ \text{26}\text{24}\\ \text{27}\text{C0}\end{array}$

$\text{28}\text{00}$

The program counter instruction register is as follows.

$\begin{array}{l}\text{20}\text{1220}\\ \text{22}\text{3230}\\ \text{24}\text{B021}\\ \text{21}\text{2032}\end{array}$

$\begin{array}{l}\text{23}\text{30B0}\\ \text{25}\text{2124}\\ \text{27}\text{C000}\end{array}$

Explanation:

• The bit pattern $12$ is stored in register $2$ because the first instruction $1220$ provided for the execution will result in the storage of the content at cell address $20$ to be placed in register $2$.
• The address $30$ bit pattern will store bit pattern $12$ because the next instruction $3230$ for the execution will result in the storage of contents of register $2$ at cell address $30$

b.

Explanation of Solution

Determine the bit pattern in register $0\times 30$ when the machine halts:

The bit pattern in register $0\times 30$ , when the machine halts is $\text{12}$

c.

Explanation of Solution

Determine the bit pattern in register $0\times \text{B}0$ when the machine halts.

The bit pattern in register $0\times \text{B}0$ , when the machine halts is $\text{32}$