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Modeling the state of a population in the face of a pandemic in C ++ assignment help

In this assignment, the behavior of a population in the face of a pandemic is modeled. For this program, our C++ assignment helpers use equations that model the number of susceptible, infected, recovered, and dead people, as time passes. In this problem, a finite population is simulated within an enclosed area (a vessel at sea) using data from a real vessel called the Diamond Princess.

Mathematical Modeling and Simulation of Diamond Princess 

The mathematical model is described by the following equations

Modeling the state of a population in the face of a pandemic in C ++ assignment help

#include #include usingnamespace std; intmain() { // files ofstream susceptible, infected; susceptible.open("Susceptible.txt"); infected.open("Infected.txt"); // Define the number of people in the curise int N =3711; // Define the time-step float dt =1.0f; // 1 day // Define the alpha value float alpha =0.1f; // so 1/alpha = 3 // Define the number of days for the study constint days =150; // 150 days // We need to calculate the value of R0 // Inside Diamond Princess cruise, the numbers of infected went from 1 to 621 in 31 days float R0 =3.231f; float beta = alpha * R0 /static_cast(N); std::cout <<"Alpha: "<< alpha << std::endl; std::cout <<"beta: "<< beta << std::endl; std::cout <<"R0: "<< R0 << std::endl << std::endl; // Now, define the initial conditions for S and I float S, I, R, T; float Snew, Inew; S = N-1; // 0 susceptible persons at day 0 I =1; // 1 infected person at day 0 R =0; T = R + I; // Now, start the simulations int day =1; /* ANSWERS: 1) For alpha = 1/3, R must be 1.6103 2) For alpha = 1/6, R0 must be 2.298 3) for alpha = 1/10, R0 must be 3.231 // For alpha = 1/3, the day at which the number of infected reached a max was at day 39 with a total of 329 infected // For alpha = 1/6, the day at which the number of infected reached a max was at day 42 with a total of 790 infected // For alpha = 1/10, the day at which the number of infected reached a max was at day 44 with a total of 1267 infected For alpha = 1/6, the earliest date that it would be safe to allow all the passengers to disembark is at day 104, when there are no more infected // The chance of being susceptible is calculated as the number of remaining susceptible persons, divided // by the total number of passengers For alpha = 1/3, the chance of being susceptible after the number if infected is zero is: 33.93% For alpha = 1/6, the chance of being susceptible after the number if infected is zero is: 12.72% For alpha = 1/10, the chance of being susceptible after the number if infected is zero is: 3.98% */ susceptible <(S) << endl; infected <(I) << endl; std::cout <<"Day\t Susceptible\t Infected\t Recovered\t\tI+R"<< std::endl; std::cout <<"------------------------------------------------------------------"<< std::endl; std::cout <<0<<"->\t|\t"<< (int)S <<"\t|\t"<< (int)I <<"\t|\t"<< (int)R <<"\t|\t"<< (int)T << std::endl; for (int i =0; i < days; i++) { Snew= S - beta * I*S; Inew = I + beta * I*S - alpha * I; R = N - Snew - I; T = R + I; std::cout << day <<"->\t|\t"<< (int)Snew <<"\t|\t"<< (int)Inew <<"\t|\t"<< (int)R <<"\t|\t"<< (int)T; if (day ==31) cout <<" **"; cout << endl; susceptible <(Snew) << endl; infected <(Inew) << endl; S = Snew; I = Inew; day++; } susceptible.close(); infected.close(); }