Solutions to sample questions

Short problems

Problem 1:

An OS uses the elevator algorithm to schedule the disk-arm. I/O requests are currently pending for blocks on tracks 1, 3, 8, 11, 15, and 16. The disk arm is currently at track 9 and moving upwards. In what order will these requests be handled?

Answer: 11, 15, 16, 8, 3, 1

Problem 2:

An OS uses a paging system with 1Kbyte pages. A given process uses a virtual address space of 128K and is assigned 16K of physical memory. How many entries does its page table contain?

Answer: 128

Problem 3:


Consider a file system that uses an FAT. The state of the directory is
 
File name | First block | Size in bytes 
---------------------------------------
  AA               5           1350
  BB               7           2200
(The other information in the directory is omitted.) The current state of the FAT is
 0  -- FREE
 1  -- FREE
 2 --  -1
 3 --   4
 4 --  -1
 5 --   2
 6 --  FREE
 7 --   3 
 8 --  FREE
The system uses a block size of 1 Kbyte. What is the last block of file BB? How many bytes of that block are used? (Useful fact: 1K=1024).

Answer: The blocks of BB are 7, 3, 4. Since BB has length 2200 and blocks 7 and 3 hold 2048 bytes, there are 152 bytes used in block 4.

Problem 4:


A particular system uses a page size of 1K bytes. A page table for a particular process begins as follows: [ 3, 4, *, 1, *, 8 ...] (Note: everything uses 0-based indexing.)

Answer:
A. 3*1024+50 = 3122.
B. Any virtual address between 2048 and 3071 or between 4096 and 5119.

Problem 5:

Consider a Web browser and Web server that are communicating in HTTP layered over TCP layered over IP. For each of the following functionalities, state whether it is (i) implemented at the level of HTTP (ii) implemented at the level of TCP, (iii) implemented at the level of IP; (iv) carried out by the DNS; or (v) none of the above.

Long Problems

Problem 6

What is swapping? Why is it used in systems with variable-length partitions? Why is it used in systems with paging?

Answer: In swapping, a process is entirely removed from main memory and its state is stored on disk. Swapping is used in an OS that uses variable-length partitions in the case there are active processes that are ready to run but do not fit into any free segment of main memory (either because the total memory requirements of the active processes is too large, or because space has been wasted in external fragmentation.)

In an OS that uses pages, swapping may be used when the sum of the working set sizes for the processes currently in memory is greater than the size of physical memory. When this happens, thrashing -- frequent page faults -- is almost inevitable, as the processes are competing for a resource (frames) that is inadequate to their combined needs. The only solution in such a case is to swap out (suspend) one or more of the processes.

Problem 7:

A user program has opened a file for reading and been reading sequentially through the program for some while. It now attempts to read the next byte, and finds that this byte belongs to a new block, and that this block is not in memory. Describe the major steps that are involved in the passage of control from the user program, which wants to read the byte, to the disk, which must carry out the read, and in the passage of information from the disk to the user program. In particular, your answer should discuss the operations of the device driver, the device-independent software, the actual disk, and the disk controller, in both directions. You may assume that the program is running on a single-process system. At any point, if there are different options for implementation, you can pick one to describe; you need not describe all the options.

Answer: (Note: this is a lot more detail than I would expect on an exam answer. Also, the exact steps and the division of the steps among components of the system can vary a lot from one system to another.)

The device-independent software passes to the device driver the number of bytes to be read, the address in user space to read them into, and the file pointer. It then traps to the kernel. The kernel blocks the process and calls the device driver from the top. The device driver determines that the next byte is on a block B not in memory. It finds the address of B from the i-node (presumably in memory). It issues a request to the device controller to read B into memory. The device controller translate the block number of B into a set of consecutive sectors on a track, and instructs the actual disk to read those sectors. The disk moves the arm to the correct track, waits for the disk to rotate to the proper sectors, and reads the information, putting the output into the disk controller's buffer. The disk controller issues a interrupt. The interrupt handler in the kernel sends control to the bottom of the disk driver. The disk driver transfers the block of data from the disk contoller's buffer to the file cache, and transfers the byte of data to the specified address in user space. The kernel unblocks the process and schedules it to run. The device-independent software finds its data where it expects, and returns control to the user process.

Problem 8

On CD-ROM, each file occupies a physically contiguous section of memory.

Answer:

Problem 9

A. Explain in 1 or 2 sentences the difference between non-preemptive and preemptive scheduling.
Answer: In a non-preemptive scheduler, a process stops running only when it blocks; it never transitions from "running" to "ready". In a preemptive scheduler, the scheduler may preempt a running process and put it on the ready queue,

B. Name one preemptive scheduling algorithm and one non-preemptive scheduling algorithm. (Just the names; you do not have to describe them.)
Answer: Of the scheduling algorithms we have discussed, shortest remaining time first, round robin, selfish round robin are preemptive; first come first served and shortest job first are non-preemptive.

C. Even if all jobs are CPU-bound, a preemptive scheduler may give a better average turnaround time than a non-preemptive scheduler. Give an example that illustrates this.
Answer: Suppose that process P, with one CPU burst of 100 sec, enters at time 0, and the process Q, with one CPU burst of 1 sec, enters at time 1. Then a non-preemptive scheduler will run first P from time 0-100 and then process Q from time 100-101 for an average turnaound time of 100. By contrast, assuming a small quantum and ignoring time lost to context switches, a preemptive scheduler will complete process Q at time 3 and process P at time 101, for an average turnaround time of 52.

Problem 10:

Consider a paging system that uses the "Not Recently Used" (NRU) page replacement algorithm.

Answer: The R bit is 1 if the page has been referenced since the last clock tick. The M bit is 1 if the page has been modified since it was read in. At each machine instruction, the R bit for any page of any virtual address referenced is set to 1. The M bit is set to 1 if the process writes to that page. The R bit is reset to 0 at each clock tick. The M bit is reset to 0 when the page is written out to disk. In the NRU algorithm, pages are divided into 4 categories:

A page is chosen for replacement from the lowest of these categories. If there is more than one page in that categories, one is chosen at random.