In a simplified file system the following operations are implemented:   mkdir() – creates a directory creat() – creates an empty file open(), write(), close() – appends a block to a file link() – creates a hard link to a file unlink() – unlinks a file   The on-disk file system is represented by the following:   inode bitmap inode table (each inode surrounded by square brackets) data bitmap data blocks (each block surrounded by square brackets)   The inode structure contains the type (d – directory, f – regular file), a pointer to a single data block (a), and a reference count (r).  A directory data block contains (filename,inode number) pairs.   The current state of the file system is as follows:   inode bitmap  11111100inodes [d a:0 r:4][d a:1 r:2][f a:-1 r:1][f a:-1 r:2][d a:2 r:2][f a:-1 r:1][][]data bitmap   11100000data [(.,0) (..,0) (g,1) (q,2) (u,3) (x,3) (t,4)][(.,1) (..,0)(c,5)] [(.,4) (..,0)][][][][][]

In data journaling, a transaction consists of a begin block, blocks for all updates to the FS, and an end block.

You see a segment with the following set of writes block 100: [(“.” 0), (“..” 0), (“foo” 1)] block 101: [size=1,ptr=100,type=d] block 102: [size=0,ptr=-,type=f] block 103: [imap: 0->XXX,1->102]    What could be the contents of XXX in the above segment?

Hardware that meets the Popek Goldberg criteria can use the trap-and-emulate model for virtualization.

Assuming we start with no state in memory, and have a LFS with that looks like the following: how many disk I/Os do we need to perform to read the contents of the directory “/”?   block 0: Checkpoint region: imap[0..2] -> 3block 1: [(“.” 0), (“..” 0)] block 2: [size=1,ptr=1,type=d] block 3: [imap: 0->2]

Direct Memory Access (DMA) orchestrates the transfer of data between main memory and  devices without much CPU intervention.

In the FFS filesystem, the system attempts to spread files within the same directory across different groups.

Suppose that in a TCP communication that uses sequence numbers, the last message that the receiver process has seen has a sequence number of K. The next message that it receives can have a sequence number of K + 2

One way to solve the load imbalance problem in MQMS is work stealing.

Which four of the following are symptoms of abnormal uterine bleeding? (Select all that apply.)