Question

This is vhdl code can you please explain how they got the answer?

How many sor following instructions are executed by the MIPS single-cycle per instruction processor from class proces cycles will it elelt take for this processor's program counter to reach the "nop" instruction? To get credit explain how the cycles are accountecd andi $3, $3,0 andi $2, $2,0 addi $2, $2, 20 : initialize to O ; clear reg. ;loop bound ;load x(i) to R15 ; load yi) to R14 Slabel: Iw $15, 4000($3) lw $14, 8000($3) add $24, $15, $14 sw $24, 8000(S3) save new y) addi $3, $3,4 bne $3, $2, Slabel ; update address reg. check loop bound loop o20 nap

Answer 1

There are three instructions before the loop.

Therefore total number of cycles taken before the loop gets executed is: 3

The total number of instructions within the loop is 6.

The loop will get executed if the values in registers $3 and $2 are not equal.

The initial values of the registers when the loop gets executed first( total 6 instructons) :

$3 = 0

$2 = 20

Hence total 6 cycles for first time of loop execution.

In the first iteration of loop, the value of register $3 gets changed to 4.

As the values of registers $2 and $3 are not equal, the loop gets executed second time.

In the second iteration of loop the value of register $3 gets changed to 8. As the values of $2 and $3 are not equal, the loop gets executed third time.

Like that the loop gets iterated 5 times. In the last fifth iteration the value of register $3 gets changed to 20. Now both the registers $3 and $2 are equal. Hence loop get exited and the instruction nop which the next instruction after the loop is executed.

Therefore the total number of instructions before nop instruction is :

Number of instructions before loop + number of times loop executed * number of instructions in the loop

3 + 5*6= 33

As it is a single cycle per instruction, the total number of cycles taken before reaching nop instruction is: 33