Question

MIPS

Assembly

Computer Organization and Design

MIPS Algorithm Design (25 pts. Throughout the course, we have learned different types of instruction sets in MIPS programming: arithmetic instructions, branch instructions, jump instructions, and memory access instructions. Let's assume, we have a 10 line of MIPS code in which we perform 5 arithmetic operations, 2 branch operations. 1 jump operation, and 2 memory operations (1 store, 1 load). In the datapath, we are given that: fetching the instruction set takes 2 states. decoding takes 1 state, • arithmetic takes 3 states, • branch takes 2 states. • jump takes 1 state. • memory access takes 1 state, • sw takes 1 state, • lw takes 2 states. a) Please calculate Cycles Per Instruction (CPI) of this datapath. For this, you may want to get the total number of cycles needed to perform the MIPS code and ther divide by total number of instructions. [10 pts.) Now, we are given a different datapath: . • fetching the instruction set takes 1 state, • decoding takes 1 state, arithmetic takes 6 states. • branch takes 1 state, - jump takes 1 state, • memory access takes 1 state, • sw takes 1 state, • lw takes 1 state, omnare updated machine with the original system and try to justify which system is better to use and why? Are these changes smart? [15 points)

Answer 1

The MIPS instruction set architecture

The MIPS has a 32 bit architecture, with 32 bit instructions, a 32 bit data word, and 32 bit addresses. It has 32 addressable internal registers requiring a 5 bit register address. Register 0 always has the the constant value 0. Addresses are for individual bytes (8 bits) but instructions must have addresses which are a multiple of 4. This is usually stated as “instructions must be word aligned in memory.”

(a)
Total cycles taken by arithmetic operations
2 + 1 + 3 = 6
Total cycles taken by branch operations
2 + 1 + 3 + 2 = 8
Total cycles taken by jump operations
2 + 1 + 1 = 4
Total cycles taken by lw operations
2 + 1 + 3 + 1 + 2 = 9
Total cycles taken by sw operations
2 + 1 + 3 + 1 + 1 = 8

Total cycles = 3*6 + 4*8 + 4 + 9 + 8 = 18 + 32 + 21 = 71
Cycles per instruction (CPI)= 71/10 = 7.1

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