I have made a selection form a huge amount of ID's, using the following query:
select ID from [tabelname] where id > 0 and id < 31
This gives me 30 ID's to work with.
What I would like to do now, is to use 3 threads, with the first one using ID 1, 4, 7, 10 etc, the second ID 2, 5, 8, 11 etc and the third one ID 3, 6, 9, 12 etc.
Up until now, I have only been able to have all threads use ID 1 through 30 parallel to each other. Would it be at all possible to do this?
Thanks in advance!
JMeter has a build-in operation that you can use in combination with a pre-processor to find the current thread number:
https://jmeter.apache.org/api/org/apache/jmeter/threads/JMeterContext.html#getThreadNum()
If you now use ctx.getThreadGroup().getNumThreads() to find the number of threads you're using, you can basically divide your testset into subsets and let each thread compute on its own subset (e.g. thread1 computes on 0..10, thread2 on 11..20, thread3 on 21..30, etc..)
Related
I am writing a model to calculate the maximum production capacity for a machine in a year based on 15-min data. As the maximum capacity is not the sum of the required capacity for all 15-min over the year, I want to write a piece of code that determines the maximum value in the list and then adds this maximum value and the three next consecutive values after this maximum value to a new variable. An simplified example would be:
fifteen_min_capacity = [10, 12, 3, 4, 8, 12, 10, 9, 2, 10, 4, 3, 15, 8, 9, 3, 4, 10]
The piece of code I want to write would be able to determine the maximum capacity in this list (15) and then add this capacity plus the three consecutive ones (8,9,3) to a new variables:
hourly_capacity = 35
Does anyone now the code that would give this output?
I have tried using the max(), the sum() and a combination of both. However, I do not get a working code. Any help would be much appreciated!
I have some activities with weights, and I would like to select non overlapping activities by maximizing the total weight. This is known problem and solution exists.
In my case, I am allowed to shift the start time of activities in some extend while duration remains same. This will give me some flexibility and I might increase my utilization.
Example scenario is something like the following where all activities are supposed to be in interval (0-200):
(start, end, profit)
a1: 10 12 120
a2: 10 13 100
a3: 14 18 150
a4: 14 20 100
a5: 120 125 100
a6: 120 140 150
a7: 126 130 100
Without shifting flexibility, I would choose (a1, a3, a6) and that is it. On the other hand I have shifting flexibility to the left/right by at most t units for any task where t is given. In that case I might come up with this schedule and all tasks can be selected except a7 since conflict cannot be avoided by shift .
t: 5
a1: 8 10 120 (shifted -2 to left)
a2: 10 13 100
a3: 14 18 150
a4: 18 24 100 (shifted +4 to right)
a5: 115 120 100 (shifted -5 to left)
a6: 120 140 150
In my problem, total time I have is very big with respect to activity duration. While activities are like 10sec on average, total time I have would even be 10000sec. However that does not mean all of activities can be selected since shifting flexibility would not be enough for some activities to non-overlap.
Also in my problem, there are clusters of activities which overlaps and very big empty space where no activities and there comes another cluster of overlapping activities i.e a1, a2, a3 and a4 are let say cluster1 and a5, a6 and a7 is cluster2. Each cluster can be expanded in time by shifting some of them to left and right. By doing that, I can select more activities than the original activity selection problem. However, I do not know how to decide which tasks to be shifted to left or right.
My expectation is to find an near-optimal solution where total profit would be somehow local optima. I do not need global optimum value. Also I do not have any criteria about cluster utilization., i.e I do not have a guarantee about a minimum number of activity per cluster etc. Actually, these clusters something I visually describe. There is not defined cluster. However, in time domain, activities are separated as clusters somehow.
Also activity start and end times are all integers since I can dismiss fractions. I would have around 50 activities whose duration would be 10 on average. And time window is like 10000.
Are there any feasible solution to this problem?
You mentioned that you can partition the activities into clusters that don't overlap even if activities within them are shifted to the extent. Each of these clusters can be considered independently, and the optimal results computed for each cluster simply summed up for the final answer. So the first step of the algorithm could be a trial run that extends all activities in both directions, finds which ones form clusters, and process each cluster independently. In the worst case, all of the activities might form a single cluster.
Depending on the maximum size of the remaining clusters, there are several approaches. If it's under 20 (or even 30, depending on whether you want your program to run in seconds or minutes), you could combine search over all subsets of activities in the given cluster with a greedy approach. In other words: if you are processing a subset of N elements, try every one of its 2^N possible subsets (okay, 2^N-1 if we forget the empty subset), check whether the activities in this specific subset can be scheduled in non-overlapping manner, and pick the subset that is eligible and has maximum sum.
How do we check that a given subset of activities can be scheduled in non-overlapping manner? Let's sort them in ascending order of their end and process them from left to right. For every activity, we try to schedule it as early as possible, making sure it does no intersect with activities we already considered. So, the first activity in the cluster is always started time t earlier than originally planned, the second one is started either when the first one ends, or t earlier than originally planned, whichever is larger, and so on. If at any point we can't schedule the next activity in a way that it does not overlap with previous one, then there is no way to schedule the activities in this subset in a non-overlapping manner. This algorithm takes O(NlogN) time, and overall each cluster is processed in O(2^N * NlogN). Once again, note that this function grows very quickly, so if you are dealing with large enough clusters, this approach goes out the window.
===
Another approach is specific to the additional restrictions you provided. If the activities' starts and ends and parameter t are all measured in integer number of seconds, and t is about 2 minutes, then the problem for each cluster is set in a small discrete space. Even though you could position a task to start at a non-integer second value, there always is an optimal solution that uses only integers. (To prove it, consider an optimal solution that does not use integers - since t is integer, you can always shift tasks, starting from the leftmost, to the left a bit so that it starts at an integer value.)
Knowing that the start and end times are discrete, you can build a DP solution: process the activities in the ascending order of their end*, and memoize the maximum possible sum of weights you can obtain from the first 1, 2, ..., N activities for each x from activity_start - t to activity_start + t if a given activity ends at time x. If we denote this memoized function as f[activity][end_time], then the recurrence relation is f[a][e] = weight[a] + max(f[i][j] over all i < a, j <= e - (end[a] - start[a]), which roughly translates to "if activity a ended at time e, the previous activity must have ended at or before start of a - so let's pick the maximum total weight over previous activities and their ends, and add the current activity's weight".
*Again, we can prove that there is at least one optimal answer where this ordering is preserved, even though there might be other optimal answers which do not possess this property
We could go further and eliminate the iteration over previous activities, instead encoding this information in f. Its definition would then change to "f[a][e] is the maximum possible total weight of the first a activities if none of them ends after e", and recurrence relation would become f[a][e] = max(f[a-1][e], weight[a] + max(f[a-1][i] over i <= e - (end[a] - start[a])])), and its computational complexity would be O(X * N), where X is the total span of the discrete space where task starts/ends are placed.
I assume you need to compute not just the maximum possible weight, but also the activities you need to select to obtain it, and possibly even the exact time each of them needs to be started. Thankfully, we can derive all of this from the values of f, or compute it at the same time as we compute f. The latter is easier to reason about, so let's introduce a second function g[activity][end]. g[activity][end] returns a pair (last_activity, last_activity_end), essentially pointing us to the exact activity and its timing that the optimal weight in f[activity][end] uses.
Let's go through the example you provided to illustrate how this works:
(start, end, profit)
a1: 10 12 120
a2: 10 13 100
a3: 14 18 150
a4: 14 20 100
a5: 120 125 100
a6: 120 140 150
a7: 126 130 100
We order the activities by their end time, thereby swapping a7 and a6.
We initialize the values of f and g for the first activity:
f[1][7] = 120, f[1][8] = 120, ..., f[1][17] = 120, meaning that the first activity could end anywhere from 7 to 17, and costs 120. f[1][i] for all other i should be set to 0.
g[1][7] = (1, 7), g[1][8] = (1, 8), ..., g[1][17] = (1, 17), meaning that the last activity that was included in f[1][i] values was a1, and it ended at i. g[1][i] for all i outside [7, 17] is undefined/irrelevant.
That's where something interesting begins. For each i such that a2 cannot end at time i, let's assign f[2][i] = f[1][i], g[2][i] = g[1][i], which essentially means that we wouldn't be using activity a2 in those answers. For all other i, namely, in [8..18] interval, we have:
f[2][8] = max(f[1][8], 100 + max(f[1][0..5])) = f[1][8]
f[2][9] = max(f[1][9], 100 + max(f[1][0..6])) = f[1][9]
f[2][10] = max(f[1][10], 100 + max(f[1][0..7])). This is the first time when the second clause is not just plain 100, as f[1][7]>0. It is, in fact, 100+f[1][7]=220, meaning that we can take activity a2, shift it in a way that puts its end at time 10, and get a total weight of 220. We continue computing f[2][i] this way for all i <= 18.
The values of g are: g[2][8]=g[1][8]=(1, 8), g[2][9]=g[1][9]=(1, 9), g[2][10]=(2, 10), because it was optimal to take activity a2 and end it at time 10 in this case.
I hope the pattern of how this continues is visible - we compute all the values of f and g through the end, and then pick the maximum f[N][e] over all possible end times e of the last activity. Armed with the auxiliary function g, we can traverse the values backwards to figure out the exact activities and times. Namely, the last activity we use and its timing is in g[N][e]. Let's call them A and T. We know that A began at T-(end[A]-start[A]). Then, the previous activity must have ended at that point or before - so let's look at g[A-1][T-(end[A]-start[A]) for it, and so on.
Note that this approach works even if you don't partition anything into clusters, but with the partitioning, the size of the space in which tasks can be scheduled is reduced, and with it the runtime.
You might notice that neither of these solutions is polynomial in the size of input. I have a feeling that your problem doesn't have a general polynomial solution, but I was unable to prove it by reducing another NP-complete problem to it. Would be really curious to read a reduction / better general solution!
In dask or even pandas how would you go about grouping an dask data frame that has a 3 columns of time / level / spread into a set of fixed ranges by time.
Time is only used to move one direction. Like a loop counting up. So the end result would be start time and end time with high of level, low of level, first value of level and last value of level over the fixed range? Example
12:00:00, 10, 1
12:00:01, 11, 1
12:00:02, 12, 1
12:00:03, 11, 1
12:00:04, 9, 1
12:00:05, 6, 1
12:00:06, 10, 1
12:00:07, 14, 1
12:00:08, 11, 1
12:00:09, 7, 1
12:00:10, 13, 1
12:00:11, 8, 1
For a fixed level range of (7). So level from start to end can not be more than 7 total distance from start to end for each bin of level. Just because first bin is only 8 difference in time and second is only 2 different in time, this dose not madder one the high to low madders that the total distance from high to low dose not go passed 7 the fixed bin size. The first bin could have been 5 not 8 for first bin and 200 for next bin not 2 in the example below. So the First few rows in dask would look something like this.
First Time, Last Time, High Level, Low Level, First Level, Last Level, Spread
12:00:00, 12:00:07, 13, 6, 10, 13, 1
12:00:07, 12:00:09, 14, 7, 13, 7, 1
12:00:09, X, 13, 7, X, X, X
How could this be aggregated in dask with a fix window of level moving forward in time binning each time level moves above X or equal too high/low with in X or below X?
I have forked many child processes and assigned priority and core to each of them. Porcess A executes at period of 3 sec and process B at a period of 6 sec. I want them to execute in such a way that the higher priority processes should preempt lower priority ones only at predefined points and tried to acheive it with semaphores. I have used this same code snippets within the 2 processes with different array values in both.
'bubblesort_desc()' sorts the array in descending order and prints it. 'bubblesort_asc()' sorts in ascending order and prints.
while(a<3)
{
printf("time in sort1.c: %d %ld\n", (int)request.tv_sec, (long int)request.tv_nsec);
int array[SIZE] = {5, 1, 6 ,7 ,9};
semaphore_wait(global_sem);
bubblesort_desc(array, SIZE);
semaphore_post(global_sem);
semaphore_wait(global_sem);
bubblesort_asc(array, SIZE);
semaphore_post(global_sem);
semaphore_wait(global_sem);
a++;
request.tv_sec = request.tv_sec + 6;
request.tv_nsec = request.tv_nsec; //if i add 1ms here like an offset to the lower priority one, it works.
semaphore_post(global_sem);
semaphore_close(global_sem); //close the semaphore file
//sleep for the rest of the time after the process finishes execution until the period of 6
clk = clock_nanosleep(CLOCK_MONOTONIC, TIMER_ABSTIME, &request, NULL);
if (clk != 0 && clk != EINTR)
printf("ERROR: clock_nanosleep\n");
}
I get the output like this whenever two processes get activated at the same time. For example at time units of 6, 12,..
time in sort1.c: 10207 316296689
time now in sort.c: 10207 316296689
9
99
7
100
131
200
256
6
256
200
5
131
100
99
1
1
5
6
7
9
One process is not supposed to preempt the other while one set of sorted list is printing. But it's working as if there are no semaphores. I defined semaphores as per this link: http://linux.die.net/man/3/pthread_mutexattr_init
Can anyone tell me what can be the reason for that? Is there a better alternative than semaphores?
Its printf that's causing the ambiguous output. If the results are printed without '\n' then we get a more accurate result. But its always better to avoid printf statements for real time applications. I used trace-cmd and kernelshark to visualize the behaviour of the processes.
Imagine 3 circles. Each circle has some numbers
Circle 1 has the following numbers
1, 4, 7, 9
Circle 2 has the following numbers
2, 5, 8, 9
Circle 3 has the following numbers
3, 6, 7, 8, 9
Circle 1 and Circle 2 share the following numbers
10, 9
Circle 1 and Circle 3 share the following numbers
7, 9
Circle 2 and Circle 3 share the following numbers
8, 9
All three circles share
9
Each number represents symptoms so in my case
Circle 1's numbers could be symptoms for a short circuit
Circle 2 could be numbers for a component failure
Circle 3 could be numbers for external issues
each of the three issues share certain symptoms
If given #9, we wouldn't be able to deduce the problem but could display a list of all issues involving #9
If given more #'s, we can attempt to show relevant issues.
My problem is how do I put this into a table so my code can look things up.
My Database of choice is SQLite3
#Vincent, the only issue I have is that there are several variables. I have variables called t1, t2, t3, a1, a2, a3. Each of these variables are symptoms. The user interface for my application allows the user to input a value for each variable then I want to check the DB. All values for each symptom can be any value in the 3 circles (mentioned in original problem)
Create 3 tables as:
symptom = (symptom_id, descr)
problem = (problem_id, descr)
problem_symptom = (problem_id, symptom_id)
e.g.
Symptom
Symptom_id Desc
1 doda
2 dado
3 dada
Problem
Problem_id Descr
1 Short Circuit
2 Component Failure
Symptom_Problem
Symptom_id Problem_id
1 1 --- doda is a symptom of Short circuit
2 1 --- dado is a symptom of short circuit
etc.
You can then query and join to determine the problems based on the symptoms.
I would suggest 3 tables: symptom, problem, and problem_symptoms. The latter would be a join table between the first two.