1. Please show only structural hazards. mov [100], [150] mov [200], [250] add [1
ID: 3570473 • Letter: 1
Question
1. Please show only structural hazards. mov [100], [150] mov [200], [250] add [150], [250], r3 sub r3, #5, r4 add r3, #2, r5 div r4, r5, IS Fi IDi Co IFo Ei W Instruction encoding: instruction opl , op2, result; [xxxx] - memory address; #x - constant; rx - register. 2. Please show only data hazards. mov [100], [150] Fi D3i CO Fo Ei mov [200], [250] add [100], [150], [300] I add [300], #100, [322] add [300], #200, [333] sub [250], [322], [326] 3. Memory is divided in 32 blocks, cache size is 8 blocks. In which cache slot should be put block nr 20 if cache organization is:Explanation / Answer
1)
mov [100][150]
Fi
Di
Co
Fo
Ei
W
mov [200][250]
Fi
Di
Co
Fo
Ei
W
add [150],[250],r3
Fi
Di
Coadd
Foadd
Ei
Wr3
sub r3,#5, r4
idle
Fi
Di
Cosub
Foadd
Ei
Wr4
add r3,#2,r5
idle
Fi
Di
Coadd
Foadd
Ei
Wr5
div r4,r5,r6
idle
Fi
Di
Codiv
Foiv
Ei
Wr6
2)
mov [100][150]
Fi
Di
Co
Fo
Ei
W[150]
mov [200][250]
Fi
Di
Co
Fo
Ei
W[250]
add [100], [150],[300]
Fi
Di
Coadd
Foadd
Ei
W[300]
add [300],#100, [322]
idle
Fi
Di
Coadd
Foadd
Ei
W[322]
add [300],#200,[333]
idle
Fi
Di
Coadd
Foadd
Ei
W[333]
sub [250,[322][326]
idle
Fi
Di
Codiv
Foiv
Ei
W[326]
3)
The size of total memory is 25=32 blocks and the cache size is 8 blocks 28-1=27 is size of the cache.
If the organization is a two way, then the nr 20 is placed in 20’Th block.
Since, there are 32 blocks in main memory, which tells that in a 32 block of main memory, if 8 blocks of cache are being placed, and then the nr 20 will fit at 20’Th block.
BLOCK 0
0
0
BLOCK 1
1
1
BLOCK 2
2
2
BLOCK 3
3
3
BLOCK 4
4
4
BLOCK 5
5
5
BLOCK 6
6
6
BLOCK 7
7
7
BLOCK 8
8
8
BLOCK 9
9
9
BLOCK 10
10
10
BLOCK 11
11
11
BLOCK 12
12
12
BLOCK 13
13
13
BLOCK 14
14
14
BLOCK 15
15
15
BLOCK 16
16
16
BLOCK 17
17
17
BLOCK 18
18
18
BLOCK 19
19
19
BLOCK 20
20
20
BLOCK 21
21
21
BLOCK 22
22
22
BLOCK 23
23
23
BLOCK 24
24
24
BLOCK 25
25
25
BLOCK 26
26
26
BLOCK 27
27
27
BLOCK 28
28
28
BLOCK 29
29
29
BLOCK 30
30
30
BLOCK 31
31
31
mov [100][150]
Fi
Di
Co
Fo
Ei
W
mov [200][250]
Fi
Di
Co
Fo
Ei
W
add [150],[250],r3
Fi
Di
Coadd
Foadd
Ei
Wr3
sub r3,#5, r4
idle
Fi
Di
Cosub
Foadd
Ei
Wr4
add r3,#2,r5
idle
Fi
Di
Coadd
Foadd
Ei
Wr5
div r4,r5,r6
idle
Fi
Di
Codiv
Foiv
Ei
Wr6
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