Models of Infectious Diseases
4858 words (19 pages) Essay
5th Sep 2017 Health And Social Care Reference this
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1.1 Introduction
As an aspiring doctor, I am always on the lookout for medical news and keeping up to date with the current affairs is something that has always interested me. Another thing which has sparked my attention over the recent years is the ability of mathematicians and scientists to model infectious disease, in order for the medical professionals to make decisions on how best to act.
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Find out moreFor instance, prior to the 2016 Rio Olympic Games there was a lot of controversy due to competitors pulling out of their events because of the scare of contracting and spreading the Zika Virus. It is times like these that processes, such as modelling infectious disease, are so important; in order to understand whether an infectious disease will spread, how it will spread and what the dangers of the spread of the diseases are.
These models allow decisions to be made such as; whether the border control procedures need to change, whether there should be research into developing new drugs or whether they need to take further action and produce a vaccination.
The SIR model
One method of modelling infectious disease is through using the SIR model. This model starts from the premise that the whole population can be divided into three distinct classes; susceptible, infectious and recovered. ‘Susceptible’ refers to anybody that is currently healthy, but could potentially contract the disease. ‘Infectious’ refers to anybody that is carrying or suffering the disease, meaning that they could spread the disease to anybody that is susceptible. Finally, the ‘recovered’ class refers to anybody that is assumed to be immune to the disease; this could be naturally or through vaccination.
The SIR model is used in epidemiology in order to understand how the disease has evolved over time, and more importantly how it will evolve over time. It is therefore known as a dynamical system; because it tells us how state variables change over time. In this case, the state variables are the population of the three classes; ‘susceptible’, ‘infectious’ or ‘recovered’.
Notation to understand:
In order to model infectious disease, two parameters must be given:
a = the recovery rate
b = the infectious rate
The following notation represents the state variables:
S_{t}= the number of susceptible individuals at time t
I_{t }= the number of infective individuals at time t
R_{t }= the number of recovered individuals at time t
T = the time, in days
The changing state variables:
If S_{t }is the number of susceptible individuals at time t thenâ€¦
(1)
Andâ€¦
Â (2)
It can therefore be deduced that the change in the number of susceptible individuals is equal to the negative number of individuals that become infected in day t. The number is negative because in order for someone to become infected they move out of the susceptible state and therefore the population of susceptible individuals decreases.
(3)
The change in the number of infective individuals is equal to the number of individuals that are infected in day t minus the number of individuals that recover in day t.
(4)
Thirdly, the change in the number of recovered individuals is equal to equation (2), the recovery rate multiplied by the number of infective individuals at time t.
(5)
A theoretical example:
If we specify some values for the recovery rate and the rate of infection, we can begin to understand how individuals move between these classes over time.
Parameters for an infectious disease:
a = 0.05
b = 0.0001
Initial state variable values:
S_{0 }= 10000
I_{0 }= 1000
R_{0 }= 0
Applying equation (3) tells us the number of individuals that will become infected on day 1:
Therefore, from day 0 to day 1 the number of susceptible individuals will change to:
The change in the number of recovered people, given in equation (5) is as follows:
(5)
So the number of recovered people after day 1 will be:
Finally, the number of infectious individuals will be:
(4)
Therefore, we can see that from day 0 to day 1 the populations have changed to:
S_{1 }= 9000
I_{1 }= 1950
R_{1 }= 50
If we repeat the process for the next 3 days, we can begin to see how the disease could spread over time:
Day 2:
(3)
(5)
(4)
Day 3:
(3)
(5)
(4)
Table to show the populations after four days:
Day 
Number of susceptible individuals 
To the nearest whole number 
0 
10000 
10000 
1 
9000 
9000 
2 
7245 
7245 
3 
4631.36625 
4631 
Number of infectious individuals 
To the nearest whole number 

0 
1000 
1000 
1 
1950 
1950 
2 
3607.5 
3608 
3 
6040.75875 
6041 
Day 
Number of recovered individuals 
To the nearest whole number 
0 
0 
0 
1 
50 
50 
2 
147.5 
148 
3 
327.875 
328 
If we plot these on a graph against time we can see how this disease has spread through the population:
Continuing this process for the next fifteen days gives the following results:
Day: 
Number of susceptible people 
Number of infectious people 
Number of recovered people 
0 
9000 
1950 
50 
1 
7245 
3607.5 
147.5 
2 
4631.36625 
6040.75875 
327.875 
3 
1833.66963 
8536.417432 
629.9129375 
4 
268.3726905 
9674.8935 
1056.733809 
5 
8.724970602 
9450.796545 
1540.478484 
6 
0.4791784 
8986.50251 
2013.018311 
7 
0.048564611 
8537.607998 
2462.343437 
8 
0.00710205 
8110.769061 
2889.223837 
9 
0.001341741 
7705.236368 
3294.76229 
10 
0.000307898 
7319.975584 
3680.024108 
11 
8.25174E05 
6953.97703 
4046.022887 
12 
2.5135E05 
6606.278236 
4393.721739 
13 
8.53011E06 
6275.964341 
4724.035651 
14 
3.17665E06 
5962.166129 
5037.833868 
15 
3.17665E06 
5962.166129 
5037.833868 
Rounding these numbers to the nearest whole numbers, and excluding negative numbers gives the following table:
Day: 
Number of susceptible people 
Number of infectious people 
Number of recovered people 
0 
9000 
1950 
50 
1 
7245 
3608 
148 
2 
4631 
6041 
328 
3 
1834 
8536 
630 
4 
268 
9675 
1057 
5 
9 
9451 
1540 
6 
0 
8987 
2013 
7 
0 
8538 
2462 
8 
0 
8111 
2889 
9 
0 
7705 
3295 
10 
0 
7320 
3680 
11 
0 
6954 
4046 
12 
0 
6606 
4394 
13 
0 
6276 
4724 
14 
0 
5962 
5038 
15 
0 
5962 
5038 
From the graph we can see that there is a point at which all of the groups level off. For example, by day 14 the recovered and infectious group level off, and be day 5 the susceptible group levels off at 0. The SIR model is therefore useful at this basic level in order to see how the three state variables interact with one another.
Using the SIR model for Ebola:
However, in order to model a disease such as Ebola, a few changes to the equations must be made. Two parameters must be introduced; the mortality rate and the duration of the disease.
The SIR model will be used to model the spread of Ebola in Guinea, according to the data provided by WHO, in 2014.
Calculating the rate of infection (b) and the rate of recovery (a) requires the mortality rate and the duration of the disease. The infection rate is equal to the mortality rate divided by the number of susceptible people. The recovery rate is equal to 1 divided by the duration of the disease.
In June 2014, there were 390 total cases of Ebola, 273 of these resulting in death. The duration of the disease is between 2 and 21 days. The duration of the disease used for the SIR model will be the midpoint; 11.5 days. The mortality rate of Ebola at this point was (273/390) 0.7. The population of susceptible people in Guinea in 2014 was 11,474,383
Therefore, the infection rate and recovery rate are as follows:
Bringing back the equations from before, the following values are going to be used, having rounded the recovery and infection rate to 1 significant figure.
a =
S_{0 }= (11474383 – 117) 11,474,266
I_{0 }= (390 – 273) 117
R_{0 }= 273
T_{0 }= the start of June 2014
The following tables show the process of calculating the changes in populations:
a 
b 
s 
i 
r 
t 
0.1 
0.00000006 
11,474,266 
117 
273 
0 
0.1 
0.00000006 
80.54934732 
68.84935 
11.7 
1 
0.1 
0.00000006 
127.9481926 
109.3633 
18.58493 
2 
0.1 
0.00000006 
122.6893008 
104.868 
17.82126 
3 
0.1 
0.00000006 
147.484557 
126.0614 
21.42313 
4 
0.1 
0.00000006 
158.9781358 
135.8852 
23.09295 
5 
0.1 
0.00000006 
180.3286874 
154.134 
26.19466 
6 
0.1 
0.00000006 
199.6512232 
170.6493 
29.00192 
7 
0.1 
0.00000006 
223.579187 
191.1009 
32.47833 
8 
0.1 
0.00000006 
249.0221104 
212.8471 
36.17502 
9 
0.1 
0.00000006 
278.0642545 
237.6695 
40.39479 
10 
0.1 
0.00000006 
310.1130082 
265.0614 
45.05166 
11 
0.1 
0.00000006 
346.0453313 
295.7723 
50.27308 
12 
0.1 
0.00000006 
386.0276536 
329.9443 
56.08336 
13 
0.1 
0.00000006 
430.6727697 
368.1011 
62.57165 
14 
0.1 
0.00000006 
480.4377735 
410.6332 
69.80454 
15 
0.1 
0.00000006 
535.9504153 
458.077 
77.87343 
16 
0.1 
0.00000006 
597.8468096 
510.9758 
86.87102 
17 
0.1 
0.00000006 
666.8678465 
569.9626 
96.90528 
18 
0.1 
0.00000006 
743.8203017 
635.7265 
108.0938 
19 
0.1 
0.00000006 
829.610506 
709.0416 
120.5689 
20 
0.1 
0.00000006 
925.2410683 
790.7643 
134.4768 
21 
0.1 
0.00000006 
1031.828496 
881.8479 
149.9806 
22 
0.1 
0.00000006 
1150.611313 
983.3501 
167.2612 
23 
0.1 
0.00000006 
1282.964805 
1096.445 
186.5198 
24 
0.1 
0.00000006 
1430.413979 
1222.434 
207.9795 
25 
0.1 
0.00000006 
1594.649251 
1362.761 
231.8879 
26 
0.1 
0.00000006 
1777.54255 
1519.023 
258.5196 
27 
0.1 
0.00000006 
1981.165088 
1692.987 
288.1784 
28 
To the nearest whole number, the following table shows the number of people in each population over 28 days.
Day: 
Number of susceptible people after day t 
Number of infectious people after day t 
Number of recovered people after day t 
0 
11474185 
186 
285 
1 
11474058 
178 
30 
2 
11473935 
214 
36 
3 
11473787 
231 
39 
4 
11473628 
262 
45 
5 
11473448 
290 
49 
6 
11473248 
325 
55 
7 
11473025 
362 
61 
8 
11472776 
404 
69 
9 
11472498 
451 
77 
10 
11472188 
503 
85 
11 
11471842 
561 
95 
12 
11471456 
626 
106 
13 
11471025 
698 
119 
14 
11470544 
779 
132 
15 
11470008 
869 
148 
16 
11469411 
969 
165 
17 
11468744 
1081 
184 
18 
11468000 
1206 
205 
19 
11467170 
1345 
229 
20 
11466245 
1500 
255 
21 
11465213 
1673 
284 
22 
11464063 
1865 
317 
23 
11462780 
2080 
354 
24 
11461349 
2319 
394 
25 
11459755 
2585 
440 
26 
11457977 
2882 
490 
27 
11455996 
3212 
547 
28 
11455996 
1693 
288 
Plotting these on graphsâ€¦
To calculate the equation of graph 1â€¦
The graph appears to be exponential, so using the form y = Ae^{kt }, I will try to calculate the equation.
Taking days 3 and 24, I can solve the equation simultaneously.
11473787 = Ae^{3k }
11461349 = Ae^{kt }
Ln (11473787) = ln A x 3k
Ln (11461349) = ln A x 24k
K = 0.000051648769
Substituting into the original equations to find Aâ€¦
Ln (11433787) = ln A + 3k
Ln A = ln (11473787) – 3k
Ln A = 16.25568â€¦
A = 11475064.9
Soâ€¦
N (of susceptible) = 11475064.9e^{0.000051648769t}
Due to the massive range of numbers, the three series cannot be plotted on the same graph, and be of any use. Therefore, three separate graphs are necessary. The graphs show that with the parameters calculated for Ebola in 2014, the population of infectious people seems appears to be increasing exponentially, with no sign of a slowing rate. A similar pattern is shown in the recovered population, whilst the number of susceptible people is appearing to decrease exponentially.
The graphs over just 28 days did not show much, so the equations were used over a longer period of time. These graphs are shown below:
As expected, and shown in the previous graphs, the number of recovered and infected people increases rapidly over approximately the first 75 days, after which point the populations of both of these populations reduce in size. The number of susceptible people decreases after approximately 45 days, because more people have become infected or recovered. The number of susceptible people is important in predicting the way that the disease will spread, as the more susceptible people there are, along with a high infection rate, the more likely it is that another epidemic will occur.
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View our servicesThe study of models of the Ebola virus are particularly important at the moment, as there are worries that the disease will spread in a different way, as it adapts to become airborne. Therefore, it is very important to look at how the disease has spread in the past, not just how it could spread in the future.
The graph below shows the number of cases of Ebola in Guinea, Liberia and Sierra Leone.
Andâ€¦
Â (2)
It can therefore be deduced that the change in the number of susceptible individuals is equal to the negative number of individuals that become infected in day t. The number is negative because in order for someone to become infected they move out of the susceptible state and therefore the population of susceptible individuals decreases.
(3)
The change in the number of infective individuals is equal to the number of individuals that are infected in day t minus the number of individuals that recover in day t.
(4)
Thirdly, the change in the number of recovered individuals is equal to equation (2), the recovery rate multiplied by the number of infective individuals at time t.
(5)
A theoretical example:
If we specify some values for the recovery rate and the rate of infection, we can begin to understand how individuals move between these classes over time.
Parameters for an infectious disease:
a = 0.05
b = 0.0001
Initial state variable values:
S_{0 }= 10000
I_{0 }= 1000
R_{0 }= 0
Applying equation (3) tells us the number of individuals that will become infected on day 1:
Therefore, from day 0 to day 1 the number of susceptible individuals will change to:
The change in the number of recovered people, given in equation (5) is as follows:
(5)
So the number of recovered people after day 1 will be:
Finally, the number of infectious individuals will be:
(4)
Therefore, we can see that from day 0 to day 1 the populations have changed to:
S_{1 }= 9000
I_{1 }= 1950
R_{1 }= 50
If we repeat the process for the next 3 days, we can begin to see how the disease could spread over time:
Day 2:
(3)
(5)
(4)
Day 3:
(3)
(5)
(4)
Table to show the populations after four days:
Day 
Number of susceptible individuals 
To the nearest whole number 
0 
10000 
10000 
1 
9000 
9000 
2 
7245 
7245 
3 
4631.36625 
4631 
Number of infectious individuals 
To the nearest whole number 

0 
1000 
1000 
1 
1950 
1950 
2 
3607.5 
3608 
3 
6040.75875 
6041 
Day 
Number of recovered individuals 
To the nearest whole number 
0 
0 
0 
1 
50 
50 
2 
147.5 
148 
3 
327.875 
328 
If we plot these on a graph against time we can see how this disease has spread through the population:
Continuing this process for the next fifteen days gives the following results:
Day: 
Number of susceptible people 
Number of infectious people 
Number of recovered people 
0 
9000 
1950 
50 
1 
7245 
3607.5 
147.5 
2 
4631.36625 
6040.75875 
327.875 
3 
1833.66963 
8536.417432 
629.9129375 
4 
268.3726905 
9674.8935 
1056.733809 
5 
8.724970602 
9450.796545 
1540.478484 
6 
0.4791784 
8986.50251 
2013.018311 
7 
0.048564611 
8537.607998 
2462.343437 
8 
0.00710205 
8110.769061 
2889.223837 
9 
0.001341741 
7705.236368 
3294.76229 
10 
0.000307898 
7319.975584 
3680.024108 
11 
8.25174E05 
6953.97703 
4046.022887 
12 
2.5135E05 
6606.278236 
4393.721739 
13 
8.53011E06 
6275.964341 
4724.035651 
14 
3.17665E06 
5962.166129 
5037.833868 
15 
3.17665E06 
5962.166129 
5037.833868 
Rounding these numbers to the nearest whole numbers, and excluding negative numbers gives the following table:
Day: 
Number of susceptible people 
Number of infectious people 
Number of recovered people 
0 
9000 
1950 
50 
1 
7245 
3608 
148 
2 
4631 
6041 
328 
3 
1834 
8536 
630 
4 
268 
9675 
1057 
5 
9 
9451 
1540 
6 
0 
8987 
2013 
7 
0 
8538 
2462 
8 
0 
8111 
2889 
9 
0 
7705 
3295 
10 
0 
7320 
3680 
11 
0 
6954 
4046 
12 
0 
6606 
4394 
13 
0 
6276 
4724 
14 
0 
5962 
5038 
15 
0 
5962 
5038 
From the graph we can see that there is a point at which all of the groups level off. For example, by day 14 the recovered and infectious group level off, and be day 5 the susceptible group levels off at 0. The SIR model is therefore useful at this basic level in order to see how the three state variables interact with one another.
Using the SIR model for Ebola:
However, in order to model a disease such as Ebola, a few changes to the equations must be made. Two parameters must be introduced; the mortality rate and the duration of the disease.
The SIR model will be used to model the spread of Ebola in Guinea, according to the data provided by WHO, in 2014.
Calculating the rate of infection (b) and the rate of recovery (a) requires the mortality rate and the duration of the disease. The infection rate is equal to the mortality rate divided by the number of susceptible people. The recovery rate is equal to 1 divided by the duration of the disease.
In June 2014, there were 390 total cases of Ebola, 273 of these resulting in death. The duration of the disease is between 2 and 21 days. The duration of the disease used for the SIR model will be the midpoint; 11.5 days. The mortality rate of Ebola at this point was (273/390) 0.7. The population of susceptible people in Guinea in 2014 was 11,474,383
Therefore, the infection rate and recovery rate are as follows:
Bringing back the equations from before, the following values are going to be used, having rounded the recovery and infection rate to 1 significant figure.
a =
S_{0 }= (11474383 – 117) 11,474,266
I_{0 }= (390 – 273) 117
R_{0 }= 273
T_{0 }= the start of June 2014
The following tables show the process of calculating the changes in populations:
a 
b 
s 
i 
r 
t 
0.1 
0.00000006 
11,474,266 
117 
273 
0 
0.1 
0.00000006 
80.54934732 
68.84935 
11.7 
1 
0.1 
0.00000006 
127.9481926 
109.3633 
18.58493 
2 
0.1 
0.00000006 
122.6893008 
104.868 
17.82126 
3 
0.1 
0.00000006 
147.484557 
126.0614 
21.42313 
4 
0.1 
0.00000006 
158.9781358 
135.8852 
23.09295 
5 
0.1 
0.00000006 
180.3286874 
154.134 
26.19466 
6 
0.1 
0.00000006 
199.6512232 
170.6493 
29.00192 
7 
0.1 
0.00000006 
223.579187 
191.1009 
32.47833 
8 
0.1 
0.00000006 
249.0221104 
212.8471 
36.17502 
9 
0.1 
0.00000006 
278.0642545 
237.6695 
40.39479 
10 
0.1 
0.00000006 
310.1130082 
265.0614 
45.05166 
11 
0.1 
0.00000006 
346.0453313 
295.7723 
50.27308 
12 
0.1 
0.00000006 
386.0276536 
329.9443 
56.08336 
13 
0.1 
0.00000006 
430.6727697 
368.1011 
62.57165 
14 
0.1 
0.00000006 
480.4377735 
410.6332 
69.80454 
15 
0.1 
0.00000006 
535.9504153 
458.077 
77.87343 
16 
0.1 
0.00000006 
597.8468096 
510.9758 
86.87102 
17 
0.1 
0.00000006 
666.8678465 
569.9626 
96.90528 
18 
0.1 
0.00000006 
743.8203017 
635.7265 
108.0938 
19 
0.1 
0.00000006 
829.610506 
709.0416 
120.5689 
20 
0.1 
0.00000006 
925.2410683 
790.7643 
134.4768 
21 
0.1 
0.00000006 
1031.828496 
881.8479 
149.9806 
22 
0.1 
0.00000006 
1150.611313 
983.3501 
167.2612 
23 
0.1 
0.00000006 
1282.964805 
1096.445 
186.5198 
24 
0.1 
0.00000006 
1430.413979 
1222.434 
207.9795 
25 
0.1 
0.00000006 
1594.649251 
1362.761 
231.8879 
26 
0.1 
0.00000006 
1777.54255 
1519.023 
258.5196 
27 
0.1 
0.00000006 
1981.165088 
1692.987 
288.1784 
28 
To the nearest whole number, the following table shows the number of people in each population over 28 days.
Day: 
Number of susceptible people after day t 
Number of infectious people after day t 
Number of recovered people after day t 
0 
11474185 
186 
285 
1 
11474058 
178 
30 
2 
11473935 
214 
36 
3 
11473787 
231 
39 
4 
11473628 
262 
45 
5 
11473448 
290 
49 
6 
11473248 
325 
55 
7 
11473025 
362 
61 
8 
11472776 
404 
69 
9 
11472498 
451 
77 
10 
11472188 
503 
85 
11 
11471842 
561 
95 
12 
11471456 
626 
106 
13 
11471025 
698 
119 
14 
11470544 
779 
132 
15 
11470008 
869 
148 
16 
11469411 
969 
165 
17 
11468744 
1081 
184 
18 
11468000 
1206 
205 
19 
11467170 
1345 
229 
20 
11466245 
1500 
255 
21 
11465213 
1673 
284 
22 
11464063 
1865 
317 
23 
11462780 
2080 
354 
24 
11461349 
2319 
394 
25 
11459755 
2585 
440 
26 
11457977 
2882 
490 
27 
11455996 
3212 
547 
28 
11455996 
1693 
288 
Plotting these on graphsâ€¦
To calculate the equation of graph 1â€¦
The graph appears to be exponential, so using the form y = Ae^{kt }, I will try to calculate the equation.
Taking days 3 and 24, I can solve the equation simultaneously.
11473787 = Ae^{3k }
11461349 = Ae^{kt }
Ln (11473787) = ln A x 3k
Ln (11461349) = ln A x 24k
K = 0.000051648769
Substituting into the original equations to find Aâ€¦
Ln (11433787) = ln A + 3k
Ln A = ln (11473787) – 3k
Ln A = 16.25568â€¦
A = 11475064.9
Soâ€¦
N (of susceptible) = 11475064.9e^{0.000051648769t}
Due to the massive range of numbers, the three series cannot be plotted on the same graph, and be of any use. Therefore, three separate graphs are necessary. The graphs show that with the parameters calculated for Ebola in 2014, the population of infectious people seems appears to be increasing exponentially, with no sign of a slowing rate. A similar pattern is shown in the recovered population, whilst the number of susceptible people is appearing to decrease exponentially.
The graphs over just 28 days did not show much, so the equations were used over a longer period of time. These graphs are shown below:
As expected, and shown in the previous graphs, the number of recovered and infected people increases rapidly over approximately the first 75 days, after which point the populations of both of these populations reduce in size. The number of susceptible people decreases after approximately 45 days, because more people have become infected or recovered. The number of susceptible people is important in predicting the way that the disease will spread, as the more susceptible people there are, along with a high infection rate, the more likely it is that another epidemic will occur.
The study of models of the Ebola virus are particularly important at the moment, as there are worries that the disease will spread in a different way, as it adapts to become airborne. Therefore, it is very important to look at how the disease has spread in the past, not just how it could spread in the future.
The graph below shows the number of cases of Ebola in Guinea, Liberia and Sierra Leone.
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