4 Illustrative example: Infected Cohort

Let us assume a situation wherein we have a closed cohort of 1000 individuals who are infected as well as infective. The duration of infectiousness will remain for 07 days. Once recovered, they remain immune throughout their lifespan.

4.1 Understanding the model

Figure 4.1: Infected Cohort Model

4.2 Evidence synthesis

As a pre-requisite to model building, we need to undertake evidence synthesis exercise to extract required parameters from the literature or by conducting a study.

initial_number_infected <- 1000
initial_number_recovered <- 0 
Duration_of_infectiousness <- 7 
recovery_rate <- 1/Duration_of_infectiousness
follow_up_duration <- 30

4.3 Create input parameters.

Once parameters are obtained, they are placed into three major parameter inputs viz initial values for each compartment, transition parameters from one compartment to another, and the time stamps for which modelling is required.

initial_state_values <- c(I = initial_number_infected, 
                          R = initial_number_recovered) 

parameters <- c(gamma = recovery_rate) 
times <- seq(from = 0, to = follow_up_duration, by = 1) 

initial_state_values

##    I    R 
## 1000    0

parameters

##     gamma 
## 0.1428571

times

##  [1]  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
## [26] 25 26 27 28 29 30

4.4 Create a model function using input values

The model is created from the above mentioned parameters by specifying differential equations. In R, a function is an object so that the researcher is able to pass control to the function, along with arguments that may be necessary for the function to accomplish the actions. For performing differential equations, we create a function with compartment as well as transition parameter at varied timestamps.

cohort_model <- function(time, state, parameters) {     
   with(as.list(c(state, parameters)), {  
   dI <- -gamma * I
   dR <- gamma * I
   return(list(c(dI, dR)))                             
   })
}

4.5 Return a dataframe as output using model equations

A data frame is constructed by application of the differential model. Herein, Odinary Differential Equations (ode) function takes initial compartment values, timestamps, parameters and the model function as arguments.

output <- as.data.frame(ode(y = initial_state_values, 
                            times = times, 
                            func = cohort_model,
                            parms = parameters))

4.6 Exploratory Data Analysis of the output

knitr::kable(output)

time	I	R
0	1000.00000	0.0000
1	866.87790	133.1221
2	751.47726	248.5227
3	651.43886	348.5611
4	564.71820	435.2818
5	489.54178	510.4582
6	424.37297	575.6270
7	367.87958	632.1204
8	318.90670	681.0933
9	276.45316	723.5468
10	239.65110	760.3489
11	207.74823	792.2518
12	180.09235	819.9076
13	156.11808	843.8819
14	135.33531	864.6647
15	117.31919	882.6808
16	101.70141	898.2986
17	88.16270	911.8373
18	76.42630	923.5737
19	66.25227	933.7477
20	57.43262	942.5674
21	49.78707	950.2129
22	43.15931	956.8407
23	37.41385	962.5861
24	32.43324	967.5668
25	28.11566	971.8843
26	24.37284	975.6272
27	21.12828	978.8717
28	18.31564	981.6844
29	15.87742	984.1226
30	13.76379	986.2362

Since, more than one column is representing the state of the person (Infected or recovered), we need to transform the data into tidy format. A tidy data format is said to present when each column represents one variable, each row has one observation and each cell has one value.

dat <- output %>% pivot_longer(cols = c(I,R),
                        names_to = "state",
                        values_to = "value")
knitr::kable(head(dat,10))

time	state	value
0	I	1000.0000
0	R	0.0000
1	I	866.8779
1	R	133.1221
2	I	751.4773
2	R	248.5227
3	I	651.4389
3	R	348.5611
4	I	564.7182
4	R	435.2818

Visualization is a recommended exploratory data analysis method. Let us see how infected and recovered compartments were changing in the model.

dat %>% ggplot(aes(x = time,
                   y = value,
                   color = state,
                   group = state))+
  geom_line()+
  xlab("Time(Days)")+
  ylab("Number of persons")+
  labs(title = "Changes in Infected and Recovered individuals")

Based on the output, at day 03 of follow up, as a Public Health specialist, you are required to provide inputs for logistic planning for the next 28 days. You are required to provide following details:-

How many people in the cohort are likely to recover by end of 4 weeks of follow- up? What is the proportion of people who are likely to be recovered in next 28 days?
If all of them were hospitalized patients, how many people are likely to be present in the hospital after 28 days?
Assuming the disease require intensive care to the extent of one ventilator requirement for every five patients, how many ventilators are required to be functional at 07 days of follow up?

Recovered after 28 days:-

dat$value[dat$time == 28 &
            dat$state == "R"]

## [1] 981.6844

Proportion of cohort recovered:-

(dat$value[dat$time== 28 & dat$state == "R"]/
  (dat$value[dat$time== 28 & dat$state == "R"] + 
     dat$value[dat$time== 28 & dat$state == "I"]))*100

## [1] 98.16844

Likely to be in the hospital after 28 days:-

dat$value[dat$time == 28 &
            dat$state == "I"]

## [1] 18.31564

Ventilators required at 7 days:-

(dat$value[dat$time == 7 &
            dat$state == "I"])/5

## [1] 73.57592

Illustration: Infected Cohort models using R

Dr Gurpreet Singh, Dr Biju Soman

2020-11-20

1 Introduction.

2 Load Library.

3 Framework for mathematical modeling