Question

A) The following code fragment takes as input a number in t1 and produces a value in register $s2. Assume that initially $s2 equals zero. Line Loop: addi $t2, $zero, 10 2 Loop2: addi$s2, $s2, 2 3 subi $t2, $t2, 1 4 bne $t2, $zero, Loop2 5 subi $t1,$t1, 1 6 bne $t1, $zero, Loop For each of the following problems, justify your answers and show your calculations clearly. Final answers without justifications are not accepted. What is the final value of $s2, after running this code with an initial value of a) $t1 = 5, b) $t1 = 10, c) $t1 = 50 iii What is the final value of $t2, after running this code with an initial value of a) $t1 = 5, b) $t1 = 10, c) $t1 = 50 What is the number of MIPS instructions executed, if initial value of $t1 equals N (N> 0)? Show your answer in terms of N. Assume that the code is run on a non-pipeline machine with 2.5 GHz clock that requires one clock cycle for each of addi, subi and bne instructions. What is the time, in seconds, it will take to execute this code if: a) $t1 = 5, b) $t1 = 10, c) $t1 = 50 iv

Answer 1

In the given program, Loop2 runs for 10 iterations and in each increment s2 by 2.

Hence after Loop2 has run s2 is incremented by 20.

Loop2 is run t1 number of times.

Hence the program return the value of s2 = 20*t1.

Based on the above facts it is easy to find the answers for the given questions as follows:

i)

a) $t1 = 5

Answer = 20*5 = 100

b) $t1 = 10

Answer = 20*10 = 200

c) $t1 = 50

Answer = 20*50 = 1000

ii) $t2 is the counter register for Loop2. Loop2 stops when $t2 = 0.

Hence for each case, the answer is 0.

a) $t1 = 5

Answer = 0

b) $t1 = 10

Answer = 0

c) $t1 = 50

Answer = 0

iii) Loop2 is executed 10 times and has 3 instructions.

Hence each execution of the subroutine Loop2 executed 30 instructions.

The Loop subroutine executes 3 instructions and the Loop2 subroutine.

Sine the Loop subroutine is executed $t1 = N no. of times, the total instructions executed = (3 + 30)N = 33N

iv) The time taken to execute 33N instructions when each instruction takes 1 clock cycle = 33N/clock rate = 33N / 2.5 ns

i) $t1 = 5

Time Taken = 33*5/2.5 = 66 ns

ii) $t1 = 10

Time Taken = 33*10/2.5 = 132 ns

iii) $t1 = 50

Time Taken = 33*50/2.5 = 660 ns

v) Here is the MIPS code for the given C Program. I have added comments for better understanding:

swap:
# $a0: base address of v
# $a1: k
sll $t0,$a1,2 # Multiply by 4
addi $t0,$t0,$a0 # FInd address of v[k]
lw $t1,0($t0) # Load v[k]
addi $t2,$t0,4
lw $t3,0($t2)
sw $t3,0($t0)
sw $t1,0($t2)

Screenshot of the code:

swap: # $a0: base address of v # $al: sll $t0, $al, 2 # Multiply by 4 addi $t0,$t0, $aD # FInd address of v[k] lw $t1,0 ($to) # Load v[k] addi $t2, $t0,4 lw $t3,0 ($t2) sw $t3,0 ($to) sw $t1,0 ( $t2)

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