Final answer:
The bit vector required for a 2 TB disk with 512-byte blocks is approximately 4096 MB. This calculation was done by converting the total disk capacity to bytes, determining the number of disk blocks, and then converting the bit vector size to megabytes.
Step-by-step explanation:
The question asks to calculate the size of a bit vector used for free space management on a disk with a capacity of 2 TB, where each disk block is 512 bytes. A bit vector is a data structure commonly used in computing to track the allocation status of disk blocks, where each bit represents the status of a single block - allocated (1) or free (0).
First, let's convert the capacity of the disk from terabytes to bytes. Since 1 terabyte is equal to 1024 gigabytes, 1 gigabyte is 1024 megabytes, and 1 megabyte is 1024 kilobytes, and finally, 1 kilobyte is 1024 bytes, we get:
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- 1 TB = 1024 GB
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- 1 GB = 1024 MB
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- 1 MB = 1024 KB
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- 1 KB = 1024 bytes
Therefore, 2 TB is:
2 TB = 2 * 1024 GB * 1024 MB * 1024 KB * 1024 bytes = 2,199,023,255,552 bytes.
Now, to calculate the number of blocks on the disk:
Number of blocks = Total bytes / Block size = 2,199,023,255,552 bytes / 512 bytes = 4,294,967,296 blocks.
To find the size of the bit vector, we need one bit per block, so the size in bytes is the same as the number of blocks:
Bit vector size in bytes = Number of blocks = 4,294,967,296 bytes.
Finally, converting the bit vector size to megabytes:
Bit vector size in MB = Bit vector size in bytes / (1024 bytes/KB) / (1024 KB/MB) = 4,294,967,296 bytes / 1024 / 1024 ≈ 4096 MB.