See below.
The subroutine uses two variables so there is eight bytes of space on the stack frame for them.
# 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 $fp,($sp)
sub $sp,$sp,4 # 3. Push register $s1
sw $s1,($sp)
sub $fp,$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,___(___) # b = " "
lw $t0,___(___) # get b
add $t0,$t0,___ # b+7
sw $t0,___(___) # c = " "
. . . . .
jr $ra # return to caller
The program is not very efficient, as written.
There is no need to store and then load b.
A non-optimizing compiler might do just that, however.