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How optimization is achieved in DNS?
Two primary optimizations used in DNS and they are: replication and caching. All root servers is replicated; various copies of the server exist around the world. While a new site joins the internet, the site configures its local DNS server along with a list of root server. The site server uses any root server is most responsive at a provided point of time. Into DNS caching each server maintains a cache of names. When this looks up a new name, the server puts a copy of the binding in its cache. Before contacting other server to request a binding, the server checks its cache; when the cache have the answer the server uses the cached answer to produce a reply.
Question: (a) Identify the four main enhancements brought along by TKIP on WPA to solve the problems in WEP. (b) The term WarDriving has been frequently used while talking
State about Dynamic modelling and its inputs Dynamic modelling is elaborated further by adding concept of time: new attributes are computed, as a function of the attribute chan
Goals - artificial intelligence: One desirable way to make perfect an agent's performance is to enable it to have some details of what it is trying to complete. If it is given
How can we pass selection and parameter data to a report? There are three options for passing selection and parameter data to the report. Using SUBMIT...WITH Using a rep
The micro-instruction cycle can comprises two basic cycles: the fetch and execute. Here in the fetch cycle address of micro-instruction is produced and this micro-instruction is pu
Specified the code segment below and that n is the problem size, answer the following queries: // . . . int sum = 0; if(x > 12){ for(int i = 1; i for( i
What do you do if you have given functionality that wasn't listed in the requirements? - If the functionality isn't essential to the purpose of the application, it should be re
Q. What is interrupt? How it is useful in controlling I/O operations. Q. Explain function of Color Graphics Adopter card and Diskette Drive Adopter Card.
Reduce the following equation using k-map Y = BC‾D‾ + A‾BC‾D + ABC‾D + A‾BCD + ABCD Ans. Multiplying the first term with (A+A') Y = A'BC'D' + ABC'D' + A'BC'D + ABC'D + A'BCD + A
Disadvantages of Address translation: Disadvantages are following: A program that is too large to be held in a part needs some special design, that called overlay
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