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Full file at http:testbank360.eusolution-manual-computer-organization-and-architecture-9thedition-william-stallings CHAPTER 1 OVERVIEW ANSWERS TO QUESTIONS 1.1 Computer architecture refers to those attributes of a system visible to a programmer or, put another way, those attributes that have a direct impact on the logical execution of a program. Computer organization refers to the operational units and their interconnections that realize the architectural specifications. Examples of architectural attributes include the instruction set, the number of bits used to represent various data types (e.g., numbers, characters), IO mechanisms, and techniques for addressing memory. Organizational attributes include those hardware details transparent to the programmer, such as control signals; interfaces between the computer and peripherals; and the memory technology used. 1.2 Computer structure refers to the way in which the components of a computer are interrelated. Computer function refers to the operation of each individual component as part of the structure. 1.3 Data processing; data storage; data movement; and control. 1.4 Central processing unit (CPU): Controls the operation of the computer and performs its data processing functions; often simply referred to as processor. Main memory: Stores data. IO: Moves data between the computer and its external environment. System interconnection: Some mechanism that provides for communication among CPU, main memory, and IO. A common example of system interconnection is by means of a system bus, consisting of a number of conducting wires to which all the other components attach. 1.5 Control unit: Controls the operation of the CPU and hence the computer Arithmetic and logic unit (ALU): Performs the computers data processing functions Registers: Provides storage internal to the CPU Full file at http:testbank360.eusolution-manual-computer-organization-and-architecture-9thedition-william-stallings CPU interconnection: Some mechanism that provides for communication among the control unit, ALU, and registers Full file at http:testbank360.eusolution-manual-computer-organization-and-architecture-9thedition-william-stallings CHAPTER 2 COMPUTER EVOLUTION AND PERFORMANCE ANSWERS TO QUESTIONS 2.1In a stored program computer, programs are represented in a form suitable for storing in memory alongside the data. The computer gets its instructions by reading them from memory, and a program can be set or altered by setting the values of a portion of memory. 2.2 A main memory, which stores both data and instructions: an arithmetic and logic unit (ALU) capable of operating on binary data; a control unit, which interprets the instructions in memory and causes them to be executed; and input and output (IO) equipment operated by the control unit. 2.3 Gates, memory cells, and interconnections among gates and memory cells. 2.4 Moore observed that the number of transistors that could be put on a single chip was doubling every year and correctly predicted that this pace would continue into the near future. 2.5 Similar or identical instruction set: In many cases, the same set of machine instructions is supported on all members of the family. Thus, a program that executes on one machine will also execute on any other. Similar or identical operating system: The same basic operating system is available for all family members. Increasing speed: The rate of instruction execution increases in going from lower to higher family members. Increasing Number of IO ports: In going from lower to higher family members. Increasing memory size: In going from lower to higher family members. Increasing cost: In going from lower to higher family members. 2.6 In a microprocessor, all of the components of the CPU are on a single chip. Full file at http:testbank360.eusolution-manual-computer-organization-and-architecture-9thedition-william-stallings Full file at http:testbank360.eusolution-manual-computer-organization-and-architecture-9thedition-william-stallings ANSWERS TO PROBLEMS 2.1 a Location 0 1 2 3 InstructionValue 1 2 0 4L 4R 5L 5R 6L 6R LOAD M(0 ADD M(1) MUL M(0) DIV M(2) STOR M(3) JUMP M(6,20:39) b. Location 0 1 2 3 InstructionValue 1 1 1 4L 4R 5L 5R 6L 6R 7L 8R 8L 8R LOAD M(0 SUB M(2) JUMP + M(6,0:19) JUMP + M(5,20:39) LOAD M(2) ADD M(1) STOR M(2) ADD M(3) STOR M(3) JUMP M(4,0:19) Comments Constant (N) initialized to some value Constant; Integer value = 1 Constant; Integer value = 2 Variable Y (initialized to integer zero); Sum(Y) N → AC AC + 1 → AC N(N+1) → AC AC2 → AC AC → Y; saving the Sum in variable Y Done; HALT Comments Constant (N) initialized to some value Constant (loop counter increment) Variable i (loop index value; current) Variable Y = Sum of X values (Initialized to One) N → AC (the max limit) Compute N–i → AC Check AC > 0 ? i < N i=N; done so HALT i Youre Reading a Free Preview Pages 6 to 12 are not shown in this preview. Academia.edu uses cookies to personalize content, tailor ads and improve the user experience.By using our site, you agree to our collection of information through the use of cookies.To learn more, view our Privacy Policy.
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