To determine the size of the Tag field in a Line Entry, we need to consider the organization used, which in this case is One Way Set Associative.

In a One Way Set Associative cache, each line in the cache can only store one block of data. The line size is given as 4 Bytes, which means each line can hold 4 Bytes of data.

To calculate the size of the Tag field, we need to determine the number of bits required to uniquely identify each block in the main memory. Since the main memory size is given as 2GB and the block size is 4 Bytes, we can calculate the number of blocks in the main memory as follows:

Number of blocks = Main memory size / Block size
= 2GB / 4 Bytes
= 2 * 1024 * 1024 * 1024 Bytes / 4 Bytes
= 536,870,912 Blocks

To uniquely identify each block, we need to use a unique tag for each block. The number of bits required to represent 536,870,912 blocks is given by:

Number of bits = log2(Number of blocks)
= log2(536,870,912)
≈ 29.9 bits

Since we cannot have fractional bits, we need to round up to the nearest whole number. Therefore, the size of the Tag field in a Line Entry would be 30 bits.

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To determine the size of the Tag field in a Line Entry, we need to consider the organization used, which in this case is One Way Set Associative.

In a One Way Set Associative cache, each line in the cache can only store one block of data. The line size is given as 4 Bytes, which means each line can hold 4 Bytes of data.

To calculate the size of the Tag field, we need to determine the number of bits required to uniquely identify each block in the main memory. Since the main memory size is given as 2GB and the block size is 4 Bytes, we can calculate the number of blocks in the main memory as follows:

Number of blocks = Main memory size / Block size
                = 2GB / 4 Bytes
                = 2 * 1024 * 1024 * 1024 Bytes / 4 Bytes
                = 536,870,912 Blocks

To uniquely identify each block, we need to use a unique tag for each block. The number of bits required to represent 536,870,912 blocks is given by:

Number of bits = log2(Number of blocks)
              = log2(536,870,912)
              ≈ 29.9 bits

Since we cannot have fractional bits, we need to round up to the nearest whole number. Therefore, the size of the Tag field in a Line Entry would be 30 bits.

Knowee AI · Accepted Answer

To determine the size of the Tag field in a Line Entry, we need to consider the organization used, which in this case is One Way Set Associative.

In a One Way Set Associative cache, each line in the cache can only store one block of data. The line size is given as 4 Bytes, which means each line can hold 4 Bytes of data.

To calculate the size of the Tag field, we need to determine the number of bits required to uniquely identify each block in the main memory. Since the main memory size is given as 2GB and the block size is 4 Bytes, we can calculate the number of blocks in the main memory as follows:

Number of blocks = Main memory size / Block size
                = 2GB / 4 Bytes
                = 2 * 1024 * 1024 * 1024 Bytes / 4 Bytes
                = 536,870,912 Blocks

To uniquely identify each block, we need to use a unique tag for each block. The number of bits required to represent 536,870,912 blocks is given by:

Number of bits = log2(Number of blocks)
              = log2(536,870,912)
              ≈ 29.9 bits

Since we cannot have fractional bits, we need to round up to the nearest whole number. Therefore, the size of the Tag field in a Line Entry would be 30 bits.

The main memory size is 2GBThe Block size in main memory is 4 BytesThe Cache memory size is 64KBytesThe Line size is 4 BytesOrganization used is One Way Set Associative.The size of the Tag field in a Line Entry is

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