(5 pts) Exercise 7-1: Direct Mapped Cache
Memory
20
7
21
3
22
27
23
32
24
101
25
78
26
59
27
24
28
56
29
87
30
36
31
98
Cache (N = 5)
Address
0
1
Processor
Data
30,25
36,78
31,26
98,59
28,23,28
56,32,56
M
M
H
M
M
M
M
H
M
2
3
4
1.
2.
3.
4.
5.
6.
7.
8.
9.
Read 30
Read 31
Read 30
Read 26
Read 25
Read 28
Read 23
Read 25
Read 28
2
7
Total hits?
Total misses?
(5 pts) Exercise 7-2: Direct Mapped Cache
N = 4)
Cache (
Memory
20
7
21
3
22
27
23
32
24
101
25
78
26
59
27
24
28
56
29
87
30
36
31
98
Address
Data
28
56
1
25
78
2
30,26
36,59
3
31,23
98,32
0
Processor
Total hits?
Total misses?
2-M
M
3-M
M
2-H
H
2-M
M
1-M
M
0-M
M
3-M
M
1-H
H
0-H
H
3
6
1.
2.
3.
4.
5.
6.
7.
8.
9.
Read 30
Read 31
Read 30
Read 26
Read 25
Read 28
Read 23
Read 25
Read 28
(5 pts) Exercise 7-6
•
Look back at Exercises 1 and 2 and
A.) identify two different kinds of reasons for why there might be a
cache miss.
B.) explain a technique for addressing each type of miss
•
Reason #1 for missing:
1. Evicted b/c of conflicts
(add associativity)
– How to address?
•
Reason #2 for missing:
2. Never accessed
(increase block size, or pre-fetching)
– How to address?
(5 pts) Exercise 7-11 – Show final cache and total hits
Block size = 2, N = 4
Memory
20
7
21
3
22
27
23
32
24
101
25
78
26
59
27
24
28
56
29
87
30
36
31
98
Cache
Address
0
1
2
Processor
Data
16,24,16,24
(16),(24),(16),(24)
(17),(25),(17),(25)
26
(26)
(27)
12
(12)
(13)
14
(14)
(15)
3
Total # hits?
0-M
3-M
0-H
2-M
0-M
0-M
3-H
0-M
1-M
1.
2.
3.
4.
5.
6.
7.
8.
9.
Read 16
Read 14
Read 17
Read 13
Read 24
Read 17
Read 15
Read 25
Read 27
2 hits
(5 pts) Exercise 7-12
Show the correct formula for calculating the cache index, given the
cache parameters below
Index = (Address/4) mod 10
1. N = 10, Block size = 4
•
2. N = 8, Block size = 1, Associativity = 4
Index = (Address/1) mod (8/4)
3. N = 16, Block size = 8, Associativity = 2
Index = (Address/8) mod (16/2)
(5 pts) Exercise 7-13 – Fill in blanks, show final cache,
hit/miss for each access, and total hits
Block size = _____, N = ______, Assoc = _____
Memory
2
20
7
Address
21
3
22
27
23
32
24
101
25
78
26
59
27
24
28
56
29
87
30
36
31
98
8
Cache
4
Data
24
(24)
(25)
20
(20)
(21)
26
(26)
(27)
0
1
Processor
0-M1.
0-H2.
1-M3.
0-H4.
0-M5.
1-H6.
0-H7.
1-H8.
1-H9.
Read 24
Read 25
Read 26
Read 24
Read 21
Read 26
Read 24
Read 26
Read 27
(15 pts) Exercise 7-16
•
•
Assume you have an empty cache with
a total of 16 blocks, with each block
having 2 bytes. In this exercise, the
cache is direct mapped.
At right is a series of byte addresses.
You should:
1. Write the appropriate FORMULA
below (filled in for value of N, etc.)
2. For each address, compute the
index and label each one hit or
miss
3. Draw the cache and show the final
contents of the cache
As always, show your work.
Address
Hit / Miss
Miss?
Index
Index
2
M
1
3
H
1
11
M
5
16
M
8
21
M
10
10
H
5
80
M
8
48
M
8
Formula:
39
M
3
Index = (Address/2) mod 16
11
H
5
3
H
1
80
M
8
•
Index =
Since the block size is 2, a load to
address 2 will also bring in address 3, or
a load to address 7 will also bring in
address 6.
Index
First address in block
(last number is actual contents,
earlier numbers were evicted)
0
1
2
2
3
38
4
5
10
6
7
8
16,80,48,80
9
10
11
12
13
14
15
20
(15 pts) Exercise 7-17
Assume you have an empty cache with a
total of 16 blocks, with each block having
2 bytes. In this exercise, the cache is twoway set associative. Assume LRU
replacement.
•
At right is a series of byte addresses.
You should:
1. Write the appropriate FORMULA
below (filled in for value of N, etc.)
2. For each address, compute the
index and label each one hit or miss
3. Draw the cache and show the final
contents of the cache
As always, show your work.
•
Formula:
Index = (Address/2) mod (16/2)
Index == (Address/2) mod 8
There is a total of 16 blocks. 2-way
associativity divides these into 16/2 =8
sets.
Address
Hit
Hit/miss
/ Miss?
Index
Index
2
M
1
3
H
1
11
M
5
16
M
0
21
M
2
10
H
5
80
M
0
48
M
0
39
M
3
11
H
5
3
H
1
80
H
0
Index
First address in block
(last number is actual contents,
earlier numbers were evicted)
0
16, 48
80
1
2
2
20
3
38
4
5
6
7
10