Taking the student’s registration number to be:
We can execute this task as follows:
We begin by low-order leaving:
With the specifications:
AR ß PC
We determine that the program counter (PC) will require 12-bits to execute this task
Number of words = 256K words = 28 *210 words Number of bits pre each word = 32 bit = 25 bit Number of registers = 64 bit = 26 Register a) operation code = 7 bits, register code = 6 bits, address part = 18 bits
Part 2: With a PC = 789 H , the CPU will fetch the instruction(binary code) from memory using common bus system by using a bus which is constructed using multiplexer of equal capacity like a 4-unit. This will enable a set of registers to share a common bus for data transfer. Data Transfer Using the Bus.