Question

A digital computer has a memory unit with 24 bits per word. The instruction set consists of 199 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. [8 marks]

  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 signed binary number that can be accommodated in one word of memory?
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Answer #1

1. bits are needed for the opcode

Number of operations = 199

--------------------- remainder

2| 199

2| 99    1

2|49 1

2| 24    1

2| 12    0

2|6 0

2|3 0

2|1 1

2|0 1

so to store 199 operations in binary format (11000111)

opcode needs = 8 bits

2. bits are left for the address part of the instruction = 24 - 8 = 16 bits

3. maximum allowable size for memory

Total instructions = 199

1 word = 24 bits

maximum memory size = 199*24 = 4776

4. largest signed binary number that can be accommodated in one word of memory

one word of memory = 24 bits

1 bit for sign

largest binary number = 0111 1111 1111 1111 1111 1111

0 to denote positive number in sign bit position

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