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Sladkaya [172]
3 years ago
7

Lief wants to keep the information of his chart, but he doesn’t want it taking up space in his page. What should he select after

going to the Move Chart tab?
Computers and Technology
2 answers:
irakobra [83]3 years ago
5 0

Answer:

New Sheet.

Explanation:

To avoid that the graph made from the available information occupies an extra space in the current worksheet, what is done is to place it on another work page. In order to perform this function, first it is selected in the graph, then in the move tab and finally new sheet. In this way, the graph will appear on a different sheet, preventing it from taking place on the sheet that contains the information.

Murljashka [212]3 years ago
4 0
If wants to keep the chart but does not want it taking up space, he can move it  to another sheet. Once he goes to the Move Chart tab, he should select New Sheet which will ask for a name of the new sheet. After doing this, the chart will be moved to a different sheet.
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Consider a set of mobile computing clients in a certain town who each
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We have n clients and k-base stations, say each client has to be connected to a base station that is located at a distance say 'r'. now the base stations doesn't have allocation for more than L clients.

To begin, let us produce a network which consists of edges and vertex

Network (N) = (V,E)

where V = [S, cl-l, - - - -  cl-n, bs-l - - - - - - bs-k, t]

given that cl-l, - - - - - cl-n represents nodes for the clients

also we have that bs-l, - - - - - bs-k represents the nodes for base station

Also

E = [ (s, cl-i), (cl-i,bs-j), (bs-j,t)]

(s, cl-i) = have capacity for all cl-i (clients)

(cl-i,bs-j) = have capacity for all cl-i  clients & bs-j (stations)

⇒ using Fond Fulkorson algorithm we  find the max flow in N

⇒ connecting cl-i clients to  bs-j stations

      like (cl-i, bs-j) = 1

   if f(cl-i, bs-j)  = 0

⇒ say any connection were to produce a valid flow, then

if cl-i (clients) connected                f(s,cl-i) = 1 (o otherwise)

if cl-i (clients) connected  to bs-j(stations)   f(cl-i,bs-j) = 1 (o otherwise)

   f(bs-j,t) = no of clients  (cl-i)  connected to bs-j

⇒ on each node, the max flow value (f) is longer than the no of clients that can be connected.

⇒ create the connection between the client and base station i.e. cl-l to base bs-j iff    f(cl-i, bs-j) = 1

⇒ when considering the capacity, we see that any client cannot directly connect to the base stations, and also the base stations cannot handle more than L clients, that is because the load allocated to the base statsion is L.

from this, we say f is the max no of clients (cl-i) that can be connected if we find the max flow, we can thus connect the client to the base stations easily.

cheers i hope this helps

5 0
3 years ago
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