CMSC 412 Midterm #2 (Spring 2003)

1.) (20 points) Define and explain the following terms:

a) Inverted Page Table

A page table organized by physical page frames rather than by virtual address. This reduces the size of the pages table to one entry per physical page frame. It often uses a hash table to provide virtual address to physical address translation.

b) Virtual Address

An address that is translated via a page table to a physical address. This allows all processes to think their memory starts a virtual address 0.

c) Rotational Latency

The time required for a disk drive to wait until the head spins under the desired sector (once the disk is at the required track).

d) Thrashing

A severe shortage of memory caused by not being able to fit the working sets of all active processes into memory at once. This results in high I/O (paging) rates and low CPU utilization.

2.) (20 points) Memory Systems: Consider a system with an average memory access time of 100 nano-seconds, a three level page table (meta-directory, directory, and page table). For full credit, your answer must be a single number and not a formula.

a) If the system had an average page fault rate of 0.01% for any page accessed (data or page table related), and an average page fault took 1msec to service, what is the effective memory access time (assume no TLB or memory cache).

EAT = 4 access to memory (3 page related + memory) X time for one access

= 4 (100 ns + 1/10,000(1,000,000 ns) = 4 x 200ns = 800ns

b) Now assume the system has no page faults, we are considering adding a TLB that will take 1 nano-second to lookup an address translation. What hit rate (to the nearest 5%) in the TLB is required to reduce the effective access time to memory by a factor of 2.5?

Need access time of 2/5 * 400 = 160 ns

Access time = 100 X + (1-x) 400 – 100 ns for hit (read memory), 400ns for miss

160 = 100 x + 400 – 400x à x = .8 à 80% TLB hit rate

3.) (20 Points) Synchronization: Given a multi-processor system that supports a single global lock via: Lock(). Unlock(), and the ability to suspend a process with a Wait(id) call, and the ability to wakeup all processes waiting on an id, with Wakup(id). Show how to implement counting semaphores.

typedef struct {

int count;

} Semaphore;

P(Semaphore *sem)

While (1) {

Lock();

If (sem->count > 0)

Sem->count--;

Unlock();

Return;

Else

Unlock();

Wait();

}

V(Semaphore *sem)

Lock();

Sem->count++;

If (sem->count == 1)

Wakeup(sem)

Unlock();

4.) (15 points) File Abstraction

a) (10 points) Explain why the write and close system calls may not result in data immediately being written to disk.

Files are often kept in a file buffer cache even after a close to allow disks writes to be optimized (i.e. process many requests in a single sweep across the drive).

b) (5 points) Does the seek system call result in a disk head moving (explain your answer)?

No, the seek system call simply moves the pointer in the open file data structure, it is a read or write than can result in the head being seeked.

5.) (10 points) Explain how you were able to use the accessed bit in the IA-32 (aka x86) processor to simulate LRU page replacement in the project. How is this a more accurate approximation than a simple second chance algorithm?

By dividing the pages into epochs. Each page fault (or other clock interval), we examine and clear the accessed bit. For those pages whose accessed bit was set, we change their clock value to the current clock. Second chance only has two epochs (accessed and not).

6.) (15 points) Filesystems

a) (5 points) List two issues that must be addressed to handle DAG directories (vs. simple tree directories)

Backups needs to be carefully written to not backup files with multiple names twice.

Need to maintain reference counts in the inodes to know when the underlying file should be deleted.

b) (5 points) What are the 12 protection bits used in a UNIX filesystem?

Read, write, and execute for each of user, group, and other/world

Setuid, set group id, sticky

c) (5 points) How can an access control list be used to represent negative rights to a file or directory?

A field in each entry in the access list could indicate that this is a negative right. As part of the checking of access rights, after checking if an access is allowed, check the negative access to see if the specific user should be disallowed.