Question

If I have a problem set like so:

Below is a list of 64-bit memory address references, given as word addresses.
3, 180, 43, 2, 191, 88, 190, 14, 181, 44, 186, 253

5.2.1

BLOCK SIZE: 1 word

CACHE SIZE: 16 1-word blocks

a) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with 16 one-word blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

b) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with two-word blocks and a total size of 8 blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

c) You are asked to optimize a cache design for the given references. There are three direct-mapped cache design possible, all with a total of eight words of data. Which cache provides best performance?

a. C1 has 1-word blocks

b. C2 has 2-word blocks

c. C3 has 4-word blocks

Answer 1

Cache À Memory address = 64 bit. Block size = l word .: Word Offset - Blocker- o bit Offset Total # of block in, 16 Index offset - log, 16 = 4 bit. S Tag= 64-4 = 60 bits. (1) 3 - pon * Block #: 3. Tago . This is a compulsary miss since it is bought for the first time in cache. Ci) 180 = 10110100 J Block # = 4. Tag: 1011= OXB. This is a miss ciji) 43 = 101011 Block # :|). Tag:1000XR. This is a miss.

4) 2: 0010 Block #:2. Tag = 0 This is a miss . (V) 1913 1011!!! y Block #=15 Тая: охв. This is a nuss. wi) 88 = 1011000 Block #-8. Tag=0x5 This is a miss. wi) 190 = 101110 > Block #-14. 1. Tag=0xB This is a miss . vii) 14 = 1110 Block # -14. Tag = 0x0 This is a miss. ist) 181 + 10110101 Block #=5 Tag= OXB. This is a miss.

x) 44 101100 I L Block # -12. Tag = 0x2. This is a miss. xi) 186 10111010 5 Block #=10 Tag=0xB. This is a miss. xii) 253 = 111 1101 5 Block #=13 OXF This is a miss the memory reference is a miss. This hit ratio is o. b) Worksite Block size = 2 word. :: Block Offset - Ibit. Total # of block inside 8 Cache :: Index offset- 3 bit. Tag: 64- (3+) = 60 bits. (i) 3 - 0011 4 Block #-1. Tag=0x0 This is a niss.

(ii) 180 = 10110100 Block #:22. Tag: OxB. This is a miss. Gji) 43 = 101011 Block # = 5, Tag=0x2. This is aniss. (iv) a = 0010 oxo. This is Block # -1 a hit. v) 191 101111 Block# = 7 , Tag : 0xB. This is a miss . vi) 88 = 1011000 Block #=, Tag0XB , This is a miss. vii) 190 = 10111110 Tag - OXB, Block #=7. This is a hit vii) 14 5 1110 This is a miss. Block #= 7, Tag=0x0. ixviii) 1815 10110101 Tag = OxB, Block #=2. This is a hit x) 44 · 101100 Tag : 0x2, Block #: 6. This is a miss. xi) 186 = 10111010 Block # -5 Tag: 0xB. This is a miss . xii) 253 = 1110 1101 Block # = 6 , lo f. This is a miss. :: Total # of hit = 2.

ON Let us first calculate the cache performance when block size = 4 word. Index - 2 bits, Block offset=2bit, Tag-60 bits. ;) 3 = 0011 Block #=0, Tag- oxo. This is amiss. ii) 180 = 10110100 -) Tag = 0XB, Blocks 1. This is a miss . ii) -43 - 10101 =) Tag=0x2, Block 2 . This is W a : 0010 =) Tag=0x0, Block# = 0 . Hi Hit. v) 191 - 10111111 =) Tag = 0xB, Block# - 3 · Miss. vi) 88 - 01011000 - Tag: ox, Block# 2. This is a miss. vii) 190 1011110 =) Tag = OxB, Block=3 · Hit: Vmi 14 : Trio -) Tag=0x0, Block # = 3 • Miss . 1x) 1815 10110101 -) Tag = 0XB, Block#=1. Hit. x) 44 = 101100 =) Tag = 0x2, Block#=3. Miss . 186 - 10111010 5) Tag=0xB, Block LOXB. Block#zd. This is a miss xii) 253 = 11111101 =) Tag = 0xF, Block#3. Miss. with C3 =) # of hit = 3. with a =) # of hit = 2. with C1 =) # of hit=0 the best ...C3 cache provides performance.

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