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goldenfox [79]
3 years ago
8

Jacobson furniture pseudo-code

Computers and Technology
1 answer:
romanna [79]3 years ago
5 0

Answer:

We need to add four labels Taxable Income, Tax rate, total tax, and Total Due. And in front of them, you can add four textBoxes, or labels or as required.

The Tax rate is 5%.

And

Explanation:

BtnCalc_Click will calculate and display in lblTax

double tax = Convert.ToDouble((textBox1.Text)) * 5/100;

double Roundedtax = Math.Round(tax, 2);

textBox2.Text = Convert.ToString(Roundedtax);

textBox2.Show();

// we can use lblTax instead of textBox2. The further code will be almost change, and we need to use lblTax label instead of textBox2.Text.

BtnCalc_Click will calculate and display in lblTotal

textBox3.Text = Convert.ToString(Convert.ToDouble(textBox1.Text) + Convert.ToDouble(textBox2.Text));

textBox3.Show();

//This is lblTotal or the total Tax. We can use lblTotal label instead of the textBox3.Text.

BtnExit_Click will end the application:

Use this code in this method.

this.Close();

// This closes the application immediately on click on the Exit button.

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Consider a set of mobile computing clients in a certain town who each
poizon [28]

Answer: answer given in the explanation

Explanation:

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

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