Below are the results
sampler_label count average median 90%_line min max
Transaction1 2 61774 61627 61921 61627 61921
Transaction2 4 82 61 190 15 190
Transaction3 4 1862 1317 3612 1141 3612
Transaction4 4 1242 915 1602 911 1602
Transaction5 4 692 608 906 423 906
Transaction6 4 2764 2122 4748 1182 4748
Transaction7 4 9369 9029 11337 7198 11337
Transaction8 4 1245 890 2168 834 2168
Transaction9 4 3475 2678 4586 2520 4586
TOTAL 34 6073 1381 9913 15 61921
My question here is how is total average response time is being calculated (which is 6073)?
Like in my result I want to exclude transaction1 response time and then want to calculate Total average response time.
How can I do that?
Total Avg Response time = ((s1*t1) + (s2*t2)...)/s
s1 = No of times transaction 1 was executed
t1 = Avg response time for transaction 1
s2 = No of times transaction 2 was executed
t2 = Avg response time for transaction 2
s = Total no of samples (s1+s2..)
In your case, except transaction1 all other transactions have been executed 4 times. So, simple avg of (82, 1862, 1242...) should give the result you wanted.
Related
I want to understand engine log of IBM ILOG CPLEX studios for a ILP model. I have checked there documentation also but could not able to get clear idea.
Example of Engine log :
Version identifier: 22.1.0.0 | 2022-03-09 | 1a383f8ce
Legacy callback pi
Tried aggregator 2 times.
MIP Presolve eliminated 139 rows and 37 columns.
MIP Presolve modified 156 coefficients.
Aggregator did 11 substitutions.
Reduced MIP has 286 rows, 533 columns, and 3479 nonzeros.
Reduced MIP has 403 binaries, 0 generals, 0 SOSs, and 129 indicators.
Presolve time = 0.05 sec. (6.16 ticks)
Found incumbent of value 233.000000 after 0.07 sec. (9.40 ticks)
Probing time = 0.00 sec. (1.47 ticks)
Tried aggregator 2 times.
Detecting symmetries...
Aggregator did 2 substitutions.
Reduced MIP has 284 rows, 531 columns, and 3473 nonzeros.
Reduced MIP has 402 binaries, 129 generals, 0 SOSs, and 129 indicators.
Presolve time = 0.01 sec. (2.87 ticks)
Probing time = 0.00 sec. (1.45 ticks)
Clique table members: 69.
MIP emphasis: balance optimality and feasibility.
MIP search method: dynamic search.
Parallel mode: deterministic, using up to 8 threads.
Root relaxation solution time = 0.00 sec. (0.50 ticks)
Nodes Cuts/
Node Left Objective IInf Best Integer Best Bound ItCnt Gap
* 0+ 0 233.0000 18.0000 92.27%
* 0+ 0 178.0000 18.0000 89.89%
* 0+ 0 39.0000 18.0000 53.85%
0 0 22.3333 117 39.0000 22.3333 4 42.74%
0 0 28.6956 222 39.0000 Cuts: 171 153 26.42%
0 0 31.1543 218 39.0000 Cuts: 123 251 20.12%
0 0 32.1544 226 39.0000 Cuts: 104 360 17.55%
0 0 32.6832 212 39.0000 Cuts: 102 456 16.20%
0 0 33.1524 190 39.0000 Cuts: 65 521 14.99%
Detecting symmetries...
0 0 33.3350 188 39.0000 Cuts: 66 566 14.53%
0 0 33.4914 200 39.0000 Cuts: 55 614 14.12%
0 0 33.6315 197 39.0000 Cuts: 47 673 13.77%
0 0 33.6500 207 39.0000 Cuts: 61 787 13.72%
0 0 33.7989 206 39.0000 Cuts: 91 882 13.34%
* 0+ 0 38.0000 33.7989 11.06%
0 0 33.9781 209 38.0000 Cuts: 74 989 10.58%
0 0 34.0074 209 38.0000 Cuts: 65 1043 10.51%
0 0 34.2041 220 38.0000 Cuts: 63 1124 9.99%
0 0 34.2594 211 38.0000 Cuts: 96 1210 9.84%
0 0 34.3032 216 38.0000 Cuts: 86 1274 9.73%
0 0 34.3411 211 38.0000 Cuts: 114 1353 9.63%
0 0 34.3420 220 38.0000 Cuts: 82 1402 9.63%
0 0 34.3709 218 38.0000 Cuts: 80 1462 9.55%
0 0 34.4494 228 38.0000 Cuts: 87 1530 9.34%
0 0 34.4882 229 38.0000 Cuts: 97 1616 9.24%
0 0 34.5173 217 38.0000 Cuts: 72 1663 9.16%
0 0 34.5545 194 38.0000 Cuts: 67 1731 9.07%
0 0 34.5918 194 38.0000 Cuts: 76 1786 8.97%
0 0 34.6094 199 38.0000 Cuts: 73 1840 8.92%
0 0 34.6226 206 38.0000 Cuts: 77 1883 8.89%
0 0 34.6421 206 38.0000 Cuts: 53 1928 8.84%
0 0 34.6427 213 38.0000 Cuts: 84 1982 8.83%
Detecting symmetries...
0 2 34.6427 213 38.0000 34.6478 1982 8.82%
Elapsed time = 0.44 sec. (235.86 ticks, tree = 0.02 MB, solutions = 4)
GUB cover cuts applied: 32
Cover cuts applied: 328
Implied bound cuts applied: 205
Flow cuts applied: 11
Mixed integer rounding cuts applied: 17
Zero-half cuts applied: 35
Gomory fractional cuts applied: 1
Root node processing (before b&c):
Real time = 0.43 sec. (235.61 ticks)
Parallel b&c, 8 threads:
Real time = 0.27 sec. (234.23 ticks)
Sync time (average) = 0.11 sec.
Wait time (average) = 0.00 sec.
------------
Total (root+branch&cut) = 0.71 sec. (469.84 ticks)
Mainly I want to understand what are nodes,left,gap,root node processing, parallel b&c.
I hope anyone of you will give a resource or explain it clearly so that it can be helpful when someone starts using IBM ILOG CPLEX studio in future
Thanks a lot in advance
I am expecting for someone to fill knowledge gaps regarding Engine log of IBMs ILOG CPLEX studio
I recommend
Progress reports: interpreting the node log
https://www.ibm.com/docs/en/icos/12.8.0.0?topic=mip-progress-reports-interpreting-node-log
Similar to
How to assign an ID to a group of variables
My dataset is sorted by ID, then timestamp. I need to create an 'order' variable, incrementing on each change in say Status, but my sort must remain time stamp, so I think I am correct in suggesting that by BY (group) will not work. The order field below illustrates what I seek...
ID Status Timestamp Order
188 3 12:15 1
188 4 12:45 2
188 4 13:10 2
188 3 14:20 3
189 10 11:00 1
189 11 13:00 2
189 10 13:30 3
189 10 13:35 3
The first and second '3's are separate, likewise the first and subsequent '10's.
You can use the NOTSORTED option to have SAS automatically set the FIRST.STATUS flag for you.
data want ;
set have ;
by id status notsorted;
if first.id then order=0;
order + first.status;
run;
As you mentioned, it is very similar to that other question. The trick here is to set the order of the first observation in each by group to zero.
data temp;
input ID $ Status $ Timestamp $;
datalines;
188 3 12:15
188 4 12:45
188 4 13:10
188 3 14:20
189 10 11:00
189 11 13:00
189 10 13:30
189 10 13:35
;
run;
data temp2;
set temp;
by id;
if first.id then order = 0;
if status ~= lag(status) then order + 1;
run;
My data looks like below and I would like to group it by zip. Along with the grouping I would like the lower, avg, upper, sum of idk, and avgDiff columns. The two calculated fields would be the Total Count by zip and sum of 'idk' (the table below this one). I would specifically like to use DataTable for this....thank you.
zip lower avg upper RISK idk diff avgDiff total
1: 12007 -170.3723 592 1354.372 676 0 84 137.2903 123
2: 12007 -170.3723 592 1354.372 828 1 236 137.2903 123
3: 12007 -170.3723 592 1354.372 627 1 35 137.2903 123
4: 12009 -150.3723 300 1200.372 770 1 178 125.2903 456
5: 12007 -170.3723 592 1354.372 770 1 178 137.2903 123
6: 12010 -100.3723 200 1100.372 893 1 301 300.2903 890
desired result
zip lower avg upper zipCount avgDiff sumidk
1: 12007 -170.3723 592 1354.372 4 137.2903 3
2: 12009 -150.3723 300 1200.372 1 125.2903 1
3: 12010 -100.3723 200 1100.372 1 300.2903 1
The lower, avg, upper, and avgDiff will be the same within the zip.
So far I have DT[, .(zipcount =.N), by = zip]....which is grouping zip and giving me the total zips (rows), but I'm getting stuck at this point.
thank you
Alright. Now this question is pretty hard. I am going to give you an example.
Now the left numbers are my algorithm classification and the right numbers are the original class numbers
177 86
177 86
177 86
177 86
177 86
177 86
177 86
177 86
177 86
177 89
177 89
177 89
177 89
177 89
177 89
177 89
So here my algorithm merged 2 different classes into 1. As you can see it merged class 86 and 89 into one class. So what would be the error at the above example ?
Or here another example
203 7
203 7
203 7
203 7
16 7
203 7
17 7
16 7
203 7
At the above example left numbers are my algorithm classification and the right numbers are original class ids. As can be seen above it miss classified 3 products (i am classifying same commercial products). So at this example what would be the error rate? How would you calculate.
This question is pretty hard and complex. We have finished the classification but we are not able to find correct algorithm for calculating success rate :D
Here's a longish example, a real confuson matrix with 10 input classes "0" - "9"
(handwritten digits),
and 10 output clusters labelled A - J.
Confusion matrix for 5620 optdigits:
True 0 - 9 down, clusters A - J across
-----------------------------------------------------
A B C D E F G H I J
-----------------------------------------------------
0: 2 4 1 546 1
1: 71 249 11 1 6 228 5
2: 13 5 64 1 13 1 460
3: 29 2 507 20 5 9
4: 33 483 4 38 5 3 2
5: 1 1 2 58 3 480 13
6: 2 1 2 294 1 1 257
7: 1 5 1 546 6 7
8: 415 15 2 5 3 12 13 87 2
9: 46 72 2 357 35 1 47 2
----------------------------------------------------
580 383 496 1002 307 670 549 557 810 266 estimates in each cluster
y class sizes: [554 571 557 572 568 558 558 566 554 562]
kmeans cluster sizes: [ 580 383 496 1002 307 670 549 557 810 266]
For example, cluster A has 580 data points, 415 of which are "8"s;
cluster B has 383 data points, 249 of which are "1"s; and so on.
The problem is that the output classes are scrambled, permuted;
they correspond in this order, with counts:
A B C D E F G H I J
8 1 4 3 6 7 0 5 2 6
415 249 483 507 294 546 546 480 460 257
One could say that the "success rate" is
75 % = (415 + 249 + 483 + 507 + 294 + 546 + 546 + 480 + 460 + 257) / 5620
but this throws away useful information —
here, that E and J both say "6", and no cluster says "9".
So, add up the biggest numbers in each column of the confusion matrix
and divide by the total.
But, how to count overlapping / missing clusters,
like the 2 "6"s, no "9"s here ?
I don't know of a commonly agreed-upon way
(doubt that the Hungarian algorithm
is used in practice).
Bottom line: don't throw away information; look at the whole confusion matrix.
NB such a "success rate" will be optimistic for new data !
It's customary to split the data into say 2/3 "training set" and 1/3 "test set",
train e.g. k-means on the 2/3 alone,
then measure confusion / success rate on the test set — generally worse than on the training set alone.
Much more can be said; see e.g.
Cross-validation.
You have to define the error criteria if you want to evaluate the performance of an algorithm, so I'm not sure exactly what you're asking. In some clustering and machine learning algorithms you define the error metric and it minimizes it.
Take a look at this
https://en.wikipedia.org/wiki/Confusion_matrix
to get some ideas
You have to define a error metric to measure yourself. In your case, a simple method should be to find the properties mapping of your product as
p = properties(id)
where id is the product id, and p is likely be a vector with each entry of different properties. Then you can define the error function e (or distance) between two products as
e = d(p1, p2)
Sure, each properties must be evaluated to a number in this function. Then this error function can be used in the classification algorithm and learning.
In your second example, it seems that you treat the pair (203 7) as successful classification, so I think you have already a metric yourself. You may be more specific to get better answer.
Classification Error Rate(CER) is 1 - Purity (http://nlp.stanford.edu/IR-book/html/htmledition/evaluation-of-clustering-1.html)
ClusterPurity <- function(clusters, classes) {
sum(apply(table(classes, clusters), 2, max)) / length(clusters)
}
Code of #john-colby
Or
CER <- function(clusters, classes) {
1- sum(apply(table(classes, clusters), 2, max)) / length(clusters)
}
Note: I am still looking for a fast solution. Two of the solutions below are wrong and the third one is terribly slow.
I have N toys from 1....N. Each toy has an associated cost with it. You have to go on a shopping spree such that on a particular day, if you buy toy i, then the next toy you can buy on the same day should be i+1 or greater. Moreover, the absolute cost difference between any two consecutively bought toys should be greater than or equal to k. What is the minimum number of days can I buy all the toys.
I tried a greedy approach by starting with toy 1 first and then seeing how many toys can I buy on day 1. Then, I find the smallest i that I have not bought and start again from there.
Example:
Toys : 1 2 3 4
Cost : 5 4 10 15
let k be 5
On day 1, buy 1,3, and 4
on day 2, buy toy 2
Thus, I can buy all toys in 2 days
Note greedy not work for below example: N = 151 and k = 42
the costs of the toys 1...N in that order are :
383 453 942 43 27 308 252 721 926 116 607 200 195 898 568 426 185 604 739 476 354 533 515 244 484 38 734 706 608 136 99 991 589 392 33 615 700 636 687 625 104 293 176 298 542 743 75 726 698 813 201 403 345 715 646 180 105 732 237 712 867 335 54 455 727 439 421 778 426 107 402 529 751 929 178 292 24 253 369 721 65 570 124 762 636 121 941 92 852 178 156 719 864 209 525 942 999 298 719 425 756 472 953 507 401 131 150 424 383 519 496 799 440 971 560 427 92 853 519 295 382 674 365 245 234 890 187 233 539 257 9 294 729 313 152 481 443 302 256 177 820 751 328 611 722 887 37 165 739 555 811
You can find the optimal solution by solving the asymmetric Travelling Salesman.
Consider each toy as a node, and build the complete directed graph (that is, add an edge between each pair of nodes). The edge has cost 1 (has to continue on next day) if the index is smaller or the cost of the target node is less than 5 plus the cost of the source node, and 0 otherwise. Now find the shortest path covering this graph without visiting a node twice - i.e., solve the Travelling Salesman.
This idea is not very fast (it is in NP), but should quickly give you a reference implementation.
This is not as difficult as ATSP. All you need to do is look for increasing subsequences.
Being a mathematician, the way I would solve the problem is to apply RSK to get a pair of Young tableaux, then the answer for how many days is the height of the tableau and the rows of the second tableau tell you what to purchase on which day.
The idea is to do Schensted insertion on the cost sequence c. For the example you gave, c = (5, 4, 10, 15), the insertion goes like this:
Step 1: Insert c[1] = 5
P = 5
Step 2: Insert c[2] = 4
5
P = 4
Step 3: Insert c[3] = 10
5
P = 4 10
Step 4: Insert c[4] = 15
5
P = 4 10 15
The idea is that you insert the entries of c into P one at a time. When inserting c[i] into row j:
if c[i] is bigger than the largest element in the row, add it to the end of the row;
otherwise, find the leftmost entry in row j that is larger than c[i], call it k, and replace k with c[i] then insert k into row j+1.
P is an array where the lengths of the rows are weakly decreasing and The entries in each of row P (these are the costs) weakly increase. The number of rows is the number of days it will take.
For a more elaborate example (made by generating 9 random numbers)
1 2 3 4 5 6 7 8 9
c = [ 5 4 16 7 11 4 13 6 5]
16
7
5 6 11
P = 4 4 5 13
So the best possible solution takes 4 days, buying 4 items on day 1, 3 on day 2, 1 on day 3, and 1 on day 4.
To handle the additional constraint that consecutive costs must increase by at least k involves redefining the (partial) order on costs. Say that c[i] <k< c[j] if and only if c[j]-c[i] >= k in the usual ordering on numbers. The above algorithm works for partial orders as well as total orders.
I somewhat feel that a greedy approach would give a fairly good result.
I think your approach is not optimal just because you always pick toy 1 to start while you should really pick the least expensive toy. Doing so would give you the most room to move to the next toy.
Each move being the least expensive one, it is just DFS problem where you always follow the least expensive path constrained by k.