Solutions to HW 5.
9.2 Discuss the hardware support required to support demand paging.
Answer: For every memory access operation, the page table needs to
be consulted to check whether the corresponding page is resident or
not and whether the program has read or write privileges for accessing
the page. These checks would have to be performed in hardware. A
TLB could serve as a cache and improve the performance of the lookup
operation.
9.3 What is the copy-on-write feature and under what circumstances is it
beneficial to use this feature? What is the hardware support required to
implement this feature?
Answer: Copy on Write allows processes to share pages rather
than each having a separate copy of the pages. However,
when one process tried to write to a shared page, then
a trap is generated and the OS makes a separate copy of the
page for each process. This is commonly used in a fork()
operation where the child is supposed to have a complete
copy of the parent address space. Rather than create
a separate copy, the OS allows the parent and child
to share the parent's pages. However, since each is supposed
to have its own private copy of the pages, the pages
are copied when one of them attemps a write.
The hardware support required to implement is simply the following:
on each memory access, the page table needs to be consulted to
checkwhether the page iswrite-protected. If it is indeedwrite-protected,
a trap would occur and the operating system could resolve the issue.
9.4 A certain computer provides its users with a virtual-memory space of
232 bytes. The computer has 218 bytes of physical memory. The virtual
memory is implemented by paging, and the page size is 4096 bytes.
A user process generates the virtual address 11123456. Explain how
the system establishes the corresponding physical location. Distinguish
between software and hardware operations.
Answer: The virtual address in binary form is
0001 0001 0001 0010 0011 0100 0101 0110
Since the page size is 2^12, the page table size is 2^20. Therefore the low
order 12 bits 0100 0101 0110 are used as the displacement into the page,
while the remaining 20 bits 0001 0001 0001 0010 0011 are used as the
displacement in the page table. The offset bits are then
concatenated to the resulting physical page number (from the page table),
to form the final address.
9.6 Assume that you are monitoring the rate at which the pointer in the
clock algorithm (which indicates the candidate page for replacement)
moves. What can you say about the system if you notice the following
behavior:
a. pointer is moving fast
b. pointer is moving slow
Answer: If the pointer is moving fast, then the program is accessing a
large number of pages simultaneously. It is most likely that during the
period between the point at which the bit corresponding to a page is
cleared and it is checked again, the page is accessed again and therefore
cannot be replaced. This results in more scanning of the pages before
a victim page is found. If the pointer is moving slow, then the virtual
memory system is finding candidate pages for replacement extremely
efficiently, indicating that many of the resident pages are not being accessed.
9.15 9.15 What is the cause of thrashing? How does the system detect thrashing?
Once it detects thrashing, what can the system do to eliminate this
problem?
Answer: Thrashing is caused by under allocation of the minimum number
of pages required by a process, forcing it to continuously page fault.
The system can detect thrashing by evaluating the level of CPU utilization
as compared to the level of multiprogramming. It can be eliminated
by reducing the level of multiprogramming.
