Question

Please help me out.. A and C

Question. 5. (30 points) Consider a computer with byte addressable main memory bytes, and the block size is 8 bytes. Assume that a direct mapping cache consisting 32 lines is used with this machine. (a) (5 points) How many bits are required to hold a memory address? (b) (5 points) How many total bytes of memory can be stored in the cache? 256 bytes (C) (10 points) How is that memory address divided into tag, line number, and byte (word) fields? (d) (10 points) Into what line would the data having following memory addresses be stored? . 0001 0001 0001 1011 as the cache line part is oooll, its decimal is encivalent is 3 • 1100 0011 0011 0100 As the cache line part is Gollo, its decima is caninant is 6 • 1101 0000 0001 1101 As the cache line part is oooll, its dea equivalat is 3 . 1010 1010 1010 1010 As the cache line part is 10101, dumal is equivalent is 21

Answer 1

Solution of part A and C is here below.

Part A.

Size of main memory (MM) = 26 B = 64 KB Size of each Block = 813 No. of Block in M.M = 216 = 2'3 Blocks No. of Lines in cache memory = 32 Lines. Total size of cache memory = 324 block size - 32x8 = 25 6 3 No. of bits required in Addressing main memory = 16 bits, ( Memory is Byte addressable il. 13-1W) ( For 216 Bytes, 16 bits are required)

Since the memory in Byte addressable i.e. 1 Byte =1 Word.

Part B

In Direct Mapping Address is divided into 3 parts. - 16 bit [tag line no, OFFSETT There are are total as Lines. . We required bits to uniquely identify each timeline. Each block contain & Byte. - 3 bits are required for addresing each byte in Line no Remaining 8 bits will be tag bit. Isbit sabits 3 bits. ] tag line no. OFFSET

The 16 bit address of Main Memory will go to cache memory. Out of these 16 bits,

3 bits --> are for offset (for identifying each word or Byte in each Line)

5 bits --> are for Line no. (Total 32 lines)

8 bits --> Rest bits are for tag information.