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Given that, A machine with a byte - addressable main memory of 2 ^ 1 6 bytes and A block size of 8 bytes. A

Given that,
A machine with a byte-addressable main memory of 2^16 bytes and
A block size of 8 bytes.
A Direct mapped cache consisting of 32 lines is used with this machine.
(a)
A 16-bit memory address is divided into the tag, line number, and byte number as follows:
The byte number is the lowest 3 bits of the address (bits 0-2).
The line number is the next 5 bits of the address (bits 3-7) since there are 32 lines in the cache.
The tag is the remaining 8 bits of the address (bits 8-15) since it uniquely identifies the block of memory being cached.
Explanation:
First, we need to understand the concept of a cache block. A cache block is a contiguous chunk of bytes in main memory that is loaded into the cache as a unit. In this case, the cache block size is 8 bytes, which means that every time the cache is loaded with data from main memory, it loads 8 bytes at a time.
Next, we need to understand the concept of a cache line. A cache line is a group of bytes in the cache that corresponds to a specific block of bytes in main memory.
In this case, there are 32 cache lines, which means that the cache can store up to 32 different blocks of data from main memory.
Now, let's look at how a 16-bit memory address is divided into tag, line number, and byte number:
The lowest 3 bits of the address (bits 0-2) represent the byte number within a cache block. Since the cache block size is 8 bytes, there are 2^3=8 possible byte numbers within a block, ranging from 000(0 in decimal) to 111(7 in decimal).
The next 5 bits of the address (bits 3-7) represent the line number within the cache. Since there are 32 cache lines, there are 2^5=32 possible line numbers, ranging from 00000(0 in decimal) to 11111(31 in decimal).
The remaining 8 bits of the address (bits 8-15) represent the tag, which uniquely identifies the cache block being accessed. Since a cache block is 8 bytes in size, the tag identifies the 8-byte block of memory that the requested byte belongs to.
To summarize, the byte number within a cache block is determined by the lowest 3 bits of the address, the line number within the cache is determined by the next 5 bits of the address, and the tag is determined by the remaining 8 bits of the address.
Adm 2
(b)
To determine the line number for each of the given addresses, we look at the bits 3-7 of the address (the 5 bits after the lowest 3 bits, which determine the byte number). The line number is the decimal value of these bits.
For the address 0001000100011011, the line number is 3.
For the address 1100001100110100, the line number is 6.
For the address 1101000000011101, the line number is 3.
For the address 1010101010101010, the line number is 21.
Explanation:
For a detailed explanation let's take the first address 0001000100011011 and determine which line it would be stored in:
The lowest 3 bits of the address are 011, which represents the byte number within a cache block.
The next 5 bits of the address are 00011, which represents the line number within the cache. In decimal, this is 1.
The remaining 8 bits of the address are 00010001, which represents the tag.
Since this is a direct-mapped cache, the byte at this address can only be stored in one line of the cache, which is line 3. Specifically, it would be stored in the byte offset within the cache line that corresponds to the lowest 3 bits of the address, which is byte 011 within the line.
To summarize, the byte at address 0001000100011011 would be stored in line 3 of the cache, with tag 00010001 and byte offset 011.
Adm 3
(c)
Since the block size is 8 bytes, the other bytes stored along with it will be the 7 bytes in the same block. To determine those byte addresses, we can simply increment the byte number from 0 to 7, while keeping the tag and line number the same:
Byte 0: 0001101000011010
Byte 1: 0001101000011011
Byte 2: 0001101000011100
Byte 3: 0001101000011101
Byte 4: 0001101000011110
Byte 5: 0001101000011111
Byte 6: 0001101000011000
Byte 7: 0001101000011001
Explanation:
In this case, the byte at address 0001101000011010 is the first byte in the block, so the other bytes are the bytes with addresses 0001101000011011,0001101000011100,0001101000011101,0001101000011110,0001101000011111,0001101000100000.(Other than 010 remaining 7 numbers in 0-8 where block size is 8 bytes)
Since all of these bytes are in the same block of memory, they are all mapped to the same line in the cache, which is line 3. Therefore, all of these bytes will be stored in the same line of the cache as the byte at address 0001101000011010.
Adm 4
(d)
In a direct mapped cache, each memory block maps to a specific cache line. The cache line size is equal to the block size.
In this case, the block size is 8 bytes, so each cache line can hold 8 bytes of memory.
The cache has 32 lines, so the total number of bytes that can be stored in the cache is:
32 lines *8 bytes per line =256 bytes.
There

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