Question

Consider a hypothetical microprocessor having 64-bit instructions composed of two fields: the first two bytes contain the opcode and the remainder the immediate operand or operand address. What is the maximum directly accessible memory capacity (in Bytes)? Provider answer in Giga Bytes.

Answer 1

In the given processor we have 64 bit instruction with two fields

1. Opcode : 2 Bytes = 16 bits ( as 1 Byte = 8 bits )

2.  Immediate operand or operand address :64 -16 = 48 bits

Now when we have 48 bits for memory then number of operand addresses possible are 2^48

So Total Memory Capacity we have = 2^48 Bytes

Now 1 Giga Byte = 2^30 Bytes

To convert the answer in Giga Bytes = 2^48 / 2^30 = 2^18 GB

We get Memory Capacity as 2^18 GB = 262144 GB

This is how we can get the memory for the Given Processor having 64 bits instruction

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