Mathematical modelling and controlling the dynamics of infectious diseases

Transcription

1 Mathematical modelling and controlling the dynamics of infectious diseases Musa Mammadov Centre for Informatics and Applied Optimisation Federation University Australia 25 August 2017, School of Science, RMIT

2 Joint work Rob Evans (University of Melbourne): R.J. Evans and M. Mammadov, Dynamics of Ebola epidemics in West Africa 2014, F1000Research 2015, 3:319 (doi: /f1000research ) (This article is included in the Ebola collection, Peter Piot) R.J. Evans and M. Mammadov, Predicting and controlling the dynamics of infectious diseases, CDC-2015: 54th IEEE Conference on Decision and Control (CDC),

3 Introduction Mathematical models SI, SIR, SEIR, SEIRS, time-delay Possible optimal control models New results Ebola epidemics in West Africa Optimal control problems capacity of beds in hospitals time to isolation (hospitalization) numerical example

4 Introduction The West African Ebola virus epidemic of 2013 to 2016 was the most widespread epidemic of Ebola virus disease in history. Country Cases Deaths Liberia 10,675 4,809 Sierra Leone 14,122 3,955 Guinea 3,814 2,543 Total 28,616 11,310

5 Mathematical models: Statistical-Based Methods Dynamical systems (State-Space Models - SI, SIR, SIRS, SEIR) Empirical/Machine Learning-Based Models

6 Model SIR : N(t) = S(t) + I(t) + R(t) N(t) - total population S(t) - susceptible population I(t) - infectious population R(t) - recovered population ds dt = β S I di dt = β S I γ I dr dt = γ I F (S, I) = β S I - force of infection N(t) = constant : dn dt = 0

7 Figure 1: SIR S(t) 0, I(t) 0, R(t) S 0

8 More Realistic Model SIR : λ - birth rate µ - natural death rate γ - recovery rate ds dt = λ µ S β S I di dt = β S I (γ + µ) I dr dt = γ I µ R

9 Analysis/Investigations on SIR : Stability of solutions: (S(t), I(t), R(t)) as t Reproduction Number: R 0 = βλ µ(µ+γ) Equilibriums: λ µ S β S I = 0 β S I (γ + µ) I = 0 γ I µ R = 0 R 0 1 : (S, I, R) = ( λ, 0, 0) - disease-free equilibrium µ R 0 > 1 : (S, I, R) = ( γ+µ β, µ β (R 0 1), γ β (R 0 1)) Theorem: If R 0 1 then lim t (S(t), I(t), R(t)) = (S, I, R) ; If R 0 > 1 then lim t (S(t), I(t), R(t)) = (S, I, R).

10 Generalizations SIRS : part of Recovered population become susceptible (R(t) S(t)) Take into account death from disease: I = β S I (γ + µ + α) I Adding new compartments; e.g. exposed population SEIR : Ṡ = λ µ S β S I + ξ R Ė = β S I σ E I = σ E (γ + µ + α) I Ṙ = γ I µ R ξ R E(exposed) becomes I(infectious) after some time (2-21 days in Ebola) Vaccination, age

11 Malaria: (S h E h I h R h ) - for human, (S m E m I m ) - for mosquito Figure 2:

12 Generalizations Pros: Good for the study particular effects (e.g. Vaccination - newborns or non-newborns) Cons: Not suitable for prediction (overfitting)! data points (weekly for I and Deaths) Ebola: many papers (Mid-2014) predicted 100,000s or millions of cases for Dec Reality: around 30,000 ( times less!) Infection control: it is crucial to have Accurate/predictive models Control parameter(s) Practical measures (realistic control parameters)

13 Time delay: a good way to simplify the model Assumption: E(exposed) becomes I(infectious) after τ 1 (days) If ξ = 0 (no R S): Ṡ(t) = λ µ S(t) β S(t) I(t) + ξ R(t) Ė(t) = β S(t) I(t) e µτ 1 β S(t τ 1 ) I(t τ 1 ) I(t) = e µτ 1 β S(t τ 1 ) I(t τ 1 ) (γ + µ + α) I(t) Ṙ(t) = γ I(t) µ R(t) ξ R(t) Ṡ(t) = λ µ S(t) β S(t) I(t) I(t) = e µτ 1 β S(t τ 1 ) I(t τ 1 ) (γ + µ + α) I(t) Do we really need S(t)? E.g. in Sierra Leone: S(t) 7, 000, 000 I(t) 14, 122 (0.2% of S(t))

14 Control models: Possible measures/interventations: distribution strategies for vaccination antibiotic programs, safe burials and community engagement travel restrictions Many papers consider β as a control parameter ( 0) Our approach Key point: it uses a second time delay to take into account isolation (hospitalization, beds)

15 I(t + 1) = β τ 1 (1 αω(i)) I(t τ 1 i). i=0 Here ω - gamma distribution, There are 2 time-delays: R 0 = β[τ α τ 1 i=0 ω(i)] τ 1 is the latent period (infected becomes infectious); τ the average time of isolation (i.e. hospitalization) There are 3 parameters: α (death rate) and β (transmission coeff.) constants; τ time dependent (τ(t) {3, 4, 5}).

16 The main features of the model: Simple (only 3 parameters) predictions more reliable; Since τ(t) {3, 4, 5} it is easy to create future scenarios; τ the average time of isolation/hospitalization: it can be well connected to preventive measures to control the spread of infection How it works in data fitting and prediction?

17 α = const, β = const, t [0, T ]; τ(t) = const in each [0, t 1 ], [t 1, t 2 ], [t 2, T ]

18 Figure 3: τ(t) {3, 4, 5} for t 11/Nov/2014

19 After rapidly building new infrastructure and increasing the capacity of beds the outbreak slowed down significantly. Starting from January 2015, the epidemic has moved to the ending phase that involves ensuring capacity for case finding, case management, safe burials and community engagement (from WHO, Ebola Situation Report, 14 Jan 2015) Each of the intense-transmission countries has sufficient capacity to isolate and treat patients, with more than 2 treatment beds per reported confirmed and probable case. However, the uneven geographical distribution of beds and cases, and the under-reporting of cases, means that not all EVD cases are isolated in several areas (from WHO, Ebola Situation Report, 28 Jan 2015)

20 Optimal control Problem: Optimal distribution of bed capacities Difficulties: Opt. dist. beds # future infecs. # future infecs. Opt. dist. beds Note: Methods/models for prediction were not accurate Main steps in our approach: Predict future infections (a few scenarios) by setting τ(t) constant over 2-3 months periods Find optimal distribution of new beds according to each scenario

21 Optimal distribution of beds (Example) Minimize (λ1,λ 2,λ 3,λ 4 ) 150 t=101 [ h1 (t : τ 1, x 1 ) b 1 (t) + h ] 2(t : τ 2, x 2 ) b 2 (t) subject to : λ k [0, 1], k = 1, 2, 3, 4; t [101, 150]; τ r (t) 1 x r (t + 1) = β rr (1 αω(i)) x r (t d i), r = 1, 2; i=0 b 1 (t) = b 1 (100) + {T b j t; j=1,2,3,4} i=1 λ i b i ; b 2 (t) = b 2 (100) + {Tj b t; j=1,2,3,4} i=1 (1 λ i ) b i

22 Here x r (t) is the # of infectious individuals and h r (t : τ, x) = σ i=τ(t) (1 αω(i)) x r (t d i) (1) where σ = 11 (ave.stay in hosp) and d = 6. # of beds at t = 100: b 1 (t) = 126, b 2 (t) = 60 Cumulative infs. at t = 100: 1259 and 675 New beds are introduced weekly: 350 new beds at t = new beds at t = new beds at t = new beds at t = 122

23 Case 1: τ 1 = τ 2 = 3. The optimal distribution of additional beds: Region1 : Region2 : T otal : The average number of bed occupancy over the time interval [100, 150] is 0.45 for Region 1 and 0.29 for Region 2. The maximum occupancy rates are: 0.83 (that is, average 0.83 patient per bed) in Region 1 and 0.55 in Region 2; that is, the demand for hospital beds is met at every point t [100, 150]. Thus the solution obtained is feasible.

24 Case 2: τ 1 = τ 2 = 4. The optimal distribution of additional beds: Region1 : Region2 : T otal : Again, the demand for hospital beds is met at every point t [100, 150] and accordingly the optimal solution obtained is feasible.

25 Case 3: τ 1 = τ 2 = 5. The optimal distribution of additional beds: Region1 : Region2 : T otal : The average number of bed occupancy over the time interval [100, 150] is 0.83 for Region 1 and 0.51 for Region 2. The maximum occupancy rates are: 1.91 (that is, 1.91 patient per bed) in Region 1 and 1.05 in Region 2.

26 Table 1: Optimal distr. add. beds (Not trivial!) τ 1 τ 2 time = week1 week2 week3 week4 τ 1 = τ 2 Region 1: Region 2: τ 1 < τ Region 1: Region 2: Region 1: Region 2: τ 1 > τ Region 1: Region 2: Region 1: Region 2:

27 Summary Consider future scenarios; in the example: τ 1 = 3, 4 or 5 and τ 2 = 3, 4 or 5 in total: 9 scenarios in fact (τ 1, τ 2 ) = (5, 5) - most likely (!?) Find optimal distribution of new beds according to each scenario; in the example: in total: 9 optimal bed distributions Analyze all optimal bed distributions/patterns; in the example: only 3 different distributions

28 Table 2: Optimal bed distributions: 3 different patterns τ 1, τ 2 time = week1 week2 week3 week4 τ 1 τ 2 Region 1: (scenario1) Region 2: τ 1 < τ 2 Region 1: (scenario2) Region 2: τ 1 > τ 2 Region 1: (scenario3) Region 2: Recall: Initial data (at t = 100): Region 1: 126 beds, 1259 cumulative-infecs Region 2: 60 beds, 675 cumulative-infecs

29 THANK YOU