4330/6310Fundamentals COSC of Operating Systems
SOLUTIONS TO THE PROBLEMS DISCUSSED IN THE THIRD REVIEW SESSION
Please try first to answer the questions yourself using the handout without the solutions.
1. True or false:
a) A process that acquires all its resources at the same time will never be involved in a deadlock
True. (The hold-and-wait condition is denied.)
b) We cannot prevent deadlocks in client/server systems.
True. (We cannot prevent deadlocks and allow messages to flow from client to server and
back.)
c) External fragmentation is often the result of using very large page sizes.
False. (Larger page sizes increase internal fragmentation but it rarely is an issue in virtual
memory systems.)
d) Global page replacement policies cannot handle real-time processes.
True. (Real-time processes must keep all their pages in memory all the time, which can only
be achieved by giving them a private partition containing all the pages they will ever need.)
e) UNIX uses file-specific passwords to control access to files.
False. (UNIX uses the file access control list to check which file access rights can be
granted to a user process.)
f) UNIX stores the name of a file in its i-node.
False. (See next answer.)
g) A UNIX file can have several names.
True. (It can have as many names as there are directory entries pointing to it.)
h) UNIX special file are not files.
True. (They are physical devicesor POSIX named semaphores.)
2. When comparing the hit ratios of two translation look-aside buffers, which question should we ask
first?
Are TLB misses handled by the firmware or the OS? If TLB misses are handled by the
firmware, the cost of a TLB miss is one extra memory reference. If TLB misses are handled by
the OS, the cost of a TLB miss is two context switches.
3. What is the purpose of the dirty bit?
The dirty bit tells whether a page has been modified since the last time it was brought into
main memory. It is used whenever a page must be expelled from main memory. If its dirty bit
is ON, the page must be saved on disk (swap area) before being removed from main memory. If
its dirty bit is OFF, there already is an exact copy of the page on disk.
1
4. How does the BSD UNIX page replacement policy simulate a missing page-referenced bit? What is
the main drawback of the technique?
Whenever the hand of the clock encounters a page that has been recently used, it marks it
INVALID instead of clearing its page-referenced bit. The main drawback of this approach is
that an interrupt will occur the next time that page will be accessed. The cost of handling this
interrupt will be roughly equal to two context switches.
5. Among the following page replacement policies:
Local FIFO, LOCAL LRU, Global FIFO, Global LRU, Original Clock, Berkeley Clock, and
Windows, which ones:
a) Are the poorest performers? Local FIFO and Global FIFO
b) Cannot be implemented at a reasonable cost? Local LRU and Global LRU
c) Provide a decent performance and do not require any special hard ware support? Berkeley
Clock and Windows (Mach would have qualified but was not in the list)
d) Support real-time processes? Windows (older exams mention VMS)
6. A virtual memory system has a virtual address space of 4 Gigabytes and a page size of 4 Kilobytes.
Each page table entry occupies 4 bytes.
a) How many bits are used for the byte offset?
Since 4K =2_12_, the byte offset will use _12_ bits.
b) How many bits are used for page the number?
Since 4G = 2_32_ we will have _32_-bit virtual addresses. Since the byte offset occupies
_12_ of these _32_ bits, _20_ bits are left for the page number.
c) What is the maximum number of page table entries in a page table?
Address space/ Page size = 2_32_ / 2_12_ = 2 _20__ PTEs.
d) What is the maximum size of a page table?
Since each PTE occupies 4 bytes, the answer is4 x 2_20_= 2_22_ bytes = _4_ MB.
7. A 32-bit UNIX file system has 16 kilobyte pages and i-nodes with 15 block addresses. How many
file bytes can be accessed
a) Directly from the i-node?
The _12_ first blocks that is _12 x 16 KB =_192 KB_.
b) With one level of indirection?
An indirect block can store _16KB_ / 4B =_4K_ block addresses. Hence we can access
the next__4K x 16K = 64_MB of the file.
c) With two levels of indirections?
With two levels of indirection we should be able to access
(2_14_ B / 4B)2 = 2_24_ blocks. Keeping in mind that the maximum file size is only __4_
GB, we will be able to access __4 GB 64 MB 192 KB_.
d) With three levels of indirections?
_None___
2
8. Redo the same problem at home with a page size of 8KB.
The answers are (a) 96 KB, (b) 16 MB, (c) 4 GB 16 MB 96 KB and (d) none.
9. Internal fragmentation is not a big problem in virtual memory systems. Is it always so in file
systems? Give a hypothetical example to support your argument.
Internal fragmentation can be a big problem whenever the disk partition contains many files
than are much smaller than the block size. Consider for instance a partition where many files
are smaller than 1 KB and the block size is 4KB. More than 75 percent of the space used by
these files will be wasted.
10. Alice, Bob and Carol want to share a few files on a UNIX system while preventing other users to
access these files. What should they do?
They should ask their system administrator to create a group abc for the three of them.
11. Give examples of tickets and access controls lists in the UNIX file system.
The file descriptor of an opened file acts as a ticket.
The protection bits of a file constitute its access control list.
12. Give examples of what could happen if metadata updates are written to disk in the wrong order. How
does the Fast File System avoid that? What is the main drawback of this approach?
Whenever we create a new file, the block containing the new i-node must be written to disk
before the block containing the directory entry pointing to that i-node. Otherwise, a crash
happening after the directory block has been written to disk but before the block containing
the new i-node has been written to disk would leave the file system in a state where a directory
entry would point to an i-node that does not exist.
The Fast File System avoids that by requiring all I/O operations involving metadata updates
to be blocking. While this solution ensures that the updates will always reach the disk in the
right order, it dramatically decreases the throughput of the file system.
13. What are the respective advantages of journaling file systems using (a) synchronous and (b)
asynchronous log updates?
Journaling file systems using synchronous log updates guarantee that no metadata data update
will be lost when the system crashes.
Journaling file systems using asynchronous log updates only guarantee that the file system will
recover to a recent consistent state but are much faster (because they do much less physical
I/Os).
3