exam1_2002 solutions - Result PC X IR X MAR X X(a) X(b) MDR...

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Department of Electrical and Computer Engineering University of Texas at Austin EE 379K Fall 2000 Y. N. Patt, Instructor TAs: Kathy Buchheit, Laura Funderburg, Chandresh Jain, Onur Multu, Danny Nold, Kameswar Subramanian, Francis Tseng, Brian Ward Solutions to Exam 1 Problem 1 Part 1 No information about MAR. MDR is 64 bits. Part 2 7 bits are needed. Largest positive number = 0111111 (63) Largest unsigned number = 1111111 (127) Part 3 Yes there is a problem. We will need 5 bits to represent each register. Add instruction takes 3 registers and needs a 4-bit opcode. Therefore we need at least 19 bits for the ADD instruction. However in the LC-2 ISA each instruction is only 16 bits. Problem 2 A B | C D Z ----|------ 0 0 | 1 1 0 0 1 | 1 0 0 1 0 | 0 1 0 1 1 | 0 0 1 Z = A AND B Problem 4 For the instruction whose opcode is 0001 (ADD): Fetch Instruction Decode Evaluate Address Fetch Data Execute Store Result PC X
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IR X MAR X MDR X For the instruction whose opcode is 0010 (LD): Fetch Instruction Decode Evaluate Address Fetch Data Execute Store
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Unformatted text preview: Result PC X IR X MAR X X(a) X(b) MDR X X There should be at least one checkmark in the two boxes marked (a), (b) For the instruction whose opcode is 0011 (ST): Fetch Instruction Decode Evaluate Address Fetch Data Execute Store Result PC X IR X MAR X X(a) X(b) X(c) X(d) MDR X X(1) X(2) X(3) There should be at least one checkmark in the boxes marked (a), (b), (c), (d) There should be at least one checkmark in the boxes marked (1), (2), (3) For the instruction whose opcode is 0000 (a control instruction): Fetch Instruction Decode Evaluate Address Fetch Data Execute Store Result PC X X IR X MAR X MDR X Problem 4 Problem 5 (1): 1001 100 001 111111 ; R4 <- NOT(R1) (2): 1001 101 010 111111 ; R5 <- NOT(R2) (3): 0101 110 100 000 101 ; R6 <- R4 AND R5 (4): 1001 011 110 111111 ; R3 <- NOT(R6) Problem 6 Sign Exponent Fraction 1 1 1 0 0 1 0 1 = -1 Sign x 2 (Exponent - 7) x 1.Fraction = -1 1 x 2 (12 - 7) x 1.101 = -52 Problem 7 Part A: M = 1, N = 1, P = 1 Part B: E changes to 0...
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exam1_2002 solutions - Result PC X IR X MAR X X(a) X(b) MDR...

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