Mid-term EXAM. 11/14/2009

Transcription

1 Mid-term EXAM. 11/14/ (15%) Data Compression a) Encode the following characters using Huffman coding with the given frequencies: A(12), B(8), C(9), D(20), E(31), F(14), G(8) (-1 point if theree is no Huffman tree) The answer is not unique. The following tree is constructed by repeating the procedure that combines the nodes that have lowest frequency. Once the encoding tree is built, we can assign the first branch with the value 0 and the second one the value 1(see the figure.) The codes can be obtained by traveling from the root node (102) to the leaves (A-G). For example, the code for character A would be 011 and the code for character B would be b) Encode the message BAABBBBAACAA using the Lempel Ziv method, and then decode the encoded message to get the original message. (-2 point if theree is no encoding process)

2 c) Encode the message xyz xyz xyz using the LZW method, and then decode the encoded message to get the original message. In the following figure, we use underline character _ to emphasize the existence of space character. Note that in LZW method, you should pass the basic dictionary to the decoder. (1 point for the compressed string; 2 points each for encoding/decoding procedure.)

3 2. (25%) Number system a) What is the binary representation of 6? b) Using an 8-bit allocation, first convert each of the following integers to two s complement, do the operation, and then convert the result to decimal. (1) (2) (1). (2) ) ) = = -58 c) What value is represented by the bit pattern when interpreted using floating-point format in which the most significant bit is the sign bit, the next three bits represent the exponent field in excess notation, and the last four bits represent the mantissa? expo: 101 = 1 (100 = 0 refer to the textbook) mantissa * 2^1 = 1.1 = 1+1/2 d) How many errors in a single code pattern could be corrected when using an error-correcting code in which each code pattern is a Hamming distance of at least five from any other code pattern? Can detect up to 4 errors and correct up to 2 errors 3. (10%) The following table shows a portion of a machine's memory containing a program written in the language described in the language description table in appendix. Answer the questions below assuming that the machine is started with its program counter containing 00. Address Content Address Content 00 B0 07 C B0 0A B0 04 0C 0B C0 0C B D 07

4 a) How many instructions will be executed before the machine halts? 4 (B003 -> B00C -> B007->C000) (The answer 3 would be graded if you wrote down the instructions flow) b) What bit pattern will be in the program counter when the machine halts? (20%) Networking: a) Please explain what are CSMA/CD and CSMA/CA, and the distinctions between them. (-2 point if you did not state the difference between them) Carrier Sense Multiple Access/ Collision Detect and Collision Avoidance CSMA/CD : 有線網路上要發送資料的 node 一旦察覺到 collision, 網路上所有 node 都要停止傳輸動作, 等待一段時間 等待的時間由公式計算 CSMA/CA : node 要發送資料時並不會馬上動作, 而會先產生一個 random backoff time 在這段時間裡, 若發現此 channel 忙碌的話, 則停止倒數 random backoof time 直到倒數完畢後, 發送端 node 會發出一個 RTS(request to send) 訊號, 目的端收到 RTS 後會回傳 CTS(clean to send) 發送端在收到 CTS 後才會發送資料 其他 node 若偵測到 CTS 或 RTS, 也會停止倒數 此協定把 collision 控制在 RTS-CTS 階段 b) Recall that the Internet Radio could be implemented by either N-unicast or multicast. Please explain what are N-unicast and multicast and the distinctions between them. (-2 point if you did not state the difference between them) Multicast: server 只丟一份 package, 由 router 來 copy 額外的份數 N-unicast: server 必須要為每個 client 丟一份 package. p (30%) Operating Systems a) (10%) Please choose the part on the operating system (file manager, memory manager, device drivers, window manager, scheduler, dispatcher) that performs the activity described. (i) Performs the switching from one process to another dispatcher (ii) Performs the actual communication with I/O units device driver (iii) Maintains a record of memory allocations memory manager (iv) Protects files from unauthorized access file manager (v) Places new entries in the process table scheduler

5 b) (5%) What conditions are necessary for deadlock to occur? 1. Mutual exclusion : non-shareable, until resources are released 2. Hold & wait : hold resources and wait for other resources to continue 3. Non-preemptive 4. *Circular Wait c) (5%) Describe the bootstrap process. (-1 point if you did not state the initialization process) Initialize software and hardware environment, and then transfer the control right to operation system. d) (10%) Suppose an operating system allocates time slices in 10 millisecond units and the time required for a context switch is negligible. (i) How many processes can obtain a time slice in one second? 1000/10 = 100 (ii) How many processes can obtain a time slice in one second if half of them use only half of their slice? 1000/ (10* *0.5) = 1000/7.5 = 133 (or 100) (Time slice is undividable in some systems, therefore, it is also correct if your answer is as same as yours in (i). But still remember that it depends on the policies of machine. ) Appendix: The following table is from Appendix C of the text. It is included here so that it can be incorporated in tests for student reference. Questions in this test bank refer to this table as the language description table. Op-code Operand Description 1 RXY 2 RXY 3 RXY 4 0RS LOAD the register R with the bit pattern found in the memory cell whose address is XY. Example: 14A3 would cause the contents of the memory cell located at address A3 to be placed in register 4. LOAD the register R with the bit pattern XY. Example: 20A3 would cause the value A3 to be placed in register 0. STORE the bit pattern found in register R in the memory cell whose address is XY. Example: 35B1 would cause the contents of register 5 to be placed in the memory cell whose address is B1. MOVE the bit pattern found in register R to register S. Example: 40A4 would cause the contents of register A to be copied into register 4.

6 5 RST 6 RST 7 RST 8 RST 9 RST A R0X B RXY C 000 ADD the bit patterns in registers S and T as though they were two s complement representations and leave the result in register R. Example: 5726 would cause the binary values in registers 2 and 6 to be added and the sum placed in register 7. ADD the bit patterns in registers S and T as though they represented values in floating-point notation and leave the floating-point result in register R. Example: 634E would cause the values in registers 4 and E to be added as floating-point values and the result to be placed in register 3. OR the bit patterns in registers S and T and place the result in register R. Example: 7CB4 would cause the result of ORing the contents of registers B and 4 to be placed in register C. AND the bit patterns in register S and T and place the result in register R. Example: 8045 would cause the result of ANDing the contents of registers 4 and 5 to be placed in register 0. EXCLUSIVE OR the bit patterns in registers S and T and place the result in register R. Example: 95F3 would cause the result of EXCLUSIVE ORing the contents of registers F and 3 to be placed in register 5. ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that started at the low-order end at the high-order end. Example: A403 would cause the contents of register 4 to be rotated 3 bits to the right in a circular fashion. JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of execution. (The jump is implemented by copying XY into the program counter during the execute phase.) Example: B43C would first compare the contents of register 4 with the contents of register 0. If the two were equal, the pattern 3C would be placed in the program counter so that the next instruction executed would be the one located at that memory address. Otherwise, nothing would be done and program execution would continue in its normal sequence. HALT execution. Example: C000 would cause program execution to stop.