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Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields : the first byte contains the opcode and the remainder is an immediate operand or an operand address. What is the maximum directly addressable memory capacity (in Megabytes)? consider memory as byte addressable.
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Instruction size = 32 bits

Opcode = 8bits (Given in question 1st byte contains the opcode)

maximum directly addressable memory capacity  = $2^{32-8} = 2^{24} = 16MB$

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