CHAPTER 1 BASIC CONCEPTS AND

Computer Organization and Architecture 10e William Stallings

(Solution Manual all Chapters)

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CHAPTER 1 BASIC CONCEPTS AND

COMPUTER EVOLUTION

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), I/O 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.

I/O: Moves data between the computer and its external environment.

System interconnection: Some mechanism that provides for

communication among CPU, main memory, and I/O. 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 computer’s data

processing functions Registers: Provides storage internal to the CPU © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 2 / 4

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CPU interconnection: Some mechanism that provides for

communication among the control unit, ALU, and registers

1.6 In 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.

1.7 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.

1.8 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 I/O 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.

1.9 In a microprocessor, all of the components of the CPU are on a single chip.

ANSWERS TO PROBLEMS

2.1 a Location Instruction/Value Comments

• <> Constant (N) [initialized to some value]
• 1 Constant; Integer value = 1
• 2 Constant; Integer value = 2
• 0 Variable Y (initialized to integer zero);

Sum(Y)

4L LOAD M(0 N → AC

4R ADD M(1) AC + 1 → AC

5L MUL M(0) N(N+1) → AC

5R DIV M(2) AC/2 → AC

6L STOR M(3) AC → Y; saving the Sum in variable Y

6R JUMP M(6,20:39) Done; HALT

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b.Location Instruction/Value Comments

• <> Constant (N) [initialized to some value]
• 1 Constant (loop counter increment)
• 1 Variable i (loop index value; current)
• 1 Variable Y = Sum of X values (Initialized to

One) 4L LOAD M(0 N → AC (the max limit) 4R SUB M(2) Compute N–i → AC

5L JUMP + M(6,0:19) Check AC > 0 ? [i < N]

5R JUMP + M(5,20:39) i=N; done so HALT

6L LOAD M(2) i

8L STOR M(3) AC → Y

8R JUMP M(4,0:19) Continue at instruction located at address

4L

2.2 a.Opcode Operand

00000001 000000000010

• First, the CPU must make access memory to fetch the instruction. The

instruction contains the address of the data we want to load. During the execute phase accesses memory to load the data value located at that address for a total of two trips to memory.

2.3 To read a value from memory, the CPU puts the address of the value it wants into the MAR. The CPU then asserts the Read control line to memory and places the address on the address bus. Memory places the contents of the memory location passed on the data bus. This data is then transferred to the MBR. To write a value to memory, the CPU puts the address of the value it wants to write into the MAR. The CPU also places the data it wants to write into the MBR. The CPU then asserts the Write control line to memory and places the address on the address bus and the data on the data bus. Memory transfers the data on the data bus into the corresponding memory location.

2.4 Address Contents 08A

08B

LOAD M(0FA)

STOR M(0FB)

LOAD M(0FA) © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

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