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Suppose that you have a computer with a memory unit of 24 bits per word. In this
computer, the assembly program’s instruction set consists of 198 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. How many additional instructions could be added to this instruction set without
exceeding the assigned number of bits? Discuss and show your calculations.
4. What is the largest unsigned binary number that the address can hold?

Respuesta :

Answer:

a )   The amount of bits required for the opcode

 8 bits

2^8=  256        

 256>198

 We get the next lower number, which is 2^7 = 128 bits, because it is greater than 198. As a result, the operation code necessitates 8 bits.

b)   The number of bits reserved for the instruction's address.

 16 bits

 24-8  =  16

c)  

 65536

2^16  =  65536

 Maximum number =  65535

  2^15 = 32768-1

 =  32767

Explanation:

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