COMP SCI 3005 Computer Architecture

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COMP SCI 3005 Computer Architecture

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COMP SCI 3005 Computer Architecture

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Course Code: COMPSCI3005
University: The University Of Adelaide

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Country: Australia

Question: 
1. A digital computer has a memory unit with 16 bits per word. The instruction set consists of 122 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. 
1. How many bits are needed for the opcode?
2. How many bits are left for the address part of the instruction?
3. what is the maximum allowable size for memory?
4. What is the largest unsigned binary number that can be accommodated in one word of memory? 
2: Consider an instruction Add 1000. Memory contents, Accumulator (AC), and R1 Register contents are depicted in the below figure: (such as memory address 1000 contains a value 1400, memory address 1100 contains a value 400 etc. All the numbers used here are decimals.) 
                                               
Assuming RI is implied in the indexed addressing mode, determine the actual value loaded into the accumulator (AC) using the following addressing modes: 
1. Immediate
2. Direct
3. Indirect
4. Indexed 
 
3. More registers appears to be a good thing, in terms of reducing the total number of memory accesses a program might require. Using the arithmetic expression S = (A+B)-(C+D), support this statement. [Hints: First, determine the number of memory accesses necessary using MARIE and the two registers for holding memory data values (AC and MBR). Then perform the same arithmetic computation for a processor that has more than three registers (for example, R1, R2, R3, R4) to hold memory data values.] [6 marks] 
 
4: Assemble the MARIE program below. 
Hex Address Label Instruction 100 Start LOAD A 101 ADD B 102 STORE D 103 CLEAR 104 OUTPUT 105 ADDI D 106 STORE B 107 HALT 108 A, HEX 00FC 109 B, DEC 14 10A C, HEX 0108 10B D, HEX 0000 
a) List the hexadecimal code for each instruction 
b) Draw the symbol table 
c) What is the value stored in the AC when the program terminates? 
Answer: 
1.

122 instructions imply that 2^7, hence 7 bits are required for the opcode.
The number of bits in a word is 16. Hence the address part is 16 – 7 = 9 bits.
Maximum allowable size of the memory is 2^9 bits.
2 ^16 – 1 or 16 ones is the highest number to be allocated in the memory.

2.
1. Immediate Addressing mode
Value added into the accumulator is 1000.
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1000 +500 =1500.
2. Direct Addressing mode
Effective address: 1000
Value added into the accumulator is 1400.
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1400 +500 = 1900.
3. Indirect Addressing mode
The effective address is the value in 1000 which is 1400.
Therefore, the value added into the accumulator is 1300
Value in accumulator is 500.
Therefore, the value loaded into the accumulator is 1300 +500 = 1800.
4. Indexed Addressing mode
The value of R1 is 200.
The effective address is 1000 + 200 = 1200. Hence the value in 1200 location would be added to the accumulator.
Value in location 1200 is 1000. Hence the value loaded into the accumulator is 1000 + 500 = 1500.
3. The following MARIE program can be utilized to realise the expression S = (A+B)-(C+D).
100 LOAD A
101 ADD B
102 STORE X
103 CLEAR
104 LOAD C
105 ADD D
106 STORE Y
107 CLEAR
108 LOAD X
109 SUBT Y
10A STORE S
10B HALT
Now for the Register memories the following program is to be used:
ADD R1, A, B
ADD R2, C, D
SUBT A, R1, R2
Hence for the first program the system would be required to store the results in three addresses and for the second program would require only a single memory unit for storing the data.
4 a.

Address

Hex

100

1108

Start

LOAD A

101

3109

ADD B

102

210B

STORE D

103

A000

CLEAR

104

F400

OUTPUT

105

B10B

ADDI D

106

2014

STORE B

107

7000

HALT

108

0200

A

HEX 00FC

109

0014

B

DEC 14

10A

0001

C

HEX 0108

10B

0000

D

HEX 0000

b.

Symbol

Location

A

108

B

109

C

10A

D

10B

Start

100

c. When the program terminates the accumulator would store the value 266 in decimal.

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