# Computer Organization Assignment PDF

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Computer Organization
Name of the Student:
Name of the University:
Author Note
1
COMPUTER ORGANIZATION
1. There are 199 instructions. Hence 2^8 instructions would be required as 2^7 = 128.
Hence 8 bits for the opcode is required.
2. Number bits left for the address part of the instruction = 24 – 8 = 16 bits.
3. The maximum allowable size of the memory = 2^16 bits.
4. 24 1s or (2^24) -1 is the highest word that can be allocated in the memory.
The value to be added to the accumulator is 900
The value already stored in the accumulator is 200.
Therefore, the value which is loaded in the accumulator = 900 + 200 = 1100
The value to be added to the accumulator is 1000.
The value already stored in the accumulator is 200.
Therefore, the value which is loaded in the accumulator = 1000 + 200 = 1200
The effective address of 900 is 1000.
The value to be added to the accumulator is 500.
Value in the accumulator is 200.
Therefore, the value which is loaded in the accumulator = 500 + 200 = 700
The value of base register is 100.
The effective address therefore = 900 + 100 = 1000.
2
COMPUTER ORGANIZATION
The value in location 1000 = 500
Value in the accumulator is 200.
Therefore, the value which is loaded in the accumulator = 500 + 200 = 700
The expression provided: F = (A-B) *(C*D+E)
PUSH C
PUSH D
MUL
PUSH E
PUSH A
PUSH B
SUB
MUL
POP F
SUB B
STORE T1

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