University Duisburg-EssenProf. Dr.-Ing. Axel Hunger SS06 Page 1 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) MULTIPLE-CHOICE QUESTIONS (MCQ) EXAMINATION DURATION : 90 MINUTES General Instructions • • This is a multiple-choice questions exam. Answer on the Multiple-Choice Solution Sheet provided. This is a closed book exam and no reference books and materials are allowed. No calculators and other electronic devices are allowed. Annotation to the Assignments and the Multiple-Choice Solution Sheet • A set of the assignment sheets, a page of the multiple-choice solution sheet and some working papers will be provided in an envelope to each student. Do not use your own paper. Only the multiple-choice solution sheet will be evaluated. Marked answers on the assignment sheet, solution approaches and workings will not be evaluated. No corrections on the markings on the multiple-choice solution sheet is allowed. In case of erroneous entries on the solution sheet, ask the examiner for a new multiple-choice solution sheet and give up the invalid copy. For each sub-task of an assignment, one or more answers may be correct. But : If the box "None of them" of one sub-task is shaded, the other shaded answers to this sub-task will be disregarded. • • • • A negative score is not attainable in any sub-task of an assignment. Hand in everything (assignment sheets, multiple-choice solution sheet and the working papers - used and unused) at the end of the examination in the envelope. Only examinations that are returned completely with a signature on the declaration page will be evaluated. Fill in your name and matriculation number on the assignment sheets and the multiplechoice solution sheets. • • • Name Matriculation No. Typ A the application programs. E: F: G: Directory organisation refers to various techniques for storing large amount of data H: I: J: K: The file system FAT16 is more reliable against bad sectors than FAT12 The file system depends on the technology of the hard disk The file system FAT32 uses the lower order 32 bits for cluster addressing None of them . The file system provides the mechanism for online storage of and access to both data and programs residing on the disks. the operating system. both programs and data. Axel Hunger SS06 Page 2 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Assignment 1 (20 Points) An operating system is an important part of almost every computer system. A computer system can be divided into four components: the hardware.University Duisburg-Essen Prof.1 A: B: C: D: Concerning operating systems and file systems. The former retain data even when the computer is switched off. Dr. 1. and the user. which statements are true? An operating system is a program that acts as an intermediary between the user of a computer and the computer hardware A soft real-time system guarantees that critical tasks be completed on time and are used for industrial control and robotics Time sharing and time-shared operating systems allow many users to use the computer system interactively at the same time Parallel systems have only one CPU with their individual computer bus and separated memory and peripheral devices A person whose job is to write programs for the level n virtual machine need not be aware of the underlying interpreters and translators Mass storage media is distinctive from main memory.-Ing. Most modern computers use disks as primary on-line storage medium for information. University Duisburg-Essen Prof. Standard Ethernet networks use CSMA/CD mechanism and enable network devices to detect a collision. The IBM Token-Ring specification has been standardized as the IEEE 802.5 standard. Token-passing networks move a small frame.2 L: Concerning CSMA/CD and Token Ring. is a set of rules determining how network devices respond when two devices attempt to use a data channel simultaneously. carrier sense multiple access/collision detection. Axel Hunger SS06 Page 3 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) CSMA/CD. called a token. the active monitor removes continuously circulating frames from the ring when its sending device fails If no station in a Token Ring wants to send data. 1. Token Ring is a type of computer network in which all the computers are arranged schematically in a circle.-Ing. the token is kept by the active monitor Token Ring is a deterministic media access method Ethernet networks are use in applications where delays must be predictable and where applications require a robust network operation None of them S: T: U: V: . which statements are true? CSMA/CD uses a token to announce that the common cable is now used M: CSMA/CD demands that a station has to wait a random time in case of collision before trying a new transmission N: CSMA/CD is a deterministic media access method O: CSMA/CD is a statistical media access method P: The CSMA/CD protocol is designed to provide fair access to the shared channel so that all the stations get a chance to use the network Q: In a Token Ring several stations can send a message simultaneously R: In a Token Ring network. Dr. When capitalized. Token Ring refers to the PC network architecture developed by IBM. around the network. University Duisburg-Essen Prof.2 G: J: What is the minimum transmitted frame length (bits)? 300 200 B: E: 500 400 C: F: 800 None of them How long (μs) has a station to wait before trying to access the channel again. 2.1 A: D: 2. 2. Axel Hunger SS06 Page 4 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Assignment 2 (9 Points) A local area network operates Ethernet with CSMA/CD mechanism and runs at a transmission rate of 8 Mbps and a one-way signal propagation time of 25 μs. after experiencing 5 successively collisions? 125 1 600 H: K: 625 3 200 I: L: 800 None of them Looking at a Token-Ring network with a Token-Holding-Time (THT) of 10ms and a transmission rate of 10 Mbps. Dr.-Ing.3 M: P: What is the longest frame size (bits)? 25 000 200 000 N: Q: 12 500 50 000 O: R: 100 000 None of them . if it would not be interrupted • • • • • The process control block is called pcb A quantum is a small unit of continuous CPU time (tQ).g. if the processes have a long runtime. if the processes have a short runtime. the scheduler changes to the next process (respectively to the method) A change of the current process (changing the pcb) takes tC = 1ms Changing the pcb is not part of a quantum The response time tR is the time between the input event for a process and its response There are 10 different processes running on a workstation.g. greater than 1s? tQ = 10ms F: tQ = 50ms G: tQ = 100ms H: tQ = 250ms .2 E: Which out of the following quantum times is the best value for small response times. Busy processes are scheduled with the Round-Robin timesharing method. After this quantum has elapsed. Axel Hunger SS06 Page 5 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Assignment 3 (16 Points) Definition: • Runtime is the time a process needs to solve a given task.University Duisburg-Essen Prof. 3.-Ing. less than 10ms? tQ = 15ms B: tQ = 40ms C: tQ = 45ms D: tQ = 50ms Which out of the following quantum times is the best value concerning CPU efficiency. e. Dr. Idle processes are waiting for an input event in the input queue.1 A: 3. e. University Duisburg-Essen Prof. Axel Hunger SS06 Page 6 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Consider the following set of processes.8ms M: None of these .-Ing. 3. and ignore the time for changing inter-processes. Dr. with the length of the CPU-burst time given in milliseconds: Process Burst Time Priority P1 10 3 P2 1 1 P3 2 3 P4 1 4 P5 5 2 Assume that P 1 is at the head of the ready queue and P5 is at the tail.2ms tav = 10.3 I: K: Using Round Robin scheduling algorithm (where the quantum q= 1ms) which is the average turnaround time? tav = 12ms tav = 7ms J: L: tav = 9. Dr. otherwise with “infinitive”. u). } } //LOOP While (set S is not Empty) { Select a node u from S with D[u] as the minimum. Quit. Initialize array D in the way.v). Axel Hunger SS06 Page 7 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Assignment 4 (20 Points) With the Dijkstra Algorithm the shortest path from a source node to any other node in the network can be calculated. } Delete u from set S.v). For each node v such that (u. If (D[u] is infinity { Error: no path exists to nodes in S. Pseudo-code for the Dijkstra Algorithm //INITIALISATION S = {all nodes except source node A}. otherwise with Null (empty). R[v] contains all nodes which lay on the shortest route from source to node v. that D[v] is the weight of the edge/link from the source to v if the edge/link exists. In D[v] the current distance from source node to the node v is stored. R[v] is initialized with the source node. if the network topology is known. if edge/link from source to v exists. R[v] = 0. } Else { D[v] = infinity. if (c < D[v]) { R[v] = merge elements (R[u]. D[v] = c. } } } } .University Duisburg-Essen Prof. The list and the arrays are initialized as follows: Initialize list S to include all nodes except the source node. The method starts with a list S of all nodes (except the source node) and calculates the arrays D[v] and R[v].v) is an edge { if ( v is still in S ) { c = D[u] + c(u.-Ing. R[v] = {A}. For all nodes v { If (v adjacent to A) { D[v] = C(A. 1 A: D: 4. Use Dijkstra’s Algorithm to define the shortest path from a to each node.3 9 What is the shortest distance from node a to node d ? N: Q: 7 10 O: 8 R: None of these M: 5 P: 4.4 S: V: 9 Using the Dijkstra algorithm. b 1 c 10 a 2 3 9 4 6 5 2 e 7 d Figure 4.University Duisburg-Essen Prof.2 What is the shortest distance from node a to node b ? 5 9 B: E: 7 10 C: 8 F: None of these What is the shortest distance from node a to node c ? H: K: 7 10 I: L: 8 None of these G: 5 J: 4. how many routes were replaced with routes of shorter distance? (Neglecting replacement of infinity values) 1 4 T: 2 U: 3 X: None of these W: 5 . Dr.-Ing. Axel Hunger SS06 Page 8 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Given is the following network topology.1: Network topology 4. E: F: G: Best Fit algorithm allocates H6. H0. H6. H: I: Worst Fit algorithm allocates H2. H3 for the mentioned request. H3. H7 for the mentioned request. Best Fit algorithm allocates H2. H6 for the mentioned request.-Ing. H3. H7 for the mentioned request.University Duisburg-Essen Prof. None of them . Worst Fit algorithm allocates H2. H6. Consider a swapping system in which memory consists of the following hole sizes in memory order: H0 H1 H2 H3 H4 H5 H6 H7 10K 4KB 20KB 18KB 7KB 9KB 12KB 15KB and a successive segment request of a) 12 KB b) 10KB c) 9KB Which of the following sentences is/are true? A: B: C: D: First Fit algorithm allocates H2. First Fit algorithm allocates H2. H5 for the mentioned request. Dr. H3 for the mentioned request. H5 for the mentioned request. Next Fit algorithm allocates H2.1. H0. H3. H7 for the mentioned request. H0. Next Fit algorithm allocates H3. Axel Hunger SS06 Page 9 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) Assignment 5 (15 Points) 5. A computer with a 16 bit address.University Duisburg-Essen Prof. 12292 MOV REG. 5000 MOV REG. Dr. 16388 MOV REG. Page table 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 000 000 000 000 111 000 101 000 000 000 011 100 000 110 001 010 0 0 0 0 1 0 1 0 0 0 1 1 1 1 1 1 Present bit Assuming that at a given time the running process issues the following instruction: MOV REG. Axel Hunger SS06 Page 10 SAMPLE EXAMINATION PAPER Operating Systems and Computer Networks (ISE) Data Processing 2 (AOS & E-Technik) 5. a 4KB page size.-Ing. has 64K of VA and 32K of PA. 12292 Which will be the outcome of the Memory Management Unit mapping? J: K: L: M: N: MOV REG. 4 None of them .2. The page table for this computer is given in the following figure.