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Allushta [10]
3 years ago
5

Suppose your company has decided that it needs to make certain busy servers faster. Processes in the workload spend 60% of their

time using the CPU and 40% on I/O. To achieve an overall system speedup of 25%: a. How much faster does the CPU need to be? b. How much faster does the disk need to be?
Computers and Technology
1 answer:
vekshin13 years ago
7 0

Answer:

CPU need 50% much faster

disk need 100% much faster

Explanation:

given data

workload spend time CPU  = 60%

workload spend time I/O = 40%

achieve overall system speedup = 25%

to find out

How much faster does CPU need and How much faster does the disk need

solution

we apply here Amdahl’s law for the overall speed of a computer that is express as

S = \frac{1}{(1-f)+ \frac{f}{k} }      .............................1

here f is fraction of work i.e 0.6 and S is overall speed  i.e 100% + 25% = 125 % and k is speed up of component

so put all value in equation 1 we get

S = \frac{1}{(1-f)+ \frac{f}{k} }  

1.25 = \frac{1}{(1-0.6)+ \frac{0.6}{k} }  

solve we get

k = 1.5

so we can say  CPU need 50% much faster

and

when f = 0.4 and S = 125 %

put the value in equation 1

S = \frac{1}{(1-f)+ \frac{f}{k} }  

1.25 = \frac{1}{(1-0.4)+ \frac{0.4}{k} }  

solve we get

k = 2

so here disk need 100% much faster

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1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
3 years ago
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