How Many Blocks Are in Main Memory

How many blocks are there in a 128 Gig memory space if each block contains 32 words. Number of lines in cache 512 lines.

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My confusion is in the second point.

. Main memory consists of 64-Mbyte16 bytes 222 blocks. Multiple memory blocks will have to map to the same line in the cache. Main memory contains 4K blocks of eight words each.

Blocks in main memory in this case are only updated when the block is to be removed from the cache. In most contemporary machines the address is at the byte level. It is said that each block is of 4 bytes that is.

That means the address is all tag and offset bits. Suppose a computer using fully associative cache has 2 24 words of main memory and a cache of 512 blocks where each cache block contains 16 words. Therefore the set plus Therefore the set plus tag lengths must be 22 bits so the tag length is 14 bits and the word field length is 4 bits.

What is the format of a memory address as seen by the cache ie what are the sizes of the tag and offset fields. Cache has 8 blocks 1 word each Main memory has 256 blocks words 8 bit address Execute the following program Use direct mapping first. Thus Number of bits in block offset 5 bits.

A computer system has an L1 cache an L2 cache and a main memory unit connected as shown below. Thus Number of bits in physical address 32 bits. To which cache block will the memory reference 170424 16 map.

Suppose a computer using fully associative cache has 220220 words of main memory and a cache of 128 blocks where each cache block contains 16 words. 32 Memory Blocks Somehow we have to map the 32 memory blocks to the 8 lines in the cache. If you have 4 blocks for each 1 KB how many blocks does it take to get 128 KB.

Which bits for the word offset. The memory access times are 2 nanoseconds. That is 224 words.

So 4M blocks of 4 bytes. 1Tag field 11 bits Number of blocks in main memory 211 2048 - View the full answer. If the tag is used to differentiate between blocks in this single set fully associative cache then there are 2 10 blocks.

The downside to this method is that the main memory and the cache are not constantly in synch. Blocks are 64 bytes in length and the cache consists of 32K blocks. Main memory consists of 64-Mbyte16 bytes 222 blocks.

Conceptually split the main memory into 2n-byte chunks too. Y 128 KB 256 Bblock 05 Kblock 210 K 512 blocks Another way of thinking about it to consider how many 256 B blocks of memory does it take to get a total of 1 KB of memory. Assume a system.

How many bits are required for addressing the main memory. Show the format for a main memory address assuming a 2-way set associative cache mapping scheme and byte addressing. The block size in L1 cache is 4 words.

The least significant w bits identify a unique word or byte within a block of main memory. To determine the block address of a byte address i you can do the integer division i 2n Our example has two-byte cache blocks so we can think of a 16-byte main memory as an 8-block main memory instead. How many bits does the memory address have.

Will need 8 addresses for a block in the cache Main memory of 64 bytes 6 bit address needed to reference 64 bytes 26 64 64 bytes 2 bytes-per-block. If this is a fully associative cache then there is no index as the cache only has one set. Access time for the cache is 22 ns and the time required to fill a cache slot from main memory is 300ns this time will allow us to determine the block is missing andbring it into cache.

S memory has 128M bytes. The cache logic interprets these s bits as. Time is thus saved since the number of writes is greatly reduced.

How many blocks of main memory are there. Therefore the set plus tag lengths must be 22 bits so the tag length is 14 bits and the word field length is 4 bits. Block size Frame size Line size 32 bytes.

Main memory size 2 32 bytes. And there is a way to answer this. Number of Bits in Block Offset- We have Block size 32 bytes 2 5 bytes.

For purposes of cache access each main memory address can be viewed as consisting of three fields. 20 nanoseconds and 200 nanoseconds for L1 cache L2 cache and main memory unit respectively. Assume a request is always started in parallel to both cache and to main memory.

On a 4-way set associative cache each set on the memory cache can hold up. How many bits are needed to represent the TAG SET and WORD fields. Keeping the 512 KB 4-way set associative example the main RAM would be divided into 2048 blocks the same number of blocks available inside the memory cache.

Number of Bits in Physical Address- We have Size of main memory 2 32 bytes. For instance memory addresses 12 and 13. The main memory has 16M.

The block size in L2 cache is 16 words. The main memory consists of 16384 blocks and each block contains 256 eight bit words. Which bits from which bit to which bit are used for the block ID portion.

The remaining s bits specify one of the 2 s blocks of main memory. This means that the cache is organized as 16K 214 lines of 4 bytes each. A How many blocks of main memory are there.

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