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.

Ad$$m $2$

$($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 $011.$

Ad$$m $3$

$($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 $0001101000011010.$

Ad$$m $4$

$($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