addressing - MIPS - How to find the address value of branch and jump instructions -

i have exam coming up, , stuck on question (see below); looking @ model answer did not help. i've tried reading our main text topic, still have no idea how this. if provide step-by-step walk through of question, grateful.

"assuming first instruction of mips snippet below located @ memory address 0x10001000. value else , exit in bne , j instruction?"

1   0x10001000:   addi $s0, $0, 5 2   0x10001004:   sub $s2, $0, $s1 3   0x10001008:   beq $s0, $s2, else 4   0x1000100c:   add $s0, $0, $0 5   0x10001010:   add $t0, $s2, $s0 6   0x10001014:   j exit 7   0x10001018:   else: addi $s1, $s0, -1 8   0x1000101c:   exit: 

model answer:

else: 0000000000000011 exit: 00000000000000010000000111 

i have included link image of original question well. http://i.imgur.com/nghpzxs.png

first, we'll deal branch. branches i-type instruction, branch target encoded in 16 bits. easiest way figure out immediate field of branch count number of instructions between branch instruction , target. in case, else label 4 instructions after beq, however, pc incremented 4 during instruction fetch stage, actual immediate field 3. of course in binary, matches sample 0000000000000011.

next, jump j-type instruction encodes target of jump using first 24 bits of address. however, because jump target must word aligned, last 2 bits 0 making them unnecessary. therefore, jump field of j instruction matches sample answer: 0x1000101c >> 2 = 0x4000407 = 00000000000000010000000111