system memory 259522560 must b

时间: 2023-05-02 09:04:56 浏览: 50
这句话需要上下文来进行理解。如果是在电脑或者手机等设备中出现,可能是指设备的内存不足,需要进行清理或者增加存储容量。如果是在程序或者代码中出现,可能是编写者在设定程序时限制了内存使用量,需要更改程序代码以适应更大的内存使用。
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ERROR: [1] bootstrap checks failed. You must address the points described in the following [1] lines before starting Elasticsearch. bootstrap check failure [1] of [1]: memory locking requested for elasticsearch process but memory is not locked ERROR: Elas

ticsearch requires memory locking in order to ensure that the Java Virtual Machine (JVM) can manage memory effectively. In order to address this error, you need to increase the maximum memory map count value of your system. To do this, you can modify the /etc/sysctl.conf file and add the following lines: ``` vm.max_map_count=262144 ``` After adding this line, you can run the following command to apply the changes: ``` sudo sysctl -p ``` This should resolve the memory locking issue and allow you to start Elasticsearch.

In the initial version of the resource provider schema in the placement API, we stuck with a simple world-view that resource providers could be related to each other only via an aggregate relationship. In other words, a resource provider “X” may provide shared resources to a set of other resource providers “S” if and only if “X” was associated with an aggregate “A” that all members of “S” were also associated with. This relationship works perfectly fine for things like shared storage or IP pools. However, certain classes of resource require a more parent->child relationship than a many-to-many relationship that the aggregate association offers. Two examples of where a parent->child relationship is more appropriate are when handling VCPU/MEMORY_MB resources on NUMA nodes on a compute host and when handling SRIOV_NET_VF resources for NICs on a compute host. In the case of NUMA nodes, the system must be able to track how many VCPU and MEMORY_MB have been allocated from each individual NUMA node on the host. Allocating memory to a guest and having that memory span address space across two banks of DIMMs attached to different NUMA nodes results in sub-optimal performance, and for certain high-performance guest workloads this penalty is not acceptable.

在placement API中的资源提供者模式的初始版本中,我们坚持了一个简单的观点,即资源提供者之间只能通过聚合关系相互关联。换句话说,如果且仅当资源提供者“X”与所有成员资源提供者“S”都关联到同一个聚合“A”时,资源提供者“X”才能向资源提供者“S”提供共享资源。 这种关系对于共享存储或IP池等情况非常适用。然而,某些类型的资源需要比聚合关联提供更多的父->子关系。两个更适合父->子关系的例子是在计算主机上处理NUMA节点上的VCPU/MEMORY_MB资源以及在计算主机上处理SRIOV_NET_VF资源的网卡。 在NUMA节点的情况下,系统必须能够跟踪从主机上的每个单独NUMA节点分配了多少VCPU和MEMORY_MB。将内存分配给虚拟机,并使该内存跨越连接到不同NUMA节点的两组DIMM的地址空间,会导致性能下降,并且对于某些高性能虚拟机工作负载来说,这种性能损失是不可接受的。

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解释一段python代码 class KalmanFilter(object): def init(self, dim_x, dim_z, dim_u=0): if dim_x < 1: raise ValueError('dim_x must be 1 or greater') if dim_z < 1: raise ValueError('dim_z must be 1 or greater') if dim_u < 0: raise ValueError('dim_u must be 0 or greater') self.dim_x = dim_x self.dim_z = dim_z self.dim_u = dim_u self.x = zeros((dim_x, 1)) # state self.P = eye(dim_x) # uncertainty covariance self.Q = eye(dim_x) # process uncertainty self.B = None # control transition matrix self.F = eye(dim_x) # state transition matrix self.H = zeros((dim_z, dim_x)) # Measurement function self.R = eye(dim_z) # state uncertainty self._alpha_sq = 1. # fading memory control self.M = np.zeros((dim_z, dim_z)) # process-measurement cross correlation self.z = np.array([[None]*self.dim_z]).T # gain and residual are computed during the innovation step. We # save them so that in case you want to inspect them for various # purposes self.K = np.zeros((dim_x, dim_z)) # kalman gain self.y = zeros((dim_z, 1)) self.S = np.zeros((dim_z, dim_z)) # system uncertainty self.SI = np.zeros((dim_z, dim_z)) # inverse system uncertainty # identity matrix. Do not alter this. self._I = np.eye(dim_x) # these will always be a copy of x,P after predict() is called self.x_prior = self.x.copy() self.P_prior = self.P.copy() # these will always be a copy of x,P after update() is called self.x_post = self.x.copy() self.P_post = self.P.copy() # Only computed only if requested via property self._log_likelihood = log(sys.float_info.min) self._likelihood = sys.float_info.min self._mahalanobis = None self.inv = np.linalg.inv

用c++解决1160. Network Time limit: 1.0 second Memory limit: 64 MB Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs). Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem - not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections. You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied. Input The first line contains two integer: N - the number of hubs in the network (2 ≤ N ≤ 1000) and M — the number of possible hub connections (1 ≤ M ≤ 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs. Output Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.

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