jedifran · December 13, 2011 02:04 · Dec 13, 2011 · Dec 13, 2011
diff --git a/upstream_challenge.R b/upstream_challenge.R
@@ -22,3 +22,24 @@ expectation <- mean(unlist(lapply(1:2000000, function(x) upstream.fxn(test.set))
 
 # the required function will tend towards 8.82 in the long run. 
 #i.e the expectation of min(sample(test.set,6)) is approximately 8.82
+
+
+
+
+#alternative method using a loop - which takes considerably longer to run!
+aa <- 0
+for(it in 1:20000){
+	aa[it] <- mean(unlist(lapply(1:it, function(x) upstream.fxn(test.set))))
+}
+
+mean(aa)
+> mean(aa)
+[1] 8.818985
+
+summary(aa)
+> summary(aa)
+   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
+  4.000   8.768   8.817   8.819   8.867  23.000 
+
+
+plot(c(1:20000),aa, type="l")  
diff --git a/upstream_challenge.R b/upstream_challenge.R
@@ -0,0 +1,24 @@
+
+
+#Challenge:
+#Given the following set of numbers {49,8,48,15,47,4,16,23,43,44,42,45,46}. A function picks a random subset of size 6, and takes the minimum, what is the expected value of this function.
+
+#put set of numbers into a container
+test.set <- c(49,8,48,15,47,4,16,23,43,44,42,45,46)
+
+#This function picks a random subset of size 6 and takes the minimum
+upstream.fxn <- function(test.set) {min(sample(test.set,6));}
+
+#to find the expectation of this function (by simulation) we resample a large number of times:
+expectation <- mean(unlist(lapply(1:2000000, function(x) upstream.fxn(test.set))))
+
+
+> system.time(expectation <- mean(unlist(lapply(1:20000000, function(x) upstream.fxn(test.set)))))
+   user  system elapsed 
+721.370  11.186 846.715 
+> expectation
+[1] 8.818875
+
+
+# the required function will tend towards 8.82 in the long run. 
+#i.e the expectation of min(sample(test.set,6)) is approximately 8.82
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