SNMP can not detect all disk space – cacti prtg

SNMP can not detect all disk space – cacti prtg | Web Hosting Talk

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  1. SNMP can not detect all disk space – cacti prtg

    The snmp service is running well, monitoring the CPU and … perfectly.
    but when I try to monitor hard disk via PRTG and cacti, it only recognizes a portion of the disk as total storage space
    / home has 22TB of storage space and 3TB are unused, but cacti only recognize 7TB of total storage space!

    centos 6 + snmp v2

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Visualization of a probability space in tungsten with two events?

Say for two events $ A $ and $ B $Suppose we have the data: $ P (A) $. $ P (B) $. $ P (A | B) $ and $ P (B | A) $Is there any way to generate something like a "Probability VENN diagram" from this information? The following online information is meant as it were:

Enter image description here

If there are other ways to visualize this in tungsten, those approaches would be of interest to me as well.

Memory – How does swap space work in this example?

Suppose I have 1 GB RAM, 10 GB HD with 250 MB swap space

Suppose I opened two files (F1 and F2) with 500 MB each on my laptop. Now both files are in memory and have consumed 1 GB of RAM.
Currently f2 is in front of me. Now I start writing f2 with 100 MB characters.

Now the operating system will swap space here and write 100 MB of F1 to disk. If I switch f1, the operating system will bring that 100MB of space back and drop it off
100 MB f2 on the hard disk.

I'll get to know many other subtleties, but I just wanted to understand the basics of how swap space works. Is my understanding correct here?

Partitioning – From the boot space, some previous kernel deleted, updated and everything seemed fine, can not boot

When I logged into Webmin, I noticed that many updates were waiting for me and a warning that my boot partition was running out of space. This was not the first time that I encountered this exact issue, and I only deleted some files after /

It seemed to be going well. I had more space and was able to install all updates.

When I restarted the system, it did not go online again. This server serves my CRM system and has brought my company to its knees.

I tried repairs, etc. used .. and I was able to get the system to start again, although it is so if it is a shell of a boot, without the "normal" abilities Server would, much less have the LAMP stack.

Apparently I have the three partitions, etc., as the link shows, that I can understand the results to the best of my ability.

More importantly, any files I could use to transfer to a new system can not (or may not) be displayed due to permission restrictions.

I did not really think much about it (apart from the time wasted on transferring the files) because I have a daily backup with Duplicati. Although I tested the backup system on the same server, I did not try to migrate to a new server and am unable to recover.
It seems to be a relatively "simple" solution, but I'm at the end …

Any help is greatly appreciated. And as someone who has generated millions of page views, I would not mind writing a post about the help received here, if that can help someone in return.

Many thanks

trees – Maximum space required by stack and queue for DFS and BFS

I am trying to determine the maximum memory usage of the Pending Nodes (Stack / Queue) data structure for both trips: BFS and (Preorder) DFS.

Since BFS and DFS have node detection control while traveling through charts (no loops), we can analyze the problem by thinking of trees instead of charts, using your starting node as usual as the root.

I first assume that the resulting travel is a complete tree (all leaves are the same depth), with a height $ h $. $ n $ Node, and thus all nodes with degrees $ d $ (except leaves).

Of course, since the tree is complete, $ n $ can be calculated from $ h $ and $ d $:

$$ n = frac {1 – d ^ {h + 1}} {1 – d} $$

Under this assumption, the worst-case scenario for DFS is when you are in the deepest non-leaf node of the first branch of the tree: since you insert a node, insert all the child nodes for each level $ D – 1 $ Pending nodes in the stack, except the last non-leaf, where all child nodes are inserted.

If you were not in the first branch but in another branch, you would have a branch with less than $ d – 1 $ pending children in the stack, so you have fewer nodes in the stack.

The maximum memory consumption of DFS for a complete graph with degree of homogeneity is ($ ms $ stands for maximum space):

$$ ms = h * (d – 1) + 1 $$

This last one + 1 Represent the additional child for the last non-leaf node. For example, for a tree with $ d = 4 $ and $ h = $ 20 Node would require a maximum DFS stack:

$$ ms_ {DFS} = 20 * 3 + 1 = 81 node $$

Considering that this graph would have $ n = 1.4660155e ^ {12} $ Node, that is a more than allowable amount. This is the advantage of logarithmic space complexity ($ h = lfloor log_d ((d-1) n) rfloor $).

However, in BFS with exponential storage complexity, the worst case scenario is that all of the sheets to be discovered were discovered while all other nodes were discovered so that all sheets are present in your queue (that is, the complete last outstanding level) but nothing else is detected:

$$ ms_ {BFS} = d ^ h node $$

which is the same in our example $ 1.0995116e ^ {12} $,

My problem now is to reduce the graph's limit to completeness $ d $ is no longer a homogeneous degree, but an average degree (which may now include decimals), and the tree may have an imbalance, even if it is a list. As a result, the number of nodes is free, so there is no connection to them d and h as previously.

If $ d $ is an average degree and $ n $ For any number of nodes, I tried to model an upper bound of space by first modeling a complete tree with a homogeneity $ lfloor d rfloor $ Degrees, and then add kids to the last sheet until they get to $ d $ (I assume that the resulting number of nodes should be the same $ n $but I'm not sure about that; I even tried to calculate some given $ d $ and $ n $, a lower and upper limit for the height of the tree).

Since $ d $ is an average when a node has more than $ d $ Children is because another node has less than $ d $ Children, and therefore, the idea was to find the worst case scenario for DFS and BFS by removing children from one node and moving the cut branch as the child of another node, or generally finding the closest possible upper memory usage limit. but I could not find a way.

The thing is, if you repeatedly apply this elevation by moving sibling branches to the lowest levels, you probably have a bunch of parent or sibling paths that only consist of lists that are quickly removed from the stack / queue, So there has to be a tree status where you can not make space consumption worse. However, I assume that the "worst tree" can be different from DFS and BFS.

Is it even possible to carry out this calculation of the upper limit (or the exact amount)? Or is the worst-case scenario just the right balance?

sharepoint online – Returns events of a calendar space after diagram and receives the error "DelegatedCalendarAccessDenied"

I face a problem when trying to get a room with graph api. I had already created two rooms in my exchange environment and the request works only if I search their events with the request below

But then I created a new room and when I try to come with the same request

the error is displayed:

{    "error": 

        {        "code": "DelegatedCalendarAccessDenied",        

                 "message": "Access is denied. Check credentials and try again.",        

                 "innerError": {            

                                         "request-id": "cb059196-02df-4186-bf35-a5fd1d184b30",            

                                         "date": "2019-08-22T20:14:48"       




Which king of permission do I miss? Because it works with other rooms. Only new ones get this error.

at.algebraic topology – Homology of a ring suspension space and action of $ mathcal {D} _1 $ -operad

If $ X $ If it is a connected connected topological space, it is known what homology is about $ Omega Sigma X $ is: According to the Bott-Samelson theorem, it is a tensor algebra about reduced homology of $ X $, (Some assumptions about coefficients should be made here).

However, it is also known that $ Omega Sigma X $ is weakly equivalent to $ D_1X $, from where $ D_1 $ is a monad connected to small intervals operad. When I go to homology, I see the following:

  1. $ H _ * ( mathcal {D} _1 (*); R) $ forms an opera in graduated $ R $modules;
  2. Homology of $ D_1X $ is an algebra about this opera.

All the evidence of the Bott-Samelson theorem that I know uses a more geometric description of $ Omega Sigma X $without using any information that comes from the action operad. My (probably rather vague) question is: Is it possible to prove B-S theorem with operadic data? The result is something like "free" $ mathcal {D} _1 $ Algebra in $ R $Modules with socket $ H _ * (X; R) $"?

Maybe this is a question to look up, but I would be grateful for any help.

General Topology – A non-banal space

Please, how to select a sequence to prove that this field is not a Banach field $ ( mathcal {C} ((0, frac12), mathbb {R}), ||. || _1) $,

In the exercise, they present this sequence of functions $ f_n (x) = frac1n frac {1} { sqrt {1-x ^ 2}} $

But this sequence is Cauchy and when it converges to 0, I can not conclude that this space is not a Banach!

Complexity theory – Injection test in o (n) space and O (n) time

The problem I want to solve is this: give a list $ A $ from $ n $ I want to check if all elements are different. If I did that "myself," I would need it $ O (n) $ room and $ O (n log n) $ Time to solve it, e.g. via a hashmap or a binary tree. Luckily, I have an untrustworthy but almighty oracle ally who is willing to give me clues to solve the problem.

The oracle may supply me $ O (n) $ of clues, and I want an algorithm that will read $ A $ and the Oracle input (both read-only) and either specify that $ A $ has no duplicates, or that the oracle gave me a bad hint in time $ O (n) $ and much smaller read / write space ($ O (1) $ or $ O ( log n) $).

Can it be done?

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