Consider the instruction formats of the basic computer in Fi
Solution
a)
 Given Instruction is :
Hexadecimal : C450
 Instruction(binary): 1100 0100 0101 0000 contains
 Instruction RTL : M[450] <- R1 ;(consider C450)
 So, xor operation between these two locations..
 0000 1001 1000 0000 = 0890
 1100 0100 0101 0000 = C450
 ----------------------------------
 1100 1101 1101 0000 = cDD0
 ------------------------------------
 In Hexadecimal: CDD0
 In binary: 1100 1101 1101 0000
------------------------------------------------------------------------------------------------------------------------------
b)
 Given data:
 Hexadecimal: 4450
 Binary instruction: 0100 0100 0101 0000
Instruction: M[450] <- R1 ;consider R1 = 4450H
 Performing XOR operation..
0000 1001 1001 0000 ----> 0990
 0100 0100 0101 0000 ----> 4450
 -----------------------------------
 0100 1101 1100 0000 ----> 4DC0
Final answer in
 Hexadecimal : 4DC0
 Binary: 0100 1101 1100 0000
 -----------------------------------------------------------------------------------------
  (c)
 Given data
 Hexadecimal : 7040
 Binary: 0111 0000 0100 0000
RTL instruction
 Ac <- 7040
 Accumulator operation, so addition performing
 0101 0000 0000 0111 ---> 5007
 0111 0000 0100 0000 ---> 7040
 ----------------------------------
 1100 0000 0100 0111 ----> C047
 -------------------------------------
 Final answer is:
 Hexadecimal : C047
 Binary : 1100 0000 0100 0111

