Homework Answers
Block Size = 32 byte => 2^5
Cache size = 64KB
Number of words = 2^16 / 2^5 => 2^11
Offset = 5
Index = 11
Tag = 32 - 5 - 11 => 16
========================================
Thanks, let me now if there is any concern.
Add Answer to:
Question 17 12 points A direct mapped cache holds 64KB of useful data (not including tag...
12 points A direct mapped cache holds 64KB of useful data (not including tag or control...
12 points A direct mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block streis 32-byte and the address is 32-bit, find the number of bits needed for tag index and byte select fields of the address Number of bits for offset bits Number of bits for index bits Number of bits for tag Moving to another question will save this response. Question 17 of 18
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that...
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
Question 17 A direct-mapped cache holds 128KB of useful data (not including tag or control bits)....
Question 17 A direct-mapped cache holds 128KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag .. bits
Question 17 12 points Save Answer A direct-mapped cache holds 32KB of useful data (not including...
Question 17 12 points Save Answer A direct-mapped cache holds 32KB of useful data (not including tag or control bits). Assuming that the block size is 16-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram....
For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram. Indicate the block size, number of blocks, and address field decomposition (block offset, index, and tag bit width) assuming a 32-bit memory address.
Design a 256KB (note the B) direct‐mapped data cache that uses a 32‐bit address and 8...
Design a 256KB (note the B) direct‐mapped data cache that uses a 32‐bit address and 8 words per block. Calculate the following: How many bits are used for the byte offset and why? How many bits are used for the set (index) field? How many bits are used for the tag? What’s the overhead for that cache?
Cache Layout: A processor has a separate D-cache and an I-cache. D-cache: 64KB, 4-way set associative,...
Cache Layout: A processor has a separate D-cache and an I-cache. D-cache: 64KB, 4-way set associative, block size of 1 word, write-back policy I-cache: 32KB, direct mapped cache, block size of 1 word The processor uses the LRU algorithm for its replacement policy. Answer the following questions. Make sure that you account for all the book -keeping bits. A word is 4 bytes (a) Calculate the number of tag, index and offset bits for the D-cache. (b) Calculate the number...
Cache question computer architecture A cache holds 128 words where each word is 4 bytes. Assuming...
Cache question computer architecture A cache holds 128 words where each word is 4 bytes. Assuming a 32-bit address, for each of the following organizations, complete the table. a.A direct-mapped cache with block size = 16words b.2-way set-associative cache with block size = 8words c.4-way set-associative cache with block size = 4words d.A fully associative cache with block size = 2words. Cache a Cache b Cache c Cache d total # bits for word & byte displacement # bits in...
For a direct-mapped cache with a 32-bit address and 32-bit words, the following address bits are...
For a direct-mapped cache with a 32-bit address and 32-bit words, the following address bits are used to access the cache. TAG INDEX OFFSET 31-15 14-8 7-0 a. What is the cache block size (in words)? [13 points] b. How many blocks does the cache have? [12 points]
The following figure shows the address that is going from a certain processor to a direct-mapped...
The following figure shows the address that is going
from a certain processor to a direct-mapped cache. The address is
divided into fields. The index of the first bit and the last bit of
each field is written below it. Calculate the size of the data that
is stored in the cache, in Kibytes, and the total number of bits
within the cache, in Kibits.
TAG = 31 - 16
INDEX = 15 - 6
BLOCK OFFSET = 5 -...
12 points A direct mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block streis 32-byte and the address is 32-bit, find the number of bits needed for tag index and byte select fields of the address Number of bits for offset bits Number of bits for index bits Number of bits for tag Moving to another question will save this response. Question 17 of 18
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
Question 17 A direct-mapped cache holds 128KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag .. bits
Question 17 12 points Save Answer A direct-mapped cache holds 32KB of useful data (not including tag or control bits). Assuming that the block size is 16-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
The following figure shows the address that is going
from a certain processor to a direct-mapped cache. The address is
divided into fields. The index of the first bit and the last bit of
each field is written below it. Calculate the size of the data that
is stored in the cache, in Kibytes, and the total number of bits
within the cache, in Kibits.
TAG = 31 - 16
INDEX = 15 - 6
BLOCK OFFSET = 5 -...
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 10 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 10 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 10 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 10 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 10 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 10 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 10 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 10 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 10 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 10 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 10 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 10 months ago