CS 450 Fall xxxx Final exam solutions. 2) a. Multiprogramming is allowing the computer to run several programs simultaneously.

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1 CS 450 Fall xxxx Final exam solutions 1) 1-The Operating System as an Extended Machine the function of the operating system is to present the user with the equivalent of an extended machine or virtual machine that is easier to program than the underlying hardware 2-The Operating System as a Resource Manager In the alternative view, the job of the operating system is to provide for an orderly and controlled allocation of the processors, memories, and I/O devices among the various programs competing for them. 2) Multiprogramming is allowing the computer to run several programs simultaneously. Kernel code can perform privileged instructions, whereas user code cannot. A system call passes control into the kernel to perform a functionality. A library call remains in user-space. d. a,c,d 3)

2 To better modularize and be able to move components out of the kernel. This raises reliability and security, at a cost of performance. 4) 1. System initialization. 2. Execution of a process creation system call by a running process. 3. A user request to create a new process. 4. Initiation of a batch jo 1.process blocks for input 2.Scheduler picks another process 3.Scheduler picks this process 4.Input becomes available 5) d, a, g, c, b, e, f, h, i User-level threads, pro: Fast context switches con: A single thread can monopolize the time-slice thus starving the other threads within the userspace process Kernel-level threads, pro: Threads can block on I/O without making all the other threads in the process block. con:

3 Slow context-switches 6) The lines beginning with * are sufficient first fit: *10 -> a 10 becomes 4, 6. *6 -> b *20 -> c best fit: *4 -> a 7 becomes 2, 5 *2 -> b *12 -> c next fit: *10 -> a 10 becomes 4, 6 *4-> b 4 becomes 2, 2 *c -> becomes 12, 8 worst fit: *20 -> a 20 becomes 4, 16 *18 -> b 18 becomes 2, 16 *16 -> c 16 becomes 12, 4 7) This means that each page has 8192 bytes. To specify any byte out of these, the off-set part of the address must be 13 bits long. Enough entries are needed to cover the remaining 35 bits of virtual address. Therefore, the number of entries needed for the page table is 2^19 Each page has 2^14 bytes. Memory has 2^28 bytes. There are 2^(28-14) = 2^14 frames. Each frame has a 14-bit address. Therefore, each entry in the page entry table will need at least 14 bits, or two bytes. We have 2^(32-14) = 2^18 pages. Page table size is thus 2 * 2^18 = 2^19 bytes = 0.5 MB N, G, H, D Choose(D)

4 8) A set of processes is deadlocked if each process in the set is waiting for an event that only another process in the set can cause. 1. Mutual exclusion condition. Each resource is either currently assigned to exactly one process or is available. 2. Hold and wait condition. Processes currently holding resources that were granted earlier can request new resources. 3. No preemption condition. Resources previously granted cannot be forcibly taken away from a process. They must be explicitly released by the process holding them. 4. Circular wait condition. There must be a circular chain of two or more processes, each of which is waiting for a resource held by the next member of the chain. 9) DMA allows I/O devices' controllers to be able to directly access memory so that the CPU need not do it instruction by instruction. 1.Sequential 2.Random 3.Tree Contiguous allocation + Read operation is fast as the file can be read at once. - Disk becomes fragmented, lowering performance drastically. Linked List Allocation: +No fragmentation. -Random access is extremely slow. 10) 1) Sharing code segments across processes 2) More dynamically growing regions.

5 3) Performance can be improved. Similarity: In memory we divide the memory into pages, whereas for the disk we divide it into blocks. Difference: It is possible to cache memory to make some accesses faster. However, there is no caching layer between memory and hard-disk. User processes device-independent software device drivers interrupt handlers hardware 11) 1- Uniform-Interfacing: I/O devices are named to maintain uniform-interfacing. The device-independent software takes care of mapping symbolic device names onto the proper driver. For example, in UNIX and MINIX 3 a device name, such as /dev/disk0, uniquely specifies the i-node for a special file, and this i-node contains the major device number, which is used to locate the appropriate driver. The i-node also contains the minor device number, which is passed as a parameter to the driver in order to specify the unit to be read or written. All devices have major and minor numbers, and all drivers are accessed by using the major device number to select the driver. 2- Buffering: Users can write data to the system faster than it can be output, necessitating buffering. Keyboard input that arrives before it is needed also requires buffering. 3- Error reporting: A typical error is caused by a disk block that has been damaged and cannot be read any more 4- Allocating and Releasing Dedicated Devices: Some devices, such as CD-ROM recorders, can be used only by a single process at any given moment. 5- Device Independent Block Size: Similarly, some character devices deliver their data one byte at a time (e.g., modems), while others deliver theirs in larger units (e.g., network interfaces). These differences may be hidden.