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The '''ski rental problem''' is the name given to a class of problems in which there is a choice between continuing to pay a repeating cost or paying a one-time cost which eliminates or reduces the repeating cost.
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==The problem==
Many [[online problem]]s have a sub-problem called the rent/buy problem. We need to decide whether to stay in the current state and pay a certain amount of cost per time unit, or switch to another state and pay some fixed large cost with no further payment.<ref name="seiden">Steven S. Seiden. A guessing game and randomized online algorithms. Annual ACM Symposium on Theory of Computing, 2000. http://portal.acm.org/citation.cfm?id=335385</ref> Ski rental <ref name = "Karlin1994">[[Anna Karlin|A. R. Karlin]], M. S. Manasse, L. A. McGeoch, and S. Owicki. Competitive randomized algorithms for non-uniform problems. In Proceedings of the First Annual ACM-SIAM Symposium on Discrete Algorithms, San Francisco, CA, 22–24 January 1990, pp. 301-309. Also in Algorithmica, 11(6): 542-571, 1994. http://theory.lcs.mit.edu/classes/6.895/fall03/handouts/papers/karlin.pdf</ref><ref>Claire Mathieu. Brown University. Lecture note: http://www.cs.brown.edu/~claire/Talks/skirental.pdf</ref> is one classical toy example, where the rent/buy is the entire problem. Its basic version is as follows:
 
You are going skiing for an unknown number of days (you do not know the exact number due to various reasons, e.g., loss of interest, accidents that break your legs, or extremely bad weather). Assume that renting skis costs $1 per day and buying skis costs $10. Every day you have to decide whether to continue renting skis for one more day or buy a pair of skis. If you know in advance how many days you will go skiing, you can decide your minimum cost. For example, if you will be skiing for more than 10 days it will be cheaper to buy skis, while if you will be skiing for fewer than 10 days it will be cheaper to rent. (If you will ski for exactly 10 days you are indifferent.) The question is what to do when you do not know in advance how many days you will ski.
 
Formally, the problem can be set up as follows. There is a number of days ''d'' (unknown to you) that you will ski. We are looking for an algorithm that minimizes the ratio between what you pay using
the algorithm (that does not know ''d'') and what you would pay optimally if you knew ''d'' in advance. The problem is generally analyzed in the worst case, where the algorithm is fixed and then we look at the worst-case performance of the algorithm over all possible ''d''. In particular, no assumptions are made regarding the distribution of ''d'' (and it is easy to see that, with knowledge of the distribution of ''d'', a different analysis as well as different solutions would be preferred).
 
==The break-even algorithm==
The break-even algorithm instructs you to rent for 9 days and buy skis on the morning of day 10 if you are still up for skiing.<ref name="Karlin1988">[[Anna Karlin|A. R. Karlin]], M. Manasse, L. Rudolph and D. Sleator. Competitive snoopy caching. Algorithmica, 3(1): 79-119, 1988</ref> If you have to stop skiing during the first 9 days, it costs the same as what you would pay if you had known the number of days you would go skiing. If you have to stop skiing after day 10, your cost is $19 which is 90% more than what you would pay if you had known the number of days you would go skiing in advance. This is the worst case for the break-even algorithm.
 
The break-even algorithm is known to be the best deterministic algorithm for this problem.
 
==Can you do better than break-even?==
Yes. For example, you can flip a coin. If it comes up head, you buy skis on day 8; otherwise, you buy skis on day 10. This is an instance of a [[randomized algorithm]]. It is easy to see that the ''expected'' cost is at most 80% more than what you would pay if you had known the number of days you would go skiing, ''regardless'' of how many days you ski. In particular, if you ski for 10 days, then your expected cost is 1/2 [7 +10] + 1/2 [9+10] = 18 dollars, only 80% excess instead of 90%.
 
The best randomized algorithm against an [[Adversary (online algorithm)|oblivious adversary]] is to choose some day i at random according to the following distribution p, rent for i-1 days and buy skis on the morning of day i if you are still up for skiing. Karlin et al.<ref name = "Karlin1994" /> first presented this algorithm with distribution
<math>
p_i = \left \{
\begin{array}{ll}
(\frac{b-1}{b})^{b-i} \frac{1}{b(1-(1-(1/b))^b)} & i \leq b \\
0 & i > b
\end{array} \right . ,
</math>
where buying skis costs $b and renting costs $1. Its expected cost is at most e/(e-1) <math>\approx</math> 1.58 times what you would pay if you had known the number of days you would go skiing. No randomized algorithm can do better.
 
==Applications==
'''Snoopy caching''':<ref name = "Karlin1994">[[Anna Karlin|A. R. Karlin]], M. S. Manasse, L. A. McGeoch, and S. Owicki. Competitive randomized algorithms for non-uniform problems. In Proceedings of the First Annual ACM-SIAM Symposium on Discrete
Algorithms, San Francisco, CA, 22–24 January 1990, pp. 301-309. Also in Algorithmica, 11(6): 542-
571, 1994. http://theory.lcs.mit.edu/classes/6.895/fall03/handouts/papers/karlin.pdf</ref> several caches share the same memory space that is partitioned into blocks. When a cache writes to a block, caches that share the block spend 1 bus cycle to get updated. These caches can invalidate the block to avoid the cost of updating. But there is a penalty of p bus cycles for invalidating a block from a cache that shortly thereafter needs access to it. We can break the write request sequences for several caches into request sequences for two caches. One cache performs a sequence of write operations to the block. The other cache needs to decide whether to get updated by paying 1 bus cycle per operation or invalidate the block by paying p bus cycles for future read request of itself. The two cache, one block snoopy caching problem is just the ski rental problem.
 
'''TCP acknowledgment''':<ref>D. R. Dooly, S. A. Goldman and S. D. Scott. TCP dynamic acknowledgment delay: theory and practice.
In Proceedings of the Thirtieth Annual ACM Symposium on Theory of Computing (STOC), Dallas, TX,
pp. 389-398, 1998.</ref> A stream of packets arrive at a destination and are required by the TCP protocol to be acknowledged upon arrival. However, we can use a single acknowledgment packet to simultaneously acknowledge multiple
outstanding packets, thereby reducing the overhead of the acknowledgments. On the other hand, delaying acknowledgments too much can interfere with the TCP's congestion control mechanisms, and thus we should not allow the latency between a packet's
arrival time and the time at which the acknowledgment is sent to increase too much. Karlin et al.<ref>[[Anna R. Karlin]] and Claire Kenyon and Dana Randall. Dynamic TCP acknowledgement and other stories about e/(e-1). Thirty-Third Annual ACM Symposium on Theory of Computing (STOC), 2001. Algorithmica. http://www.cs.brown.edu/people/claire/Publis/ACKpaper.ps</ref> described a one-parameter family of inputs, called the basis inputs, and showed that when restricted to these basis inputs, the TCP acknowledgement problem behaves the same as the ski rental problem.
 
'''Total completion time scheduling''':<ref name="seiden" />
We wish to schedule jobs with fixed processing times on m identical machines. The processing time of job j is p<sub>j</sub>.
Further, each job has a release time r<sub>j</sub>, before which the job is unknown to the scheduler. The goal is to minimize the
sum of completion times over all jobs. A simplified problem is one single machine with the following input: at time 0, a job with processing time 1 arrives; k jobs with processing time 0 arrive at some unknown time. We need to decide the time to start the first job. Waiting incurs cost 1 per time unit, yet starting the first job before the later k jobs may incur an extra cost of k in the worst case. This simplified problem may be viewed as a continuous version of the ski rental problem.
 
==See also==
*[[Adversary (online algorithm)]]
* [[Competitive analysis (online algorithm)]]
*[[Online algorithm]]
 
==References==
<references/>
 
[[Category:Online algorithms]]

Latest revision as of 17:43, 28 September 2014

A healthy mouth is an critical portion of your overall well-getting gums and teeth can indicate illness lengthy prior to you are aware that something is incorrect. The following write-up will offer you professional guidance on keeping your teeth and gums in tip-prime shape. Study on for valuable info that will hold your smile gorgeous.

Do not wait to see a dentist. If you have began feeling even the slightest bit of discomfort in your teeth, make an appointment with a dentist. Learn more on our affiliated URL - Click here: partner sites. If you wait as well long the difficulty could get even worse. If you get in correct away, you could only need a fast and easy remedy.

If you"re nervous about going to the dentist, study some of them ahead of time. Choose a dentist with a good personality who enjoys functioning with sufferers. A great bedside manner will support you feel comfy on your subsequent go to.

If you run out of toothpaste, baking soda and water can be an efficient substitute. Simply mix in some water with a modest amount of baking soda and use it just as you would toothpaste. An added benefit to making use of baking soda is that, along with neutralizing mouth odors, its abrasive nature can support with stain removal.

By no means assume that skipping your dental appointment is okay. Absolutely everyone should start off receiving dental care as soon as they are six months old. Dental verify-ups should be carried out every single six months following. This holds correct for children, teenagers, adults and the elderly. Everybody requirements to see a dentist twice a year previous their initial birthday.

To guarantee your teeth are acquiring appropriately cleaned, make it a point to brush your teeth for a minimum of two minutes. Browsing To dental discount plans in colorado probably provides lessons you might give to your uncle. It"s really easy to get bored when brushing your teeth, and boredom can lead to a rush job. Even so, thinking of a favorite song or some other pleasant thought for the duration of brushing can help pass the time. Should you need to get further about Oakley Matthews - Tips About How To Consider Better Care Of The Teeth | about.me, there are many databases people should think about pursuing. It is critical to give your teeth the focus they deserve.

Oral hygiene is important, even if you do not have any organic teeth. Just as you would brush your teeth, make confident that you brush your dentures. Don"t neglect to clean your tongue too, with the brush or a scraper, as failure to do so can lead to accumulated bacteria, or persistent negative breath.

The importance of regularly brushing and flossing your teeth can not be stressed adequate. Even though, no matter how nicely you brush, you will often leave a particular quantity of bacteria behind. Therefore, you should use mouthwash every single time you attend to your oral hygiene to make sure that your mouth is totally clean.

Attempt your greatest to stop employing tobacco products. Most individuals are not aware that smoking can trigger gum disease, which leads to tooth loss in a lot of. If you want to raise the possibilities of your teeth lasting considerably longer, try your best to get rid of your addiction to tobacco products.

Teeth and gums say a lot about a person"s wellness. Of course you want to have the best hunting set of teeth feasible but it really is critical to go beyond that and comprehend the implications of oral difficulties and what you can do about them. Keep your smile vibrant and healthier for a lengthy time to come!.Direct Dental Plans of America
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