1 00:00:00,000 --> 00:00:04,000 Now, let's look at the solution for a problem concerning waning immunity. 2 00:00:04,000 --> 00:00:10,000 First things first, the constant waning time should be defined as 2 times the infectious time. 3 00:00:10,000 --> 00:00:13,000 We can show this mathematically using the equations we have for the 4 00:00:13,000 --> 00:00:16,000 derivatives of S, I, and R with their respective time. 5 00:00:16,000 --> 00:00:20,000 We know that after a long period of time you want to attain a steady state situation 6 00:00:20,000 --> 00:00:23,000 and there's a number of people in each portion to the population-- 7 00:00:23,000 --> 00:00:27,000 susceptible, infected, and recover stays constant. 8 00:00:27,000 --> 00:00:30,000 Since they want to find out how long people should spend in the recovered stage, 9 00:00:30,000 --> 00:00:34,000 we start with the time derivative of R and set that equal to zero. 10 00:00:34,000 --> 00:00:40,000 Since you know that Rdot now has an extra term added to it or actually subtracted from it 11 00:00:40,000 --> 00:00:43,000 showing the number of people that are leaving the recovered population 12 00:00:43,000 --> 00:00:45,000 and going back to the infected population. 13 00:00:45,000 --> 00:00:49,000 We can set these to terms right here equal to zero as well. 14 00:00:49,000 --> 00:00:53,000 Just to note, I've used CINF to stand for the infectious time 15 00:00:53,000 --> 00:00:56,000 and CWAN to stand for the waning time. 16 00:00:56,000 --> 00:01:01,000 Now with just a little bit of Algebra, we come up with the answer that R=2I. 17 00:01:01,000 --> 00:01:04,000 Since we know that we want the number of recovered people to be twice the number 18 00:01:04,000 --> 00:01:09,000 of infected people, we can plug in this extra information to the equation above 19 00:01:09,000 --> 00:01:13,000 and end up with the answer that the waning time is equal to twice the infectious time. 20 00:01:13,000 --> 00:01:16,000 This put us directly into the next part of the problem. 21 00:01:16,000 --> 00:01:22,000 We defined R to S in the same way that I to R is defined except that we replace infectious time 22 00:01:22,000 --> 00:01:27,000 as the waning time and I step with R step as you can see right here. 23 00:01:27,000 --> 00:01:32,000 Then moving on to our recursive relations, for each element in a step plus one position, 24 00:01:32,000 --> 00:01:36,000 we need to take into account the value of the previous element in the number of people 25 00:01:36,000 --> 00:01:39,000 added and subtracted from the population during each time step. 26 00:01:39,000 --> 00:01:43,000 We know that the one thing that has changed in this model from the standard SIR model 27 00:01:43,000 --> 00:01:48,000 is that people are now moving from the recovered population back to the susceptible population. 28 00:01:48,000 --> 00:01:55,000 This is why we've needed to add in this extra term R to S, which we subtract from R and add to S. 29 00:01:55,000 --> 00:01:57,000 Now, let's run the program and see what we get. 30 00:01:57,000 --> 00:02:00,000 Here we see we end up with this fancy graph, 31 00:02:00,000 --> 00:02:04,000 which has three different series for the three different parts of the population we're looking at. 32 00:02:04,000 --> 00:02:08,000 Remembering how we set the initial values for S, I, and R. 33 00:02:08,000 --> 00:02:12,000 Remember that the blue line here stands for the susceptible population. 34 00:02:12,000 --> 00:02:15,000 The green line stands for the infected population, 35 00:02:15,000 --> 00:02:18,000 and the red line stands for the recovered population. 36 00:02:18,000 --> 00:02:20,000 And the maximum time that we're looking at 60 days, 37 00:02:20,000 --> 00:02:25,000 you can see that the red line is graphing twice as many people over here 38 00:02:25,000 --> 00:02:30,000 as the green line is marking, which is exactly the answer that we wanted to end up with. 39 00:02:30,000 --> 99:59:59,999 Congratulations on successfully completing the first problem of Unit 3.