resource-scheduling | 易学教程

kubernetes scheduling for expensive resources

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问题 We have a Kubernetes cluster. Now we want to expand that with GPU nodes (so that would be the only nodes in the Kubernetes cluster that have GPUs). We'd like to avoid Kubernetes to schedule pods on those nodes unless they require GPUs. Not all of our pipelines can use GPUs. The absolute majority are still CPU-heavy only. The servers with GPUs could be very expensive (for example, Nvidia DGX could be as much as $150/k per server). If we just add DGX nodes to Kubernetes cluster, then Kubernetes

Algorithm to find maximum coverage of non-overlapping sequences. (I.e., the Weighted Interval Scheduling Prob.)

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问题 I have a question that is very similar to algorithm to find longest non-overlapping sequences. The only difference to the linked question is that instead of finding the set of non-overlapping tuples that represent the longest sequence , I need to find the set of non-overlapping tuples that represent the maximum coverage , by which I mean the sum of the tuple lengths is maximum (a tuple length being last - first + 1 given the definition of tuple in the next sentence). I represent my tuples

Changing start dates of schedules to optimize resources

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问题 I have a bunch of work that needs to be performed at specific time intervals. However, we have limited resources to do that work, each day. Therefore, I am trying to optimize the start time dates (start time dates can only be moved forward not backward) so that resources used everyday are more less similar to what we have budgeted for. These functions are used in example below:: # Function to shift/rotate a vector shifter <- function(x, n = 1) { if (n == 0) x else c(tail(x, -n), head(x, n)) }

Optaplanner List Planning variable

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问题 Downloaded OptaPlanner examples, testing them. In the Meeting Scheduling example provided, a single planning variable is returned in the solved solution. for example, for a meeting scheduling, one Room is returned. I am planning to allocate multiple rooms for a single meeting schedule if the locations are different (assuming VC enabled for remote rooms). Current Meeting Scheduling example does not show this. Do I need to change the planning variable room to a List? 回答1: Planning variables can

kubernetes scheduling for expensive resources

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We have a Kubernetes cluster. Now we want to expand that with GPU nodes (so that would be the only nodes in the Kubernetes cluster that have GPUs). We'd like to avoid Kubernetes to schedule pods on those nodes unless they require GPUs. Not all of our pipelines can use GPUs. The absolute majority are still CPU-heavy only. The servers with GPUs could be very expensive (for example, Nvidia DGX could be as much as $150/k per server). If we just add DGX nodes to Kubernetes cluster, then Kubernetes would schedule non-GPU workloads there too, which would be a waste of resources (e.g. other jobs that

Prolog: foreach or forall for constraint solving?

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问题 I'm attempting project scheduling with SWI prolog and CLP. I managed to support sequential dependencies but I'm struggling with avoiding double booking people. I have a list called Schedule containing elements like [taskname, starttime] where starttime is a free variable for the constraint solver. They're already constrained by sequential dependencies. I'm trying to write a loop like this to rule out double bookings: forall /* or maybe foreach*/ (isa(P,person), ( % Filter scheduled tasks on

Solving task scheduling or bin-packing optimizations in R

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I have an optimisation issue. It's about a product that contains 20 parts (the order of producing doesn't matter). I've got 3 similar machine that can produce all 20 parts. I've got the 20 parts represented in minutes (ie. it takes 3min to produce the first part and 75min to produce the second part, etc) ItemTime<-c(3,75,55,12,45,55,11,8,21,16,65,28,84,3,58,46,5,84,8,48) So to produce 1 product it takes 730 min. sum(ItemTime) The aim is to minimise the production of one product by allocating the good item to the three machines. sum(ItemTime/3) So actually I need to be as close as 243.333 min

Solving task scheduling or bin-packing optimizations in R

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问题 I have an optimisation issue. It's about a product that contains 20 parts (the order of producing doesn't matter). I've got 3 similar machine that can produce all 20 parts. I've got the 20 parts represented in minutes (ie. it takes 3min to produce the first part and 75min to produce the second part, etc) ItemTime<-c(3,75,55,12,45,55,11,8,21,16,65,28,84,3,58,46,5,84,8,48) So to produce 1 product it takes 730 min. sum(ItemTime) The aim is to minimise the production of one product by allocating

Algorithm to find maximum coverage of non-overlapping sequences. (I.e., the Weighted Interval Scheduling Prob.)

阅读更多关于 Algorithm to find maximum coverage of non-overlapping sequences. (I.e., the Weighted Interval Scheduling Prob.)

I have a question that is very similar to algorithm to find longest non-overlapping sequences . The only difference to the linked question is that instead of finding the set of non-overlapping tuples that represent the longest sequence , I need to find the set of non-overlapping tuples that represent the maximum coverage , by which I mean the sum of the tuple lengths is maximum (a tuple length being last - first + 1 given the definition of tuple in the next sentence). I represent my tuples differently than the linked problem. Instead of (starting index, length) , I represent my tuples as

Are all scheduling problems NP-Hard?

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I know there are some scheduling problems out there that are NP-hard/NP-complete ... however, none of them are stated in such a way to show this situation is also NP. If you have a set of tasks constrained to a startAfter , startBy , and duration all trying to use a single resource ... can you resolve a schedule or identify that it cannot be resolved without an exhaustive search? If the answer is "sorry pal, but this is NP-complete" what would be the best heuristic(s?) to use and are there ways to decrease the time it takes to a) resolve a schedule and b) to identify an unresolvable schedule.