Question

Assume that a benchmark program executes in 480 seconds on a reference machine A. The same program executes on systems B, C, and D in 360, 540, and 210 seconds, respectively.

a. Show the speedup of each of the three systems under test relative to A.

b. Now show the relative speedup of the three systems. Comment on the three ways of comparing machines (execution time, speedup, relative speedup).

 

 

Answer 1

A)

Given a program, that is executed in the different machine i.e A,B,C,D.

 

The execution time of the system A,B,C,D are given as 480,360,540,210 respectively.

 

Speedup of each of the three systems under test relative to system A :-

 

(i) Speedup(B) = Execution Time of A / Execution Time of B

= 480 / 360

= 1.33

 

(ii) Speedup(C) = Execution Time of A / Execution Time of C

= 480 / 540

= 0.82

 

(iii) Speedup(D) = Execution Time of A / Execution Time of D

= 480 / 210

= 2.28

 

Relative Speedup :-

Relative Speedup = (480+480 480) / (360+ 540+ 210)

= 2.3

 

B)

Comparing the machines on different parameters :-

 

(1) Execution Time :- It is defined as the time taken by a machine for execution of the program.

 Execution Time of System A = 480 sec

Execution Time of System B= 360 sec

Execution Time of System C = 540 sec

Execution Time of System D = 210 sec

 

(2) Speedup :- It is defined as the measure of how better a system work as compare to the other system.

Speedup of B relative to A = 1.33

Speedup of C relative to A = 0.82

Speedup of D relative to A = 2.28

 

(3) Relative Speedup :- It is defined as the measure of all system work as compared to other system.

Relative Speedup of all system relative to A = 2.3

A)

 

Given a program, that is executed in the different machine i.e A,B,C,D.