Question

In: Statistics and Probability

Compare the execution times of the three equivalent R commands (a) y=c();for (t in 1:Nsim) y[t]=sin(t)...

Compare the execution times of the three equivalent R commands
(a) y=c();for (t in 1:Nsim) y[t]=sin(t)
(b) y=sin(1:Nsim)
(c) y=sapply(1:Nsim,sin)
using system.time() with Nsim = 104; 105; 106; 107.

Solutions

Expert Solution

1)

Nsim = 104

start.time <- Sys.time()

y=c();for (t in 1:Nsim) y[t]=sin(t)

end.time <- Sys.time()

time.taken.a <- end.time - start.time

time.taken.a

Time difference of 0.0003597736 secs

start.time <- Sys.time()

y=sin(1:Nsim)

end.time <- Sys.time()

time.taken.b<- end.time - start.time

time.taken.b

Time difference of 0.0001106262 secs

start.time <- Sys.time()

y=sapply(1:Nsim,sin)

end.time <- Sys.time()

time.taken.c <- end.time - start.time

time.taken.c

Time difference of 0.0003988743 secs

time.taken.c>time.taken.a& time.taken.a>time.taken.b

[1] TRUE

#exection time is for (c) is greater than execution time for (a) and (c)

2)
Nsim = 105
start.time <- Sys.time()
y=c();for (t in 1:Nsim) y[t]=sin(t)
end.time <- Sys.time()
time.taken.a <- end.time - start.time
time.taken.a

Time difference of 0.0003886223 secs

start.time <- Sys.time()
y=sin(1:Nsim)
end.time <- Sys.time()
time.taken.b<- end.time - start.time
time.taken.b

Time difference of 0.0001151562 secs
start.time <- Sys.time()

y=sapply(1:Nsim,sin)
end.time <- Sys.time()
time.taken.c <- end.time - start.time
time.taken.c

Time difference of 0.0004687309 secs

time.taken.c>time.taken.a& time.taken.a>time.taken.b

[1] TRUE

#exection time is for (c) is greater than execution time for (a) and (c)

3)

Nsim = 106

start.time <- Sys.time()

y=c();for (t in 1:Nsim) y[t]=sin(t)

end.time <- Sys.time()

time.taken.a <- end.time - start.time

time.taken.a

Time difference of 0.0004012585 secs

start.time <- Sys.time()

y=sin(1:Nsim)

end.time <- Sys.time()

time.taken.b<- end.time - start.time

time.taken.b

Time difference of 0.0001111031 secs

start.time <- Sys.time()

y=sapply(1:Nsim,sin)

end.time <- Sys.time()

time.taken.c <- end.time - start.time

time.taken.c

Time difference of 0.0005280972 secs

time.taken.c>time.taken.a& time.taken.a>time.taken.b

[1] TRUE

#exection time is for (c) is greater than execution time for (a) and (c)

4)
Nsim = 107
start.time <- Sys.time()
y=c();for (t in 1:Nsim) y[t]=sin(t)
end.time <- Sys.time()
time.taken.a <- end.time - start.time
time.taken.a

Time difference of 0.0003688335 secs

start.time <- Sys.time()
y=sin(1:Nsim)
end.time <- Sys.time()
time.taken.b<- end.time - start.time
time.taken.b

Time difference of 0.0001432896 secs
start.time <- Sys.time()

y=sapply(1:Nsim,sin)
end.time <- Sys.time()
time.taken.c <- end.time - start.time
time.taken.c

Time difference of 0.0003988743 secs

time.taken.c>time.taken.a& time.taken.a>time.taken.b

[1] TRUE

#exection time is for (c) is greater than execution time for (a) and (c)

# From values Nsim = 104; 105; 106; 107 it is conclude that

The execution time required for (c) y=sapply(1:Nsim,sin) is greater than (a) y=c();for (t in 1:Nsim) y[t]=sin(t) is greater than (b) y=sin(1:Nsim)

i.e. The minimum executiion time is required for (b) y=sin(1:Nsim) other than (a) & (c)

orchestra answered 1 year ago
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