option B ) 6 cycles xy+xyz+yz = y(x+xz+z)

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A non-pipelined CPU has $12$ general purpose registers $(R0,R1,R2, \dots ,R12)$. Following operations are supported

- $\begin{array}{ll} \text{ADD Ra, Rb, Rr} & \text{Add Ra to Rb and store the result in Rr} \end{array}$
- $\begin{array}{ll} \text{MUL Ra, Rb, Rr} & \text{Multiply Ra to Rb and store the result in Rr} \end{array}$

$\text{MUL}$ operation takes two clock cycles, $\text{ADD}$ takes one clock cycle.

Calculate minimum number of clock cycles required to compute the value of the expression $XY+XYZ+YZ$. The variable $X,Y,Z$ are initially available in registers $R0,R1$ and $R2$ and contents of these registers must not be modified.

- $5$
- $6$
- $7$
- $8$

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