InClassActivity14-sol

InClassActivity14-sol - 11/26/07 7:37 PM SOLUTION In-Class...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Activity 14.1 Cache organization 11/26/07 7:37 PM SOLUTION In-Class Activity #14 11/26/07 7:37 PM We consider a cache for a computer with 32-bit byte addresses. The cache holds 16 Kbytes of data. Please calculate the information needed to fill the following table. Cache A Cache B Cache C Number of bits in a tag 18 18 19 Number of bits in cache index 12 9 8 Number of blocks cache can hold 4K 512 512 Number of sets cache can hold 4K 512 128K Total number of bytes for cache (bytes) 26,112 17,600 17,728 1 (15) Cache A is direct mapped, and has a block size of 4 bytes. Solution: The cache holds 4K blocks = 4K words of data. In Memory: Byte address = 32 bits Word address = 30 bits Block address = 30 bits In Cache: Cache index for 4K blocks = 12 bits. Tag = 30 – 12 = 18 bits Size = 16K x 8 + 4K x (18 tag bits + 1 valid bit) = 208,896 bits = 26,112 bytes. 2 (15) Cache B is direct mapped, and has a block size of 32 bytes. Solution: The cache holds 512 blocks = 4K words of data.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/08/2008 for the course ECE 562 taught by Professor Zhou during the Fall '07 term at New Hampshire.

Page1 / 2

InClassActivity14-sol - 11/26/07 7:37 PM SOLUTION In-Class...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online