Question

3. Application: Telephone Switching Networks In a directly connected telephone network, all telephones are connected and don't require a central switching station to establish calls between two telephones. For example, financial institutions on Wall Street use this type of network to maintain continuously open phone lines between firms. The number of lines needed to maintain a directly connected network for n telephones is given by this formula: lines n(n-1)/2 For example, connecting four telephones requires six separate lines, as shown in Figure CS2B. 1. Adding a fifth telephone to this network requires an additional 4 lines for a total of 10 lines. Telephone #4 Telephone #3 Line 3 Line 2 Line 1 Telephone #1 Telephone #2 Figure CS2B.1 Directly connecting four telephones Using the given formula, you're to write a C++ program that determines the number of direct lines required for connecting 100 telephones and the additional lines required if 10 telephones are added to the network Use the software development procedure, as described in the following sections Step 1: Analyze the Problem For this program, two outputs are required the number of direct lines for 100 telephones and the additional number of lines needed when 10 telephones are added to the existing network. The input item required for this problem is the number of telephones, which is denoted as n in the formula

full coding with answer

Answer 1

Solution :-

Logic :

Here, to connect n telephones we use (n*(n-1))/2 lines.

When we add an additional telephones, then to connect it to all telephone we need exactly n additional lines. So, total number of telephones becomes (n+1).

Similarly, to add an additional telephones, we need exactly (n+1) additional lines. So, total number of telephones becomes (n+2). And, so on.

eg. To add 1 additional telephone in 10 telephone network, we require 10 additional lines as to connect to all telephones.

Screenshot: -

- o х TDM-GCC 4.9.2 €4-bit Release F:\Documents\test.cpp - Dev-C++ 5.11 File Edit Search View Project Execute Tools AStyle Window Help 99 9 * - 2.5 110 :: 0 :: * 0 (globals) Project Classes Debug [*] test.cpp 1 #include<bits/stdc++.h> 2 using namespace std; 39 int main() { int n; cout<<"Enter the number of telephones : "; cin>>n; int no_direct_lines ; //Number of direct lines to connect no_direct_lines = ((n*(n-1))/2); cout<<"\nNumber of direct lines required for connecting" «n «" telephones : "<<no_direct_lines<<endl; int added_telephone; //Number of extra added telephones cout<<"\nEnter number of added addditional telephone : "; cin>>added_telephone; int additional_lines=0; //Number of additional lines required to connect for(int i=1;i<=added telephone; i++) { additional_lines = additional lines + (n); | n=n+1; // Updating no. of telephones as we are adding one telephone B! Compiler Line: 21 a 24 cout<<"\nNumber of additional lines to be added : "<<additional_lines; 25 } Resources dh Compile Log Debug Find Results Col: 52 Sel: 0 Lines: 25 Length: 807 Insert Done parsing in 0.016 seconds Type here to search A O / ENG 02:40 06-02-20208

Output :

We show two output : - a) For n=4 and additional telephones =1 (Left hand figure)

b) For n=100 and additional telephones = 10 (Right hand figure)

- O X F:\Documents\test.exe Enter the number of telephones : 4 - O X T Select F: Documents\test.exe Enter the number of telephones : 100 Number of direct lines required for connecting 100 telephones : 4950 Number of direct lines required for connecting 4 telephones : 6 Enter number of added addditional telephone : 1 Enter number of added addditional telephone : 10 Number of additional lines to be added : 1045 Number of additional lines to be added : 4 --------- === Process exited after 5.653 seconds with return value Press any key to continue ... Process exited after 21.24 seconds with return value o Press any key to continue ... Type here to search A O 2 ENG 06-02-2020 ENG 8

Code : -

#include<bits/stdc++.h>
using namespace std;
int main(){
   int n;
   cout<<"Enter the number of telephones : ";
   cin>>n;
  
   int no_direct_lines ; //Number of direct lines to connect
   no_direct_lines = ((n*(n-1))/2);
  
   cout<<"\n Number of direct lines required for connecting "
       << n <<" telephones : "<<no_direct_lines<<endl;
      
   int added_telephone;   //Number of extra added telephones
   cout<<"\nEnter number of added additional telephone : ";
   cin>>added_telephone;
  
   int additional_lines=0; //Number of additional lines required to connect
  
   for(int i=1;i<=added_telephone;i++){
       additional_lines = additional_lines + (n);
       n=n+1; //Updating no. of telephones as we are adding one telephone
   }
   cout<<"\n Number of additional lines to be added : "<<additional_lines;
}

Answer :

When n = 100, we require 4950 connecting lines.

If we add 10 additional telephones then we require 1045 additional lines.

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