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Estimate the average time when the value of a given parameter is increasing


Estimate the average time when the value of a given parameter is increasing

By : user2948509
Date : November 14 2020, 04:51 PM
I hope this helps you . There is the following array track which indicates the increase of some parameter. Each row refers to 5 days: , change the inner while to this:
code :
while ( curr< track.length && track[curr]>track[curr-1])
{
     temptime += 5;
     stop = false;
     curr++;
}


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How to estimate Inference time from average forward pass time in caffe?

How to estimate Inference time from average forward pass time in caffe?


By : dennice martinez
Date : March 29 2020, 07:55 AM
should help you out The average forward pass time is the time it takes to propagate one batch of inputs from the input ("data") layer to the output layer. The batch size specified in your models/own_xx/deploy.prototxt file will determine how many images are processed per batch.
For instance, if I run the default command that comes with Caffe:
code :
build/tools/caffe time --model=models/bvlc_alexnet/deploy.prototxt --gpu=0
...
I0426 13:07:32.701490 30417 layer_factory.hpp:77] Creating layer data
I0426 13:07:32.701513 30417 net.cpp:91] Creating Layer data
I0426 13:07:32.701529 30417 net.cpp:399] data -> data
I0426 13:07:32.709048 30417 net.cpp:141] Setting up data
I0426 13:07:32.709079 30417 net.cpp:148] Top shape: 10 3 227 227 (1545870)
I0426 13:07:32.709084 30417 net.cpp:156] Memory required for data: 6183480
...
I0426 13:07:34.390281 30417 caffe.cpp:377] Average Forward pass: 16.7818 ms.
I0426 13:07:34.390290 30417 caffe.cpp:379] Average Backward pass: 12.923 ms.
I0426 13:07:34.390296 30417 caffe.cpp:381] Average Forward-Backward: 29.7969 ms.
I0426 13:07:32.709079 30417 net.cpp:148] Top shape: 10 3 227 227 (1545870)
name: "AlexNet"
layer {
  name: "data"
  type: "Input"
  top: "data"
  input_param { shape: { dim: 10 dim: 3 dim: 227 dim: 227 } }
}
layer { ...
Bootstrap parameter estimate of non-linear optimization in R: Why is it different than the regular parameter estimate?

Bootstrap parameter estimate of non-linear optimization in R: Why is it different than the regular parameter estimate?


By : Siva Prasad
Date : March 29 2020, 07:55 AM
seems to work fine First of all, you have a very small number of values, possibly too few to trust the bootstrap method. Then a high proportion of fits fails for the classic bootstrap, because due to the resampling you often have not enough distinct x values.
Here is an implementation using nls with a selfstarting model and the boot package.
code :
doy <- c(156,205,228,276,319,380)
len <- c(36,56,60,68,68,71)
data06 <- data.frame(doy,len)

plot(len ~ doy, data = data06)

fit <- nls(len ~ SSasympOff(doy, Asym, lrc, c0), data = data06)
summary(fit)
#profiling CI
proCI <- confint(fit)
#          2.5%      97.5%
#Asym 68.290477  75.922174
#lrc  -4.453895  -3.779994
#c0   94.777335 126.112523

curve(predict(fit, newdata = data.frame(doy = x)), add = TRUE)
#classic bootstrap
library(boot)
set.seed(42)
boot1 <- boot(data06, function(DF, i) {
  tryCatch(coef(nls(len ~ SSasympOff(doy, Asym, lrc, c0), data = DF[i,])),
           error = function(e) c(Asym = NA, lrc = NA, c0 = NA))
}, R = 1e3)

#proportion of unsuccessful fits
mean(is.na(boot1$t[, 1]))
#[1] 0.256

#bootstrap CI
boot1CI <- apply(boot1$t, 2, quantile, probs = c(0.025, 0.5, 0.975), na.rm = TRUE)
#          [,1]      [,2]      [,3]
#2.5%  69.70360 -4.562608  67.60152
#50%   71.56527 -4.100148 113.9287
#97.5% 74.79921 -3.697461 151.03541


#bootstrap of the residuals
data06$res <- residuals(fit)
data06$fit <- fitted(fit)

set.seed(42)
boot2 <- boot(data06, function(DF, i) {
  DF$lenboot <- DF$fit + DF[i, "res"]
  tryCatch(coef(nls(lenboot ~ SSasympOff(doy, Asym, lrc, c0), data = DF)),
           error = function(e) c(Asym = NA, lrc = NA, c0 = NA))
}, R = 1e3)

#proportion of unsuccessful fits
mean(is.na(boot2$t[, 1]))
#[1] 0

#(residuals) bootstrap CI
boot2CI <- apply(boot2$t, 2, quantile, probs = c(0.025, 0.5, 0.975), na.rm = TRUE)
#          [,1]      [,2]     [,3]
#2.5%  70.19380 -4.255165 106.3136
#50%   71.56527 -4.100148 113.9287
#97.5% 73.37461 -3.969012 119.2380
proCI[2,1]

CIs_k <- data.frame(lwr = c(exp(proCI[2, 1]),
                            exp(boot1CI[1, 2]),
                            exp(boot2CI[1, 2])),
                    upr = c(exp(proCI[2, 2]),
                            exp(boot1CI[3, 2]),
                            exp(boot2CI[3, 2])),
                    med = c(NA,
                            exp(boot1CI[2, 2]),
                            exp(boot2CI[2, 2])),
                    estimate = exp(coef(fit)[2]),
                    method = c("profile", "boot", "boot res"))

library(ggplot2)
ggplot(CIs_k, aes(y = estimate, ymin = lwr, ymax = upr, x = method)) +
  geom_errorbar() +
  geom_point(aes(color = "estimate"), size = 5) +
  geom_point(aes(y = med, color = "boot median"), size = 5) +
  ylab("k") + xlab("") +
  scale_color_brewer(name = "", type = "qual", palette = 2) +
  theme_bw(base_size = 22)
Excel, Increasing Time Series find the average time between adjacent points

Excel, Increasing Time Series find the average time between adjacent points


By : Vittawat Sangphung
Date : March 29 2020, 07:55 AM
should help you out I have a list as follows , One formula:
code :
=SUMPRODUCT(AVERAGE(J2:INDEX(J:J,MATCH(1E+99,J:J))-J1:INDEX(J:J,MATCH(1E+99,J:J)-1)))
How to estimate the average-case complexity given the input size and average time?

How to estimate the average-case complexity given the input size and average time?


By : swissjava
Date : March 29 2020, 07:55 AM
To fix this issue No it is not possible to find the time complexity with just this information. Without looking at the code, determining the time complexity is not possible in this case.
One (of the many) simple examples where this time and input analysis would not hold good is if you have certain if statements in your code that cause something to execute that the inputs you have provided do not take into consideration.
how to estimate best, worst and average cases for time complexity?

how to estimate best, worst and average cases for time complexity?


By : Pravin Gosavi
Date : March 29 2020, 07:55 AM
this will help first note : t(n) = 2n^2 + 3n -1 will always be a big O(n^2) in worst, best and average case.
In some cases the complexity depends on the input of your algorithm, in these cases usually people calculate the complexity in worst case.
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