### A good answer might be:

See Below

# Subroutine Epilog

The epilog of a subroutine is the part that prepares the values that are returned to the caller and restores the registers and stack to where they were when the caller made the call.

    #  int mysub( int arg )
    #  {
    #    int b,c;                     // b: 0($fp)
    #                                 // c: 4($fp)
    #    b = arg*2;
    #    c = b + 7;
    #
    #    return c;
    #  }
             .text
             .globl  mysub
    mysub:
                                      # prolog
             sub     $sp,$sp,4        #   1. Push return address
             sw      $ra,($sp)
             sub     $sp,$sp,4        #   2. Push caller's frame pointer
             sw      $ra,($sp)

             sub     $sp,$sp,4        #   3. Push register $s1
             sw      $s1,($sp)

             addiu   $fp,$sp,8        #   4. $fp = $sp + space_for_variables

             move    $sp,$fp          #   5. $sp = $fp

                                      # body of subroutine
             mul     $s1,$a0,2        #     arg*2
             sw      $s1,0($fp)       # b = "   "

             lw      $t0,0($fp)       # get b
             add     $t0,$t0,7        #      b+7
             sw      $t0,4($fp)       # c = "  "

                                      # epilog
             lw      $v0,___(___)     #   1. Put return value in $v0

             add     $sp,$fp,____     #   2. $sp = $fp +space_for_variables

             lw      $s1,(____)       #   3. Pop register $s1
             add     ___,___,___      #

             lw      $fp,(____)       #   4. Pop $fp
             add     ___,___,___      #

             lw      $ra,(____)       #   5. Pop $ra
             add     ___,___,___      #

             jr      $ra              #   6. return to caller

### QUESTION 13:

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