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zhuklara [117]
3 years ago
5

What is the bandwidth-delay product for a 50 Mbps channel on a geostationary satellite (assume a RTT of 540 ms)? If the packets

are all 1500 bytes (including headers), how big should the window be in packets to maximize the packet delivery rate?
Computers and Technology
1 answer:
gogolik [260]3 years ago
8 0

Answer:

The answer is 2250 packets.

Explanation:

The round trip delay is about 540 millisec, so with a 50 Mbps channel the bandwidth product delay is 27 megabits or 3.375.000 bytes. With packets of 1500 bytes, it takes 2250 packets to fill hte pipe, so the window should be at least 2250 packets.

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The Polish mathematician Wacław Sierpiński described the pattern in 1915, but it has appeared in Italian art since the 13th cent
soldi70 [24.7K]

Answer:

/ Sierpinski.java

public class Sierpinski {

     // method to find the height of an equilateral triangle with side length =

     // length

     public static double height(double length) {

           // formula= length*sqrt(3)/2

           double h = (length * Math.sqrt(3)) / 2.0;

           return h;

     }

     // method to draw a filled triangle (pointed downwards) with bottom vertex

     // x,y

     public static void filledTriangle(double x, double y, double length) {

           // finding height

           double h = height(length);

           // filling triangle as a polygon

           // passing x, x-length/2, x+length/2 as x coordinates

           // and y, y+h, y+h as y coordinates

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     // method to draw an n level sierpinski triangle

     public static void sierpinski(int n, double x, double y, double length) {

           // checking if n is above 0 (base condition)

           if (n > 0) {

                 // drawing a filled triangle with x, y as bottom coordinate, length

                 // as side length

                 filledTriangle(x, y, length);

                 // drawing the triangle(s) on the top

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                 // drawing the triangle(s) on the right side

                 sierpinski(n - 1, x + (length / 2.0), y, length / 2);

           }

     }

     public static void main(String[] args) {

           // parsing first command line argument as integer, if you dont provide

           // the value while running the program, this program will cause

           // exception.

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           // length of the outline triangle

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           // finding height

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           // drawing a triangle (pointed upwards) to represent the outline.

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     }

}

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