Question

Using a table similar to that shown in Figure 3.6, calculate the product of the octal unsigned 6-bit integers 62 and 12 using the hardware described in Figure 3.3. You should show the contents of each register on each step.

Short answer

Final product${\left(111110100\right)}_2={\left(764\right)}_8$

Step-by-step solution

Registers required to solve the problem

Multiplier, multiplicand, and product registers are the three registers we will be using to solve this particular problem.

Initializing the product register to 0, and the multiplier and multiplicand values of course to the multiplier and multiplicand.

We will check the least significant bit in the multiplier register and add the value of the multiplicand register to the product register if the least significant bit is 0.

Then we will shift the multiplier register right 1 and the multiplicand register left 1.

Now we will repeat this entire process until the multiplier register holds the value 0.

Convert 62 and 12 into binary number

$$\begin{gathered} {\left(62\right)}_8={(110010)}_2 \\ {\left(12\right)}_8={\left(001010\right)}_2 \end{gathered}$$

Iteration	Step	Multiplier	Multiplicand	Product
0	Initial values	001010	0000 0011 0010	0000 0000 0000
1	Shift left multiplicand shift right multiplier	001010 000 101	0000 0110 0100 0000 0110 0100	0000 0000 0000 0000 0000 0000
2	Prod= Prod + Multiplicand Shift left multiplicand Shift right multiplier	000 101 000 101 000 010	0000 01100 0100 0000 1100 1000 0000 1100 1000	0000 0110 0100 0000 0110 0100 0000 0110 0100
3	Shift left multiplicand Shift right multiplier	000 010 000 001	0001 1001 0000 0001 1001 0000	0001 0110 0100 0001 0110 0100
4	Prod= Prod + Multiplicand Shift left Multiplicand Shift right multiplier	000 001 000 001 000 000	0001 1001 0000 0011 0010 0000 0011 0010 0000	0001 1111 0100 0001 1111 0100 0001 1111 0100

Now the final product is (111110100)₂= (764)₈