MULTIPLE-CHOICE QUESTIONS (MCQ) EXAMINATION DURATION: 105 MINUTES. Annotations to the assignments and the solution sheet. Note the following points

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1 Special Exercise for Operating Systems and Computer Networks Page: 1 MULTIPLE-CHOICE QUESTIONS (MCQ) EXAMINATION DURATION: 105 MINUTES Annotations to the assignments and the solution sheet This is a multiple choice examination, that means: Solution approaches are not assessed. For each subtask there is only one correct solution. For each subtask you are allowed to mark one box only. If two or more boxes for a subtask are marked, no points will be given for this subtask. Note the following points In addition to the assignment sheets there is a solution sheet. Mark the appropriate box(es) on the solution sheet! Any answer on the ASSIGNMENT SHEETS will NOT be considered. It is not possible to correct markings in the solution sheet. In case of errors request a new solution sheet. Any earlier solution sheet is substituted only as immediate exchange old against new. The previous erroneous or invalid sheet will be destroyed by the invigilating staff. In case of submission of multiple sheets neither will be evaluated. Any substitution of solution sheets is only possible up to 5 minutes before the official end of the examination time. Only use the sheets enclosed in the envelope or otherwise provided by the supervisors. Don't use any other paper. If you need more paper ask the supervisors. Return everything, i.e. assignment sheets, solution sheet and any additional sheets - used and unused. Only exams that are returned completely will be assessed. FILL-IN YOUR NAME AND MATRICULATION NUMBER ON THE ASSIGNMENT SHEET AND THE SOLUTION SHEET!

2 Special Exercise for Operating Systems and Computer Networks Page: 2 Assignment 1 (6 Points) 1.1 Concerning operating systems, which of the following statements is true? A computer system can only have one operating system. The largest advantage of an open layered model is that when application developers change an aspect in a layer, all layers of the model s specification changes with it. An operating system controls the hardware and gives a predefined set of functions to the applications. In lower level programming the software must be compatible to the top level operating system to function correctly 1.2 Concerning computer networks, which of the following statements is true? In non-persistent CSMA, when a collision occurred, the transmitting device waits for a random period of time before listening to the channel again to determine if the media is free. In p-persistent CSMA, when a collision occurred, the transmitting device waits for p time slots before listening to the channel again to determine if the media is free. The Data Link Layer in ISO/OSI Model provides a logical addressing for paths within the network. In Selective Repeat Sliding Window protocol, the sender re-transmits the unacknowledged frames only, if an erroneous frame arrived at the receiver.

3 Special Exercise for Operating Systems and Computer Networks Page: 3 Assignment 2 (16 Points) Given is a magnetic Hard disk with 6 heads, 3 platters, cylinders, 250 sectors/track on the innermost cylinders, 450 on the outermost cylinders and a sector size of 1 KiB. 2.1 Which of the following equals the total hard disk capacity of the before mentioned disk (Rounded to the second decimal place) GiB GiB GiB GiB GiB If you couldn t calculate the disk capacity in 2.1, use SC = 205 GiB for the following question. 2.2 What is the minimal cluster size that is chosen, if you want to format the hard disk from assignment 1.1 with the standard formatting using FAT32? 1 KiB 2 KiB 4 KiB 8 KiB 16 KiB 2.3 The Disk from Question 2.2 is now organized in Blocks with a length of 1KiB. What is the length of the address necessary to index all blocks of the hard drive 20 Bits 28 Bits 33 Bits 37 Bits 47 Bits

4 Special Exercise for Operating Systems and Computer Networks Page: 4 Consider the magnetic platter of a HDD shown in figure 2-1. Technical details: Fig. 2-1 Magnetic plattern It takes the head 2 ms to change its position by one track The average reposition time of the disk to the correct sector on one track is 2ms. This operation can be done parallel to the change of the track. One sector has the same size as one cluster and can be read and interpreted in 9.7ms In case of sequential files, no time delay through repositioning occurs. All other time consuming operations can be ignored. 2.4 A read-request for the data fragmented to the parts P1-P7 arrives at the hard disk at t=0s. The header is positioned on the sector with the index 2. How long after the request arrived will the data be read? 83,9 ms 85,9 ms 87,9 ms 89,9 ms 91,9 ms

5 Special Exercise for Operating Systems and Computer Networks Page: 5 Assignment 3 (13 Points) A computer with 15 bit address has virtual address space of 32 KiB and physical memory of 16 KiB. The size of a page is 2 KiB. Table 3-1 depicts the page table belonging to the virtual memory of the computer. Table 3-1 Page table 3.1 Which of the following instructions is the resulting instruction for the physical memory block if the virtual memory instruction MOVE REG is executed by a program in the computer under consideration of the page table given in Table 3-1? MOVE REG 6 MOVE REG 2054 MOVE REG 4102 MOVE REG 8198 MOVE REG MOVE REG Which of the following instructions is the original instruction for the virtual memory block if the physical memory instruction MOVE REG is executed by a program in the computer under consideration of the page table given in Table 3-1? MOVE REG 9 MOVE REG 2057 MOVE REG MOVE REG MOVE REG MOVE REG 38921

6 Special Exercise for Operating Systems and Computer Networks Page: 6 Assignment 4 (24 Points) Consider the following set of processes, with the length of CPU-burst time given in milliseconds. The processing begins at tstart = 0ms. The arrival time describes the time after tstart when a process arrives at the CPU (in milliseconds). All time delays due to context change and memory access can be ignored. The waiting time and turnaround times are determined from the point on when a process arrives at the CPU! Process Burst Time (ms) Arrival time (ms) Priority Priority Description P Highest P Medium P Highest P High P Low P Highest Table 4-1 CPU Burst-time and Process Priority Note: Assume that P1 is at the head of the ready queue and P6 is at the tail. P1 to P3 arrive at the same time. 4.1 Using non-preemptive shortest job first scheduling algorithm, determine the average turnaround time for processes described in Table 4-1! (Rounded to the second decimal place) tav,wait = 8.33ms tav,wait = 11.66ms tav,wait = 12.0ms tav,wait = 12.5ms tav,wait = 14.17ms 4.2 Using non-preemptive priority scheduling algorithm, determine the average waiting time for processes described in 4-1! In case of equal priorities, Round Robin scheduling algorithm with a quantum time of 1ms is to be applied. (Rounded to the second decimal place) tav,wait = 7.00ms tav,wait = 7.83ms tav,wait = 8.17ms tav,wait = 12.00ms tav,wait = 12.33ms

7 Special Exercise for Operating Systems and Computer Networks Page: 7 Multiprogramming is used to improve CPU utilization. Looking at it from a probabilistic point of view, it is assumed that a process spends a fraction p of its time waiting for I/O to complete. With n processes in the memory at the same time, the CPU utilization for all processes can be calculated by the probability of a busy CPU in dependence of the number of active processes, shown in Table 4-2. CPU idle p n CPU busy 1 - p n Table 4-2 Probabilities of Idle and Busy CPU Four processes are received by one CPU in a computer center as shown in table 4-3. The computer center was idle prior to the arrival of process 1 and transition times can be ignored. Process CPU mins Arrival Time (min) (hh:mm:ss) :00:00 2 0,5 06:01: :03: :04:00 Table 4-3 Job arrival and execution time Each process from table 4-3 needs 50% of its time for memory access (p=0.5) Using the information given in Table 4-2 and Table 4-3, calculate the following: 4.3 Which job from Table 4-3 is first to finish? Process 1 Process 2 Process 3 Process 4 None of them 4.4 In which time frame is the last (unfinished) Process from Table 4-3 finished? 06:07:00 06:09:00 06:09:00 06:11:00 06:11:00 06:13:00 06:13:00 06:15:00 None of them

8 Special Exercise for Operating Systems and Computer Networks Page: 8 Assignment 5 (20 Points) Two stations: A (sender) and B (receiver) are connected via a point-to-point link to exchange messages. There are 10 frames ready for buffering at Station A and the starting time for every algorithm is at 0 ms. Fig. 5-1 Network transmission depiction Consider the following statements for this assignment: The transmission and process times between the two stations and two processes are: tsend = tanswer = tprocess = 10ms The timeout-time for an unacknowledged frame is: ttimeout = 25ms The window sizes for the Selective Repeat and Go-back-n algorithms are defined with n = 4. A retransmission of a frame has a higher priority than the transmission of the next frame in the window (for Selective Repeat algorithm). Station B knows as soon as a frame arrives if an error occurred or not, without a delay in time. An ACK or NAK is processed instantaneously at Station A and B meaning that as soon as the information arrives it can be used. If the ACK- or NAK-frame cannot be interpreted, the frame connected to the response will be timed out.

9 Special Exercise for Operating Systems and Computer Networks Page: Considering an error free transmission, which of the following statements is true? The ACK for Frame 4 using Go-back-n algorithm arrives 40ms after starting time. The ACK for Frame 4 using Go-back-n algorithm arrives 50ms after starting time. The ACK for Frame 4 using Go-back-n algorithm arrives 60ms after starting time. The ACK for Frame 4 using Go-back-n algorithm arrives 70ms after starting time. The ACK for Frame 4 using Go-back-n algorithm arrives 80ms after starting time. Now consider the following error scenario for the assignments 5.2 to 5.4: 25ms after the beginning of the transmission an error occurs, that is repeated every 40ms thereafter (At 25ms, 65ms, 105ms, etc.). This error affects the frames that are currently transmitted, making the message, that they carry unreadable. 5.2 How many ms after the start of the algorithm will the ACK for frame 4 arrive at Station A using the Stop-and-wait algorithm? 80 ms 100 ms 120 ms 140 ms 160 ms 5.3 How many ms after the start of the algorithm will the ACK for frame 4 arrive at Station A using the Go-back-n algorithm? 50 ms 80 ms 110 ms 140 ms 170 ms 5.4 How many ms after the start of the algorithm will the ACK for frame 5 arrive at Station A using the Selective Repeat algorithm? 50 ms 80 ms 110 ms 140 ms 170 ms

10 Special Exercise for Operating Systems and Computer Networks Page: 10 Assignment 6 (16 Points) For network transmission, the following network frame structure depicted in Fig. 7-1 is used. The payload has a variable length of n bytes. The length (BL) of the whole network frame (block) cannot be larger than BL, max = 8 KiB. SOT SOH 6 Charachters EOH DLE STX n Byte DLE ETX BCC BCC EOT Header Payload Trailer Fig. 7-1 Network Frame Structure The header and trailer control bytes are: SOT Start of Transmission, indicates the start of the frame SOH Start of Header 6 Bytes Header with the total length of 6 Bytes EOH End of Header STX Start of Text, indicates the beginning of the payload transmission ETX End of Text, indicates the end of the payload transmission BCC Block Check, used for error detection mechanism EOT End of Transmission, indicates the end of the frame The transmission rate (bandwith) TR is Mbit/s. For this assignment the average number of additional ESC-Bytes appearing in the payload is 0 Byte/s. 6.1 Calculate the code efficiency for the case of an error-free transmission for a fixed size of 128 Byte payload (Rounded to the second decimal place). ηcode= % ηcode= % ηcode= % ηcode= % ηcode= % 6.2 Calculate the transmission rate of pure payload with an average number of additional ESC sequences within the payload is 10 3 Byte/s. Each additional ESC sequence in the payload has a total size of 2 Bytes, the payload has a fixed size of 128 Bytes and the transmission rate (bandwith) TR is Mbit/s. TR,eff = Mbit s TR,eff = Mbit s TR,eff = Mbit s TR,eff = Mbit s TR,eff = Mbit s

11 Special Exercise for Operating Systems and Computer Networks Page: 11 Now consider an erroneous transmission. The error bit rates are given as follows: Single Errors: q1 = 10-4 per Byte. Double Errors: q2 = 10-5 per Byte. Triple Errors: q3 = 10-6 per Byte. Higher error rates are not considered. 6.3 Calculate the optimal block length that does not violate the rule for BL,Max, given in this assignment, if the error correction mechanism is able to recognize all errors and correct single and double errors (Results are to be rounded to the next full Byte [ < 0]). 16 Byte B L,Opt = Byte Byte B L,Opt = Byte B L,Opt = Byte 2 B L,Opt = Byte 2 B L,Opt = 16 Byte Byte 126 Byte2 B L,Opt = B L,Opt = Byte 2 B L,Opt = Byte 2 B L,Opt = Byte 2 Hint: Optimal Block length is the block length that results in a maximum payload transfer rate (given below) under the errorneous transmission. net tare preamble payload trailer gross Payload transfer rate

12 Special Exercise for Operating Systems and Computer Networks Page: 12 Assignment 7 10 Points Given is a set of payload with a total length of 1 MiB. This data needs to be transmitted via asynchronous communication to its destination. The transmission rate of the connection is 30 Mbps. The following attributes need to be implemented in the preamble: 3 Synchronization Bytes to inform the receiver of the transmission clock The indicators, when the header begins and ends Header information of a total length of 7 Bytes 1 Byte that indicates when the payload (Text) begins The following attributes need to be implemented in the trailer: 1 Byte that indicates when the payload (Text) ends 1 Cyclic Redundancy Check -Byte for each 1KiB of payload The indicator when the transmission ends 7.1 Which of the following examples represent the demanded preamble? SYN SYN DLE SOH 7 Bytes EOH DLE STX Preamble SOH SYN SYN SYN 7 Bytes EOH STX Preamble SOH SYN SYN SYN 7 Bytes STX EOH Preamble SYN SYN SYN SOH 7 Bytes EOH STX Preamble SYN SYN SYN STX SOH 7 Bytes EOH Preamble

13 Special Exercise for Operating Systems and Computer Networks Page: Which of the following examples represent the demanded trailer? EOT CRC #1 CRC #2 CRC #1024 ETX Trailer CRC #1 CRC #2 CRC #3 CRC #1024 ETX EOT Trailer ETX CRC #1 CRC #2 CRC #1024 EOT Trailer ETX CRC #1 CRC #2 CRC #1000 EOT Trailer ETX CRC #1 CRC #2 CRC #1000 DLE EOT Trailer If you couldn t solve the task 7.1 and/or 7.2 assume the tare to be 1100 Bytes for Task How many complete frames can be transmitted with the presented asynchronous transmission during 1 second of pure data transmission? (Consider the line to be error free) 1 Frames 2 Frames 3 Frames 4 Frames 5 Frames