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lbvjy [14]
3 years ago
7

Suppose that a computer has three types of floating point operations: add, multiply, and divide. By performing optimizations to

the design, we can improve the floating point multiply performance by a factor of 10 (i.e., floating point multiply runs 10 times faster on this new machine). Similarly, we can improve the performance of floating point divide by a factor of 15 (i.e., floating point divide runs 15 times faster on this new machine). If an application consists of 50% floating point add instructions, 30% floating point multiply instructions, and 20% floating point divide instructions, what is the speedup achieved by the new machine for this application compared to the old machine
Computers and Technology
1 answer:
marishachu [46]3 years ago
6 0

Answer:

1.84

Explanation:

Operation on old system

Add operation = 50% = 0.5

Multiply = 30% = 0.3

Divide = 20% = 0.2

T = total execution time

For add = 0.5T

For multiplication = 0.3T

For division = 0.2T

0.5T + 0.3T + 0.2T = T

For new computer

Add operation is unchanged = 0.5T

Multiply is 10 times faster = 0.3T/10 = 0.03T

Divide is 15 times faster = 0.2T/15= 0.0133T

Total time = 0.5T + 0.03T + 0.0133T

= 0.54333T

Speed up = Old time/ new time

= T/0.54333T

= 1/0.54333

= 1.84

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Following is defined the required class having accessor methods and test program:

Explanation:

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