**Cache Memory Mapping Problem****Problem Statement:**Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache set would the hexadecimal address 0xBE map if the computer uses 2-way set associative mapping?**Detailed Explanation:**1. **Memory Address Bits:** - The memory address size is 8 bits.2. **Cache Organization:** - The cache holds 16 blocks. - Each block holds 4 bytes. - Cache is organized as 2-way set associative.3. **Block Calculation:** - Number of sets in the cache = (Number of cache blocks) / (Associativity). - For a 2-way set associative cache with 16 blocks: Number of sets = 16 / 2 = 8 sets. - We need 3 bits to address the 8 sets (since 2^3 = 8).4. **Address Breakdown:** - For a set-associative cache, the address is broken down into three main parts: Tag, Index (Set), and Block Offset. - Block Offset: With each block holding 4 bytes (2^2), 2 bits are needed for the block offset. - Index: With 8 sets, 3 bits are needed for the set index. - Tag: The remaining bits after accounting for the block offset and index will represent the tag. Therefore, for an 8-bit address: - Block Offset (2 bits). - Index (3 bits). - Tag (remaining bits, i.e., 3 bits).5. **Hexadecimal to Binary Conversion:** - Hexadecimal address 0xBE needs to be converted to binary. - Hexadecimal B = 1011, E = 1110, hence 0xBE = 10111110 in binary.6. **Binary Address Breakdown:** - Binary: 10111110 - Tag = First 3 bits: 101 - Index = Next 3 bits: 111 - Block Offset = Last 2 bits: 10 The index part (111 in binary) gives the set number in the cache.7. **Set Calculation:** - Converting binary 111 to decimal gives 7.**Conclusion:**

Name: **Cache Memory Mapping Problem****Problem Statement:**Suppose we have
Uploaded: 2024-03-20T06:51:23.000Z
Channel: Bartleby
Description: **Cache Memory Mapping Problem****Problem Statement:**Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache set would the hexadecimal address 0xBE

The given system is a byte addressable computer. Number of blocks = 16 Block Size = 4 Bytes Length…

Engineering

Computer Science

Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming address has 8 bits, to which cache set would the hexadecimal address OxBE map if the computer mapping? that each memory uses 2-way set associative

Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming address has 8 bits, to which cache set would the hexadecimal address OxBE map if the computer mapping? that each memory uses 2-way set associative

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

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Q: A byte offset of 2 in an address means that each set in a multiway set associative (or in the…

A: answer is given below

Q: A cache memory system with capacity of N words and block size of B words is to be designed. If it is…

A: For a 4-way set associative cache, the length of the TAG field is given below:

Q: Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main…

A: Question from cache memory mapping. In this we are talking about set-associative cache mapping. We…

Q: Suppose a computer using direct mapped cache has 23232 bytes of byte-addressable main memory, and a…

A: I have answered this question in step 2.

Q: Here is the question: A direct-mapped cache consists of 8 blocks. A byte-addressable main memory…

A: Calculate the misses and hits for the given program that loops 3 times from addresses 0 to 75 (base…

Q: A cache memory has a line size of eight 64-bit words and a capacity of 4K words. The main memory…

A: A 64-bit word means 8 byte.Line size: 8 words in a line, means 8 x 8 bytes = 64 bytes in a line = 26…

Q: Suppose a byte-addressable computer using set associative cache has 4Mbyes of main memory and a…

A: Set-associative caches are an essential component of modern computer systems, aiming to improve data…

Q: A cache is set up with a block size of 32 words. There are 64 blocks in cache and set up to be 4-way…

A: Each block has 32 words, so the block offset = 25 words. So Block offset field can be represented by…

Q: In a microprocessor of 32 bit addresses, the tag length will change if we design a two-way…

A: two-way set-associate cache versus a four-way set- associative cache: Two way set associative cache…

Q: Consider a 64-bit word-based four-way set associative cache. The number of words each line is 8, and…

A: Given: Consider a 64-bit word-based four-way set associative cache. The number of words each line is…

Q: nas a TLB able entries minimum

Topic Video

Question

**Cache Memory Mapping Problem**

**Problem Statement:**

Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache set would the hexadecimal address 0xBE map if the computer uses 2-way set associative mapping?

**Detailed Explanation:**

1. **Memory Address Bits:**
- The memory address size is 8 bits.

2. **Cache Organization:**
- The cache holds 16 blocks.
- Each block holds 4 bytes.
- Cache is organized as 2-way set associative.

3. **Block Calculation:**
- Number of sets in the cache = (Number of cache blocks) / (Associativity).
- For a 2-way set associative cache with 16 blocks: Number of sets = 16 / 2 = 8 sets.
- We need 3 bits to address the 8 sets (since 2^3 = 8).

4. **Address Breakdown:**
- For a set-associative cache, the address is broken down into three main parts: Tag, Index (Set), and Block Offset.
- Block Offset: With each block holding 4 bytes (2^2), 2 bits are needed for the block offset.
- Index: With 8 sets, 3 bits are needed for the set index.
- Tag: The remaining bits after accounting for the block offset and index will represent the tag.

Therefore, for an 8-bit address:
- Block Offset (2 bits).
- Index (3 bits).
- Tag (remaining bits, i.e., 3 bits).

5. **Hexadecimal to Binary Conversion:**
- Hexadecimal address 0xBE needs to be converted to binary.
- Hexadecimal B = 1011, E = 1110, hence 0xBE = 10111110 in binary.

6. **Binary Address Breakdown:**
- Binary: 10111110
- Tag = First 3 bits: 101
- Index = Next 3 bits: 111
- Block Offset = Last 2 bits: 10

The index part (111 in binary) gives the set number in the cache.

7. **Set Calculation:**
- Converting binary 111 to decimal gives 7.

**Conclusion:**

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions

A byte offset of 2 in an address means that each set in a multiway set associative (or in the directly mapped) cache has 4 bytes per set, per cache line. Keeping this in mind, answer the questions given the following details. A cache is organized as a 4 way set associative cache Each set's cache line consists of 4 words (meaning there are 16 bytes per line, for each set of the cache). Each set individually has one Valid bit, and one Dirty bit, for each line. The tag field of this cache is 8 bits wide. The address is 32 bits wide.. Question 1c) What is the cache size? (Meaning the number of kilobytes of data being stored in this cache, ignoring the tag, dirty, and valid bits) Question 1d) If this same sized cache (meaning the number of kilobytes of the data part from the question 1c above) is instead organized as a direct-mapped cache, what is the total number of 'memory bit cells needed to design this cache instead (meaning how many bitcells including the tag, dirty, and valid bits?…
Suppose we have a byte-addressable computer using direct mapping with 16-bit main memory addresses and 32 blocks of cache. If each block contains 8 bytes. a. Determine the number of bits of the tag field. b. To which cache block would the hexadecimal address 0x2468 map? c. What is the tag of the hexadecimal address 0x2468
Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, set, and offset fields?
Suppose a computer using fully associative cache has 224 words of main memory and a cache of 512 blocks, where each cache block contains 16 words. How many blocks of main memory are there? What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? To which cache block will the memory reference 17042416 map?
Suppose a computer using direct mapped cache has 236 bytes of byte-addressable main memory and a cache size of 1024 bytes, and each cache block contains 64 bytes. ⦁ How many blocks of main memory are there? ⦁ What is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields? ⦁ To which cache block will the memory address 0x13A4576B map?
Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache?
Suppose a computer using 8-way set associative cache has 1 M words of main memory, and a cache of 16 K words, where each cache block contains 8 words. What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, set, and ?word fields Tag = 9-bit, Set = 8-bit, Word = 3-bit Tag = 9-bit, Set = 7-bit, Word =4-bit Tag = 9-bit, Set = 6-bit, Word = 5-bit Tag = 9-bit, Set = 5-bit, Word = 6-bit
In a certain computer, the virtual addresses are 32 bits long and the physical addresses are 48 bits long. The memory is word addressable. The page size is 16 kB and the word size is 2 bytes. The Translation Look-aside Buffer (TLB) in the address translation path has 64 valid entries. Hit ratio of TLB is 100% then maximum number of distinct virtual addresses that can be translated is K.
A byte offset of 2 in an address means that each set in a multiway set associative (or in the directly mapped) cache has 4 bytes per set, per cache line. Keeping this in mind, answer the questions given the following details. A cache is organized as a 4 way set associative cache Each set's cache line consists of 4 words (meaning there are 16 bytes per line, for each set of the cache). Each set individually has one Valid bit, and one Dirty bit, for each line. The tag field of this cache is 8 bits wide. The address is 32 bits wide. Question 1g) If we have a write through cache coherency policy, will either of the above organizations make a difference in the amount of power used to do one write through operation (specifically above this cache layer, in the upper level memory structures to maintain coherency.
Suppose we have a byte-addressable computer with a cache that holds 8 blocks of 16 bytes each. Assuming that each memory address has 16 bits, to which cache block would the hexadecimal address 0x13F map if the computer uses direct mapping? Blocks are numbered starting at 0.
Consider a direct-mapped cache memory with 12-bit addresses. The cache is byte-addressable. We have B = 16 bytes per block and S = 8 sets. For the address shown below. Indicate which bits correspond to the cache set index, tag bits, and block offset. Bit 11 Bit 10 Bit 9 Bit 8 Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
Fill in blank Suppose that linear page table is used where the memory addresses are 12-bit binary numbers and the page size is 256 bytes. If a virtual address in binary format is 101000011100, then the VPN (virtual page number) in binary format will be ---------