Question

As discussed in the text, one possible performance enhancement is to do a shift and add instead of actual multiplication. Since $9\times 6$, for example, can be written $\left(2\times 2\times 2+1\right)\times 6$ , we can calculate$9\times 6$ by shifting 6 to the left 3 times and then adding 6 to that result. Show the best way to calculate $0\times 33\times 0\times 55$ using shifts and adds/subtracts. Assume both inputs are 8 bit unsigned integers.

Short answer

The value of the $0\times 33\times 0\times 55$ by adding and shifting is .

Step-by-step solution

Determine the MIPS Multiplication by adding and shifting.

The multiplication in the register is performed by shifting and adding the digits. Traditional multiplication will consume more time and space. In order to increase the performance, addition and shifts are used. For each digit of multiplicand , multiplier will be shifted and added accordingly.

Determine the value of multiplication by add and shift operations.

Given values are$0\times 33\times 0\times 55$

Convert 0x33 and 0x55 to the decimal values.

$$\begin{gathered} 0\times 33={51}_{10} \\ 0\times 55={85}_{10} \end{gathered}$$

51 can be written as 51 =32+16+2+1

Now we will shift $0\times 55$ , 5 places left and then add the shift places.

While shifting left 5 places the value of $0\times 55$becomes 0 x AA0. Then add the shifted 4 places.

Now, add the $0\times 55$ shifted left once, then the result becomes .

$0\times AA$

$0\times AA0+0\times 550+0\times AA+0\times 55=0\times 10EF$

This takes totally 3 shifts and 3 adds