An example of a graph that is not a unit disk graph is the star K 1,7 with one central node connected to seven leaves: if each of seven unit disks touches a common unit disk, some two of the seven disks must touch each other (as the kissing number in the plane is 6). Therefore, unit disk graphs cannot contain an induced K 1,7 subgraph.
Percentage Formula. Although the percentage formula can be written in different forms, it is essentially an algebraic equation involving three values. P is the percentage, V 1 is the first value that the percentage will modify, and V 2 is the result of the percentage operating on V 1. The calculator provided automatically. Why set up classes for each width% on the bar chart? I’ve always just stuck an inline style on the row, means you don’t clog up your stylesheet with unnecessary styles and gives you more flexibility (e.g. You can use 83.46217% if you so wish). Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. It’s waiting on the disk. In the example above, disk access took 700 ms, so I/O wait is 70%. You can check your I/O wait percentage via top, a command available on every flavor of Linux. If your I/O wait percentage is greater than (1/# of CPU cores) then your CPUs are waiting a significant amount of time for the disk subsystem to catch up.
Updated version of an article first published on February 10th, 2011.
If you’re old enough to remember floppy drives, you’ve heard the symptoms of a disk I/O bottleneck. For example, while Oregon Trail loaded the next scene, you’d hear the drive grinding away, reading data from the disk. The CPU would sit idle during this time, twiddling its fingers waiting for data. If that floppy drive was faster, you'd be running the Columbia River rapids by now.
It’s more difficult to detect an I/O bottleneck if the disk isn’t on your desktop. Unrar archiver mac download. I’ll look at four important
Disk I/O encompasses the input/output operations on a physical disk. If you’re reading data from a file on a disk, the processor needs to wait for the file to be read (the same goes for writing). Torrent helix 3 pro.
The killer when working with a disk? Access time. This is the time required for a computer to process a data request from the processor and then retrieve the required data from the storage device. Since hard disks are mechanical, you need to wait for the disk to rotate to the required disk sector.
Aiseesoft mac pdf to image converter 3 1 50 inch. Disk latency is around 13ms, but it depends on the quality and rotational speed of the hard drive. RAM latency is around 83 nanoseconds. How big is the difference? If RAM was an F-18 Hornet with a max speed of 1,190 mph (more than 1.5x the speed of sound), disk access speed is a banana slug with a top speed of 0.007 mph.
This is why caching data in memory is so important for performance – the difference in latency between RAM and a hard drive is enormous*.
Your I/O wait measurement is the canary for an I/O bottleneck. I/O Wait is the percentage of time your processors are waiting on the disk.
For example,
The disk is being accessed while the rows are
You can check your I/O
If your I/O wait percentage is greater than (1/# of CPU cores) then your CPUs are waiting a significant amount of time for the disk subsystem to catch up.
In the output above, I/O wait is 12.1%. This server has 8 cores (via cat /proc/cpuinfo). This is very close to (1/8 cores = 0.125). Disk access may be slowing the application down if I/O wait is consistently around this threshold.
For random disk access (a database, mail server, file server, etc), you should focus on how many input/output operations can be performed per-second (IOPS).
Four primary factors impact IOPS:
A more exact way to determine just how close you are to your maximum I/O throughput is to calculate your theoretical IOPS and compare it to your actual IOPS. If the numbers are close, there may be an I/O issue.
You can determine theoretical IOPS via
| I/O Operations Per-Sec = | number of disks * Average I/O Operations on 1 disk per-sec |
| % of |
All but one of the pieces in this equation can be determined from your hardware specs. You’ll need to figure out the read/write workload though – it’s application dependent. For this, use a tool like
Once you’ve calculated your theoretical IOPS, compare it to
Check out a more detailed calculation of IOPS.
Even if a banana slug follows all of the tips in The 4 Hour Body, it will never be as fast as an F-18 Hornet. Likewise, you can tune your disk hardware for better performance, but it’s complicated and will not approach the speed of RAM.
If you’re hitting a disk I/O bottleneck now, tuning your hardware likely isn’t the fastest remedy. Hardware changes likely involve significant testing, data migration, and communication between application developers and
When we see I/O bottlenecks at the Blue Box Group, we first try to tweak the service that’s using the most I/O and cache more of its data in RAM. For example, we usually configure our database servers to have as much RAM as possible (up to 64 GB or so), and then have MySQL cache as much as possible in memory.
Adding servers can be a band-aid for slow code. Scout APM helps you find and fix your inefficient and costly code. We automatically identify N+1 SQL calls, memory bloat, and other code-related issues so you can spend less time debugging and more time programming.
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* - If the difference between RAM and disk latency seems striking to you, it should -- it's a massive difference in performance. If it doesn't jibe with your day-to-day experience with your computer, keep in mind three things: 1) 13ms is the time for a hard drive seek -- if you're reading contiguous data from an unfragmented file, then the hard drive's *transfer rate* (not the seek speed) comes to bear. 2) 13ms is still extraordinarily fast by human standards. 3) There are multiple mechanisms in place to ameliorate the impact of hard drive's dismal performance, like paging, write-queuing, and HD-resident cache.
In geometric graph theory, a unit disk graph is the intersection graph of a family of unit disks in the Euclidean plane. That is, it is a graph with one vertex for each disk in the family, and with an edge between two vertices whenever the corresponding vertices lie within a unit distance of each other.
They are commonly formed from a Poisson point process, making them a simple example of a random structure.
There are several possible definitions of the unit disk graph, equivalent to each other up to a choice of scale factor:
Every induced subgraph of a unit disk graph is also a unit disk graph. An example of a graph that is not a unit disk graph is the star K1,7 with one central node connected to seven leaves: if each of seven unit disks touches a common unit disk, some two of the seven disks must touch each other (as the kissing number in the plane is 6). Therefore, unit disk graphs cannot contain an induced K1,7 subgraph.
Beginning with the work of Huson & Sen (1995), unit disk graphs have been used in computer science to model the topology of ad hoc wireless communication networks. In this application, nodes are connected through a direct wireless connection without a base station. It is assumed that all nodes are homogeneous and equipped with omnidirectional antennas. Node locations are modelled as Euclidean points, and the area within which a signal from one node can be received by another node is modelled as a circle. If all nodes have transmitters of equal power, these circles are all equal. Random geometric graphs, formed as unit disk graphs with randomly generated disk centres, have also been used as a model of percolation and various other phenomena.[1]
If one is given a collection of unit disks (or their centres) in a space of any fixed dimension, it is possible to construct the corresponding unit disk graph in linear time, by rounding the centres to nearby integer grid points, using a hash table to find all pairs of centres within constant distance of each other, and filtering the resulting list of pairs for the ones whose circles intersect. The ratio of the number of pairs considered by this algorithm to the number of edges in the eventual graph is a constant, giving the linear time bound. However, this constant grows exponentially as a function of the dimension (Bentley, Stanat & Williams 1977).
It is NP-hard (more specifically, complete for the existential theory of the reals) to determine whether a graph, given without geometry, can be represented as a unit disk graph.[2] Additionally, it is probably impossible in polynomial time to output explicit coordinates of a unit disk graph representation: there exist unit disk graphs that require exponentially many bits of precision in any such representation.[3]
However, many important and difficult graph optimization problems such as maximum independent set, graph coloring, and minimum dominating set can be approximated efficiently by using the geometric structure of these graphs,[4] and the maximum clique problem can be solved exactly for these graphs in polynomial time, given a disk representation.[5] Even if a disk representation is not known, and an abstract graph is given as input, it is possible in polynomial time to produce either a maximum clique or a proof that the graph is not a unit disk graph,[6] and to 3-approximate the optimum coloring by using a greedy coloring algorithm.[7]
Jungle of the giants star coins. When a given vertex set forms a subset of a triangular lattice, a necessary and sufficient condition for the perfectness of a unit graph is known.[8] For the perfect graphs, a number of NP-complete optimization problems (graph coloring problem, maximum clique problem, and maximum independent set problem) are polynomially solvable.
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