What is displayed as a result of executing the following code segment?

A company requires an Ethernet connection from the north end of its office building to the south end. The distance between conne

BlackZzzverrR [31]

Answer:

A company requires an Ethernet connection from the north end of its office building to the south end. The distance between connections is 161 meters and will need to provide speeds of up to 10 Gbps in full duplex mode. Which of the following cable combinations will meet these requirements?

ANSWER: Use multi-mode fiber optic cable

Explanation:

MULTI-MODE FIBER OPTIC CABLE

For Ethernet distances up to 100 meters, the Copper CATX cable will do just fine. In the question, we are dealing with a distance of up to 161 meters, so we need an Ethernet extension. LAN extension over fiber optic cable with media converter can be used to convert the Ethernet cable runs from copper to fiber. Multi-mode fiber has a range of 550 meters for 10/100/1000 Ethernet links.

Multi-mode optical fiber is commonly used for communication over short distances, like within a building or on a campus. They are capable of data rate of up to 100 Gbps, which surpasses the requirements in this question. They are more economical in this case as they are not as expensive a the single-mode optical fiber cable. Fiber optic cable has the advantage of being immune to electromagnetic interferences, spikes, ground loops, and surges. This makes it more suited for this purpose.

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4 years ago

Al and Bill are arguing about the performance of their sorting algorithms. Al claims that his O(N log N)-time algorithm is alway

lisabon 2012 [21]

Answer:

See in Explanation

Explanation:

All you know is that an $O(n log n)$ algorithm is eventually (a lot) faster than an $O(n^{2} )$ algorithm, but for any specific value of n, you can't say anything. Indeed, the asymptotic growth of the function doesn't depend on its value on $n$ . This means that if $f(n)=O(n^{2} )$ then for all $c_{1} ,....\ , c_{99}$ , the following function is also $O(n^{2} )$ :

$f^{'} (n)=[\ cnf(n)\ if \ n$

So asymptotic notation is completely uninformative regarding the performance of algorithms on specific values of n.

You could say that functions such as $f^{'}(n)$ are artificial and never really occur in practice. This is not quite true (you can hard-code some auxiliary information for small n, for example), but even if you consider "natural" functions, asymptotic notation doesn't help you to determine anything other than asymptotic's: consider for example $n^{2}$ against $10^{100}n\ log \ n$ (such examples definitely happen in practice, for example in fast matrix multiplication).