Solve ( b) number only

(a) Consider a hypothetical machine containing a single data register, called an accumulator (AC). Both instructions and data are 16 bits long. The instruction format is shown below: 3.4 15 Opcode Address Here, the 12-bit address identifies a particular 1/0 device (e.g., device 1, device 2, device 3...and so on). The opcode for the two I/O instructions are: 3H = 0011= Load AC from 1/0 (moves data contained in I/O device to AC) 7H = 0111= Store AC to I/O (moves data contained in AC to the 1/0 device 5H = 0101 = Add to AC from Memory Hint: The first 4 bits (first hexadecimal digit) in the IR indicate the type of the operation. That means if IR contains 3H, it will be a load operation and if IR contains 7H it is a store operation. We will assume that the memory (contents in hex) as the following table: PC Address 300 301 302 Memory Content 3005H 5940H 7006H CPU MEMORY System Bus Address Data, Controls ALL Register Control Una - Primary Memory FROM / RAM Secondary Memory Cache Memory 0002H 1/0 940 941 +input Devices (Keyboard Mouse etc.) + Output Devices Monter, Printer Show the steps involved in the execution of the following program [ 6 marks] . Load AC from device 5 (Assume that the next value received from device 5 is 3) Add contents of memory location 940 to the content of (Assume that the location 940 contains value of 2). Store AC to device 6 (b) Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields: the first byte contains the opcode and the remainder the immediate operand or an operand address (1) What is the maximum directly addressable memory capacity (in bytes)? [ 2 marks] (ii) How many bits are needed for the program counter and the instruction register? [2 marks]