Question

1. (10 pts) What is the largest file size if the inode is as follows: size of the inode is 128 Bytes out of which 64 Bytes are used for layout information, . each pointer is of size 8 Bytes and the file system block size is 4 KBytes, the inode has 2 direct pointers, 2 single indirect pointers, 2 double indirect pointers, and 2 triple indirect pointers 2. (10 pts) Find the average seek time (under the assumptions we made for a similar problem in HW4) for the following access sequence: 15, 82, 33, 156, 199, 0, 35, 77. Assume the SCAN (or Elevator) scheduling policy. Assume the disk arm is on track 50 to begin with and will commence servicing these requests by moving towards track 199 (there are 200 tracks numbered 0,, 199)

Answer 1

1)

Size of a pointer = 8 bytes

Size of a block = 4 KBytes

Number of pointers in 1 block is

4 K Bytes 4x 1024 512 8 bytes

Data accessed through 2 direct pointers

2×4KB=8KB

Data accessed through 2 single indirect pointers

2 × 512 × 4 KB-4096 KB

Data accessed through 2 double indirect pointers

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2 × 512 × 512 × 4KB-2097152 KB

Data accessed through 2 triple indirect pointers

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2 × 512 × 512 × 512 × 4KB-107374 1824KB

So total file size is

8 + 4096 + 2097152 + 1073741824 = 1075843080 KB = 1050628 MB = 1026 GB = 1 TB

So the largest file size is approx 1 TB.

2)

Total Seeks

|50-77| +|77-82| + |82-156| + |156-199| + |199-35| + |35-33| + |33-15| + |15-0|

27 + 5 + 74 + 43 + 164 + 2 + 18 + 15 = 348

Average seek time is

348 - 43.5
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