Question 1) For a given CPU, the cycle latency for the a set of operations are g

Question

Question 1) For a given CPU, the cycle latency for the a set of operations are given as follows:

-Addition:         4

-Subtraction:     8

-Multiplication: 64

-Division:         128

If the CPU clock of this CPU runs at 2048 MHz, find the following

a) How many operations of each of the list above can this CPU perform in 5 minutes.

b) If we have 1024M of each operation in the list, compute the required time in seconds to execute all operations.

Question 2) Assuming a capacity of a RAM chip is 1MB, its physical dimensions are 1”x4” and it costs $1,000,000 to build a production facility to produce this chip. According to Moore’s and Rock’s laws compute the new capacity and the cost to produce the new chip 32 years from today.

Question 3) What is the fundamental problem of Von Neumann model? Discus –briefly three attempts to improve upon it.

Question 4) Given the following java instructions, use the block diagram of Von Neumann model to trace the machine execution cycle for the given instructions.

a) k = 11 + 99;

b) z = y / x;

c) a = b || false;

Explanation / Answer

Solution:

1)

a)

for number of additions

Average access time= Instruction count*cycles/CPU frequency

5*60= n*4/2^11*2^20 (let's say number of instructions are n and we have converted the time into seconds)

n= (2^31) * 5 * 60 / 4= 161061273600= 16.10612736*10^10

for number of subtractions

Average access time= Instruction count*cycles/CPU frequency

5*60= n*8/2^11*2^20 (let's say number of instructions are n and we have converted the time into seconds)

n= (2^31) * 5 * 60 / 8= 80530636800= 8.0530636800*10^10

for number of multiplications

Average access time= Instruction count*cycles/CPU frequency

5*60= n*64/2^11*2^20 (let's say number of instructions are n and we have converted the time into seconds)

n= (2^31) * 5 * 60 / 64= 10066329600= 1.0066329600*10^10

for number of divisions

Average access time= Instruction count*cycles/CPU frequency

5*60= n*128/2^11*2^20 (let's say number of instructions are n and we have converted the time into seconds)

n= (2^31) * 5 * 60 / 128= 5033164800= 5.033164800*10^9

b)

If we are executing all the operations at once then we need to find the CPI of the given

CPI= 4+8+64+128/4= 51

Average access time= 51 * (2^10) * (2^20) / ((2^11) * (2^20))= 25.5 seconds

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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