Question: 32 bytes of memory. 16 bytes of set-associative cache, where blocks can go anywhere within the set. Block is 4 bytes, set in cache is two blocks. Populate memory starting with 0-9, then upper case letters. The following address requests will load the cache: • 00111 • 00001 • 10000 . 10111 • 00011 The CPU then generates address 10110. If this address is in the cache, respond with the requested byte.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Question: 32 bytes of memory. 16 bytes of set-associative cache, where blocks can go
anywhere within the set. Block is 4 bytes, set in cache is two blocks. Populate memory starting
with 0-9, then upper case letters.
The following address requests will load the cache:
. 00111
00001
10000
10111
.
• 00011
The CPU then generates address 10110. If this address is in the cache, respond with the
requested byte.
Draw a diagram like the below showing how you came to your answer.
Cache
Set Tag
00 01 10 11
0 10 GHIJ
1 10 K
L MN
Memory
Tag Set 00
01 10 11
00 0 0
1 2 3
01 08
9
10 0 G
H
11 0 0 P
A B
I
J
QR
B
00 01 10 11 offset
00 01 2 3
00 4
5
6 7
Tag Set 00
00 1 4
01 1
с
10 1
K
11 1
S
01 10
5 6 7
DE
F
M N
U V
11
L
T
offset
8 blocks in
memory
= 32 bytes
Transcribed Image Text:Question: 32 bytes of memory. 16 bytes of set-associative cache, where blocks can go anywhere within the set. Block is 4 bytes, set in cache is two blocks. Populate memory starting with 0-9, then upper case letters. The following address requests will load the cache: . 00111 00001 10000 10111 . • 00011 The CPU then generates address 10110. If this address is in the cache, respond with the requested byte. Draw a diagram like the below showing how you came to your answer. Cache Set Tag 00 01 10 11 0 10 GHIJ 1 10 K L MN Memory Tag Set 00 01 10 11 00 0 0 1 2 3 01 08 9 10 0 G H 11 0 0 P A B I J QR B 00 01 10 11 offset 00 01 2 3 00 4 5 6 7 Tag Set 00 00 1 4 01 1 с 10 1 K 11 1 S 01 10 5 6 7 DE F M N U V 11 L T offset 8 blocks in memory = 32 bytes
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