How do u figure out the 16 bit offset of a branch? [Resovled]

[voidflux]

Hello everyone.

I'm studying for my exam and i'm stuck on this problem..

For the asm code given below, what is the value of the 16-bit offset in the instruction 'beq'
beq $t0, $t1, label
sub $t2, $t0, $t1
addi $t2, $t2, 4
sw $t2, 0($s0)
label:.....

the answer is 0x0010, I thought u start counting from the beq instruction but yesterday the profesor showed me on the sample test and she said u don't start counting at beq, but right after beq, so u would count down till u get to the label:...


This would make sense to me if u start counting at beq, then it would be
0
4
8
12
16

if u convert 16 from decimal to hex u get 10, so is that how they got 0x0010 ?


But the professor explained this problem to me which doesn't follow that pattern:

loop: beq $s0, $s2, exit
add $s0, $s0, $s1
j loop
exit: lui $s0, 48

whats the machine code?
answer:
beq [4][16][18][2]

2 is the 16 bit offest in this case, i asked her how did u get 2?
well she said, u start one right after the branch instruction and start counting...
so if u did that, u would count
0
4
8 and at 8, u would hit the label exit:
she said they got 2 because ur jumping 2 words, so its 2.

can anyone help me out?