cmpe 200 assignment 1

1 Name - Ajai Krishna VelayuthamSID - 010026328

San Jose State University

CMPE 200

Assignment -1

1.

a) To calculate which processor has highest performance in instructions per second -

For P1 – 4 GHz / 2.5 CPI = 1.6 x 109 instructions per second. For P2 – 3.4 GHz / 2.0 CPI = 1.7 x 109 instructions per second. For P3 – 2.4 GHz / 1.2 CPI = 2 x 109 instructions per second.

It is evident from the above that P3 executes more instructions than other processors, hence it has the highest performance.

b) To find number and clock cycles and instructions executed by each processor for executing a program which has taken 10 seconds to complete.

P1 takes 1.6 x 109 instructions per second , for 10 seconds – 1.6 x 109 x 10i.e. 16 x 109 instructions/second.Clock cycles taken – 16 x 2.5 x 109 = 40 x 109 clock cycles.

Similarly, P2 – 1.7 x 109 instructions/second, for 10seconds – 17 x 109

instructions/second.Clock cycles taken – 17 x 2.0 x 109 = 34 x 109 clock cycles.

P3 – 2.0 x 109 instructions/second, for 10seconds – 20 x 109 instructions/second.Clock cycles taken – 20 x 1.2 x 109 = 24 x 109 clock cycles.

c) Given execution time is reduced by 30% (i.e. 0.7 x execution time). And this leads to 20% increase in CPI (i.e. 1.2 x CPI).

Execution time is determined by –

Execution time = CPI * Clock cycles * Number of Instructions1

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Keeping the number of instructions constant as per the problem scenario – 0.7 x Execution time = 1.2 x CPI * Clock cycles * number of instructions.

To maintain the balance of the equation – New clock cycles is = 0.7 / 1.2 of old clock cycles = 0.5833 x old clock cycles Hence, New Clock rate = 1/ 0.5833 = 1.71428 x old clock rate.

2. Code - C = D + B B = A - C

a) For Accumulator Based – Load DAdd BStore CLoad ASub CStore B

Code Size = # instructions x size of each instruction = 6 * ( 1 + 3 ) = 24 bytes.

Data Memory Traffic = # data addresses to memory * data address size +# data received from memory * data size.

= (6 * 4) + (6 * 4) = 48 bytes.

Instruction Traffic = # instruction address to memory * address size + # instruction received from memory * instruction size.

= (6 * 4) + (6 * 4) = 48 bytes.

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3b) For Stack Based –

Push DPush BAddPop CPush APush CSub Pop B

Code Size = # instructions x size of each instruction = 6 * ( 1 + 3 ) + 2 = 26 bytes.

Data Memory Traffic = # data addresses to memory * data address size + # data received from memory * data size.

= (6 * 4) + (6 * 4) = 48 bytes.

Instruction Traffic = # instruction address to memory * address size + # instruction received from memory * instruction size.

= (4 * 8) + [(6 * 4) +2] = 58 bytes.

c) For Memory – Memory

ADD C, D, BSUB B, A, C Code Size = # instructions x size of each instruction = 2 * (3 + 3 + 1) = 20 bytes.

Data Memory Traffic = # data addresses to memory * data address size + # data received from memory * data size.

= (6 * 4) + (6 * 4) = 48 bytes.

Instruction Traffic = # instruction address to memory * address size + # instruction received from memory * instruction size.

= (2 * 4) + (6 * (3 + 3+ 1)) = 28 bytes.

4d) For Register based

Load R1,DLoad R2,BAdd R3,R1,R2Store R3,C

Code Size = # instructions x size of each instruction = (5 * 5) + (4 * 2) = 33 bytes.

Data Memory Traffic = # data addresses to memory * data address size + # data received from memory * data size.

= (5 * 4) + (5 * 4) = 40 bytes.

Instruction Traffic = # instruction address to memory * address size + # instruction received from memory * instruction size.

= (7 * 4) + [(5 * 5) + (4 * 2)] = 61 bytes.

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Ajai Krishna Velayutham010026328