A CPU has 12 registers and uses 6 addressing modes. RAM is 64K × 32. What is the maximum size of the op-code field if the instruction has a register operand and a memory address operand?a)8 bitsb)9 bitsc)10 bitsd)11 bitsCorrect answer is option 'B'. Can you explain this answer?

Question

A CPU has 12 registers and uses 6 addressing modes. RAM is 64K &times; 32. What is the maximum size of the op-code field if the instruction has a register operand and a memory address operand?a)8 bitsb)9 bitsc)10 bitsd)11 bitsCorrect answer is option 'B'. Can you explain this answer?

Answer

A CPU with 12 registers means that it has 12 small storage units (registers) that can hold data. These registers are used for storing and manipulating data during processing.

The 6 addressing modes refer to the different ways in which the CPU can access data from RAM (Random Access Memory). These addressing modes determine how the CPU calculates the memory address of the data it wants to access.

The RAM size of 64K means that the CPU can access a maximum of 64 kilobytes (or 64,000 bytes) of data from the RAM. This memory is used for storing data and instructions that the CPU needs to execute.

Overall, these specifications provide information about the capabilities of the CPU in terms of data storage and access. The number of registers and addressing modes determine the efficiency and flexibility of the CPU in performing computations and accessing data, while the RAM size determines the amount of data that can be stored and accessed by the CPU.

Answer

Data:
number of registers = 12
addessing mode = 6
RAM size =64K × 32 = 216 × 25

Formula:
Memory Capacity is of the form = 2m × 2n
Address lines required = m
Instrcution size = Data Lines = 2n
number of bits = ⌈log₂ n⌉
number of register or number addressing modes

Calculation:
Instrcution size = Data Lines = 32
Address lines required = 16 bits
number of bits for a addressing mode = ⌈log₂ 6⌉ = 3
number of bits for a register field = ⌈log₂ 12⌉ = 4

3 + x + 4 + 16 = 32
∴ x = 9
op-code field = 9 bits

Similar GATE Doubts

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