2010 Summer Answers [OS I]

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1 CS2503 A-Z Accumulator o Register where CPU stores intermediate arithmetic results. o Speeds up process by not having to store these results in main memory. Addition o Carried out by the ALU. o ADD AX, BX - - > AX = AX + BX The Complement o Refers to 2 s complement o (1) 0001 complement 1110, then add one, 1111 (-1) o (6)0110 complement 1001, then add one, 1010 (-6) Subtraction o To subtract A from B, calculate the sum of A and the complement of B o = = 0001 The Shifter o SHR(Shifter Right) Divide by 2 o 0001_1000 = 0000_1100 (SHL (24) = 12) o SHL(Shifter Left) Multiply by 2 o 0000_0101 = 0000_1010 (SHL (5) = 10) The Instruction Register o Register storing the current running instruction executing or decoding. o Holds the instruction to be executed while it is decoded by the instruction decoder. The Instruction Decoder o Instructions need to be decoded as they are stores as opcode inside the IR. o Step 1 : What type of instruction o Step 2: Identify memory address of the operands. o Step 3: Retrieve the operands from memory. The Program Counter (PC) o AKA : Instruction Pointer and Instruction Address Pointer o PC is a CPU Register which holds the address of next instruction to be executed. Stored Program Counter o Program is a list of instructions stored in main memory. The properties of a memory location o Address Length 16bits, 20bits, 32bits o Contents value stored at that address Width of the main data path o This is into the ALU and is determined by the number of bits to available to address the main memory. o 16bits data path, biggest number we can send is 2 16, which is (64k) Instruction Set o Represents the instruction block to execute. o Made up of the Instruction itself and its Operand(s).

2 Data Movement o Data movement can have many directions: o move value from register to a register o move value from register to memory o move value from memory to register Flow of Control o Instruction set is analysed, executed. o Depending on the type of instruction, flags (bit registers) may be set: o MOV AX, 5 o MOV BX, AX will result in zero flag being set o SUB BX, AX The Carry Flag o Some instructions, like the shift operations cause the carry flag to be set. o Set when there are not enough bits available to contain the result. o Adding two of the biggest 16-bit numbers : o 1111_ _1111 ( ) = 1_0000_0000 o BUT we only have 16-bits so 0000_0000 is stored and the carry flag is flipped to alert the low-level programmer that the result may be incorrect Summer Answers [OS I] QUESTION 1 (20 Marks) (a) (i) How many bits are necessary to address 32 bytes, 1K byte and 256 bytes? [3 Marks] Note: 8 bits = 1 byte 2 ^ 5 = 32bytes 2 ^ 10 = 1024 bytes 2 ^ 8 = 256bytes (ii) What are the maximum and minimum unsigned integers that can be represented by 7 bits, 9 bits and 10 bits? [3 Marks] unsigned 2 ^ 7 = 128 so (127 is maximum) 2 ^ 9 = 512 so ^ 10 = 1024 so 0 to 1023 signed

3 -64 to to to 511 (b) Convert the following decimal numbers to octal, showing all workings: 74D, 7549D, 255D, 15D. [6 Marks] 74D 74/8 = 9 remainder 2 9/8 = 1 remainder 1 1/8 = 0 remainder 1 Thus 74D = 112Q 7549D 7549/8 = 943 remainder 5 943/8 = 117 remainder 7 117/8 = 14 remainder 5 14/8 = 1 remainder 6 1/8 = 0 remainder 1 Thus 7549D = 16575Q 255D 255/8 = 31 remainder 7 31/8 = 3 remainder 7 3/8 = 0 remainder 3 Thus 255D = 377Q 15D 15/8 = 1 remainder 7 1/8 = 0 remainder 1 Thus 15D = 17Q (c) Convert the following decimal numbers to binary, showing all workings: 74D, 7549D, 255D, 15D. You may continue from part (b) if you wish. [4 Marks] 74D 74/2 = 37 remainder 0 37/2 = 18 remainder 1 18/2 = 9 remainder 0 9/2 = 4 remainder 1 4/2 = 2 remainder 0 2/2 = 1 remainder 0 1/2 = 0 remainder 1 Thus 74D = B 7549D 7549/2 = 3774 remainder /2 = 1887 remainder /2 = 943 remainder 1 943/2 = 471 remainder 1 471/2 = 235 remainder 1 235/2 = 117 remainder 1

4 117/2 = 58 remainder 1 58/2 = 29 remainder 0 29/2 = 14 remainder 1 14/2 = 7 remainder 0 7/2 = 3 remainder 1 3/2 = 1 remainder 1 1/2 = 0 remainder 1 Thus 7549D = B 15D 15/2 = 7 remainder 1 7/2 = 3 remainder 1 3/2 = 1 remainder 1 1/2 = 0 remainder 1 Thus 15D = 1111B (d) Convert the following decimal numbers to hexadecimal, showing all workings: 74D, 7549D, 255D, 15D. You may continue from part (c) if you wish. [4 Marks] 74D 74/16 = 4 remainder 10 = hex digit A 4/16 = 0 remainder 4 Thus 74D = 4AH 7549D 7549/16 = 471 remainder 13 = hex digit B 471/16 = 29 remainder 7 29/16 = 1 remainder 13 = hex digit B 1/16 = 0 remainder 1 Thus 7549D = 1B7BH 255D 255/16 = 15 remainder 15 = hex digit F 15/16 = 0 remainder 15 =hex digit F Thus 255D = FFH 15D 15/16 = 0 remainder 15 = hex digit F Thus 15D = FH QUESTION 2 (20 Marks) (a) What is a process? How is it represented in the operating system? Draw a 3-state process diagram and briefly explain the function of each state and state transition. [8 Marks] A process is an instance of a running program, consisting of executable instructions, data and management information A process is given control of the CPU to execute its instructions. Generally, multiple processes can be executing the same piece of code. The context of a process distinguishes it from other processes. The context is information on the instructions the process is currently or will next execute, the data it is working on, and administrative information such as a unique process ID.

5 READY when a process is created, it goes into the READY state, which means it is ready for the CPU RUNNING refers to processes that are currently running BLOCKED refers to processes that were halted (b) State and explain briefly the three conditions for a correct solution to the critical section problem. [3 Marks] 1. Mutual Exclusion: no two processes in their critical section at the same time 2. Progress: a process outside of its CS (critical section) can not interfere with the entry of other process into the critical section. [In other words, if one of the processes stops, the other one must keep running uninterrupted.] 3. Bounded waiting: a process that indicates its desire to enter the CS is guaranteed to do so, in _nite time. [In other words, there's no way that the two processes will get into a deadlock situation where each process is waiting for other processes to run so they're both stalling in the end.] (c) Describe the structure of a semaphore and the primitives that operate on it. [4 Marks] Before semaphores can be used a uniquely identified data structure and semaphore set (array) must be created. The semaphore set contains a predefined number of structures in an array, one structure for each semaphore in the set. The following members are in each structure within a semaphore set: semaphore value PID performing last operation number of processes waiting for the semaphore value to become greater than its current value number of processes waiting for the semaphore value to equal zero

6 Also, the primitives which operate on a semaphore consist: pointer to first semaphore in the set (array) number of semaphores in the set last semaphore operation time last semaphore change time (d) Give semaphore-based outline code for a producer and a consumer process in the bounded-buffer producer-consumer problem. You may assume the existence of high-level operations Produce Item, Consume Item, Place Item in Buffer and Take Item from Buffer. [5 Marks] Producer P(space_available) P(mutex) Put item in buffer V(mutex) V(item_available) Consumer P(item_available) P(mutex) Take Item from buffer V(mutex) V(space_available) Note: Mutex(established to ) = 1 Space_available(established to) = BUFFER_SIZE Item_available(established to) = 0 Note [P(S) = v V(S) = v++] Not needed in answer QUESTION 3 (20 Marks) The program below is written in Intel 80x86 assembly language. Assume that an opcode plus one register address can be contained in a single byte. Any immediate data requires at least one additional byte, depending on the value of the data. Assume that the offset of a memory address requires 16 bits. Ignore segmentation. (a) Describe how a two-pass assembler would process the source code file containing this program, mentioning all tables used and output files produced. [5 Marks] (b) Show the addresses generated by the first pass of the assembler, in the form of a table with the addresses in Column 1 and the corresponding contents in Column 2. It is not necessary to translate the contents. [5 Marks]

7 (c) List the contents of the relocatable file generated by the assembly process. Comment on each feature introduced by relocatability and pay attention to endian-ness. You must give everything here in numeric form, with the exception of opcodes and register addresses. For example, you can write MOV AX, 1 in exactly this form throughout your answer. [10 Marks] CSEG MOV AX, 1 ADD AX, 0 MOV RESULT, AX HLT RESULT DW 0 END