Explain the 7 aspects of the \"Instruction Set Architecture\" Explain the highli

Question

Explain the 7 aspects of the "Instruction Set Architecture" Explain the highlight terms in the following problems: A computer with a 1 GHz Pentium IV processor takes 4 ms to run a program with 600,000 instructions. Another computer with a 400 MHz UltraSparc processor takes 3.8 ms to run the same program with 300,000 instructions. (a) What are the clock rate and clock period for each processor? (b) Which processor is faster and by how much? (c) What are the CPI and IPC for each processor for this program? (d) How much time does it take for each processor to run another program with 1 million instructions? (a) What is the average CPI for a program with 1 million instructions that uses only 4 operations as follows? (b) What is the percentage of time used only in ALU operations?

Explanation / Answer

1
7 Aspects of Instruction Set Architecture are:
   1. Implicit structure of all modern ISA are VonNeumann model
   2. Format: Length and encoding of instructions. i.e. Fixed Length ISA Variable Length ISA
   3. Operand model: Where (other than memory) are operands stored?
   4. Datatypes and Operations (Data Transfer, Arithmetic operations, Logic operations)
   5. Control (procedure calls in C adn return)
   6. Memory addressing: Byte addressing scheme is most widely used
   7. Addressing Mode: Specify the address of a M object apart from register and constant
operands.

2
Clock Rate of a processor is the frequency of the processor's base clock at that given moment of time.
CPI: CPI is also known as clocks per instruction. It is generally calculated by dividing total clock cycles required by total number of instructions executed in that amount of time. And hence we get an average of how many clock cycle time each instruction would take
IPC: IPC stands for Instructions per clock. It is the multiplicative inverse of cycles per instruction.

3
a
Clockrate of intel pentium 4: 1*109 clocks per second
Clockrate of Ultra Sparc is 400*106 clocks per second

b
Ultra Sparc is faster when it comes to the program given in the problem.
Speed up = 4/3.8 = 1.053

c
Intel Pentium processor:
       time required: 4sec
       therefore clocks: 4sec * 1*109 = 4*109 clocks
       total instrucions: 600000
       therefore CPI: 4*109 / 600000 = 6666.67
       IPC = 0.00015

Ultra Sparc:
      time required: 3.8sec
       therefore clocks: .3.8sec * 0.4*109 = 15.2*108 clocks
       total instrucions: 300000
       therefore CPI: 15.2*108 / 600000 = 5066.67
       IPC = 0.000197

d
       Intel: 1Million * 6666.67 clocks
              or 10^6 * 6666.67 / 10^9 sec
              or 6.67 sec

       Ultra Sparc: 1Million * 5066.67 clocks
              or 10^6 * 5066.67 / (0.4*10^9) sec
              or 12.67 sec

Champ I was suppose to answer 1 question according to chegg protocol but managed to answer 2. Please co-operate

             
     

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