last two digits to be converted to binary are 8 and 0

8- 1000

0- 0

RO 2. Microarchitecture: 20 marks total Assume that the 2 least siginificant digits of your student ID are hexadecimal. Convert these two digits to binary and label the bits as 67, b6, 65, 64, 63, 62, 61, bo. Obtain the binary value given by C302C160 = b7b6b5b4 + b3b2b1bo(Drop the carry bit if any). Convert C302C1Co to hexadecimal. Denote the hexadecimal value by Z. Now consider the following register and memory values, determine the contents of MAR, MDR, IR, T1, T2, IR, RO, R1, R2, R3, R4, and R15 in the states: F0, F1, F2, F3 (Decode), EO, E1 and E2. You will need the operations table and the format of the IR register to answer this question. Put your results in a table format. Only record values of the registers when they change. Register Value Memory Address Value (0x0000 000Z) +0xF 0x0000 0000 Ox0101 1000 R1 (0x0000 000Z)+OxE 0x0000 0001 0x0101 2000 (0x0000 000Z) +OxD 0x0000 0002 Ox0101 3000 (0x0000 000Z)+OxC 0x0000 0003 0x0101 4000 R4 (0x0000 000Z)+OxB 0x0000 0004 0x0111 1000 R5 (0x0000 000Z)+OxA Ox0000 0005 Ox0111 2000 R6 (0x0000 000Z)+0x9 0x0000 0006 0x0111 3000 R7 (Ox0000 000Z)+0x8 Ox0000 0007 0x0111 4000 R8 (0x0000 000Z)+0x7 Ox0000 0008 Ox0121 1000 (0x0000 000Z) +0x6 Ox0000 0009 0x0121 2000 R10 (Ox0000 000Z)+0x5 Ox0000 000A 0x0121 3000 RII (0x0000 00OZ)+0x 4 Ox0000 000B Ox0121 4000 R12 (0x0000 000Z)+0x3 0x0000 0000 0x0122 1000 R13 (0x0000 000Z)+0x2 Ox0000 000D Ox0122 2000 R14 (0x0000 000Z)+0x1 0x0000 000E Ox0122 3000 R15 Ox0000 000Z Ox0000 000F 0x0122 4000 R2 R3 R9 RO 2. Microarchitecture: 20 marks total Assume that the 2 least siginificant digits of your student ID are hexadecimal. Convert these two digits to binary and label the bits as 67, b6, 65, 64, 63, 62, 61, bo. Obtain the binary value given by C302C160 = b7b6b5b4 + b3b2b1bo(Drop the carry bit if any). Convert C302C1Co to hexadecimal. Denote the hexadecimal value by Z. Now consider the following register and memory values, determine the contents of MAR, MDR, IR, T1, T2, IR, RO, R1, R2, R3, R4, and R15 in the states: F0, F1, F2, F3 (Decode), EO, E1 and E2. You will need the operations table and the format of the IR register to answer this question. Put your results in a table format. Only record values of the registers when they change. Register Value Memory Address Value (0x0000 000Z) +0xF 0x0000 0000 Ox0101 1000 R1 (0x0000 000Z)+OxE 0x0000 0001 0x0101 2000 (0x0000 000Z) +OxD 0x0000 0002 Ox0101 3000 (0x0000 000Z)+OxC 0x0000 0003 0x0101 4000 R4 (0x0000 000Z)+OxB 0x0000 0004 0x0111 1000 R5 (0x0000 000Z)+OxA Ox0000 0005 Ox0111 2000 R6 (0x0000 000Z)+0x9 0x0000 0006 0x0111 3000 R7 (Ox0000 000Z)+0x8 Ox0000 0007 0x0111 4000 R8 (0x0000 000Z)+0x7 Ox0000 0008 Ox0121 1000 (0x0000 000Z) +0x6 Ox0000 0009 0x0121 2000 R10 (Ox0000 000Z)+0x5 Ox0000 000A 0x0121 3000 RII (0x0000 00OZ)+0x 4 Ox0000 000B Ox0121 4000 R12 (0x0000 000Z)+0x3 0x0000 0000 0x0122 1000 R13 (0x0000 000Z)+0x2 Ox0000 000D Ox0122 2000 R14 (0x0000 000Z)+0x1 0x0000 000E Ox0122 3000 R15 Ox0000 000Z Ox0000 000F 0x0122 4000 R2 R3 R9