We built four models. First, we established a pharmacokinetic model to simulate the distribution and metabolism of tetracycline in vivo. At the same time, we modeled and predicted the efficiency of the tetracycline control system in regulating the transcription of downstream genes into mRNA, which was then processed into siRNA by Drosha enzymes and Dicer enzymes. In addition, we modeled the exosome-mediated entry of siRNA into cells and predicted the changes in the number of intracellular siRNA molecules. Finally, we developed a model to predict affordability of NSCLC treatment based on disposable income and Engel's coefficient of residents in China, the United States and the United Kingdom.
PKEN
Abstract
Pharmacokinetics pharmacodynamics for short is the science of quantitative description of the dynamics of drug absorption, distribution, metabolism and excretion after entering the body through various routes of drug administration by using the principles of kinetics and mathematical processing methods, that is, to study the relationship between the position, quantity, efficacy and time of drugs in the body after administration, and to propose mathematical equations to explain these relationships. In this project, we measured the changes in blood levels in mice after oral administration of tetracycline and used pharmacokinetic simulations to investigate a number of questions.
Introduction
In this project, the expression of plasmid elements is regulated using the tet - on system, which is expressed in the presence of tetracycline, thus choosing the right tetracycline concentration to fully activate expression without affecting the normal life of the mice became an important goal of our project. Here, we fit curves and calculate parameters by pharmacokinetic methods to reach our final conclusions. In this wiki, we will describe in detail the whole process of selecting the model, constructing the model, performing the corresponding parameter calculations and deriving our conclusions.
Methods
1 Introduction to pharmacokinetic models
1.1 Basic concepts of pharmacokinetics
(1) Atrial model : Atrial is an abstract concept in pharmacokinetics, where parts of the body with the same rate of drug transport are grouped into one atrial compartment, which can be divided into one - compartment model, two - compartment model and multi - compartment model according to the differences in the rate of drug transport in the body. And according to the different modes of drug delivery, it can be further divided into intravenous one chamber, two chambers, multi - compartment and oral one chamber, two chambers, multi - compartment, and how to choose the appropriate atrial model is an important step in pharmacokinetics.
(2) Drug elimination kinetics $ \frac{dc}{dt}=-kc^n $ where c is the drug concentration and k is the elimination rate constant. When n = 1, primary kinetics, the drug is eliminated at a constant ratio, the rate of elimination is proportional to its concentration, and most drugs follow primary elimination kinetics. When n = 0, it is a zero - level kinetic, with constant drug elimination and constant drug elimination rate, and when the drug concentration is high, it is a constant elimination rate.
1.2 Pharmacokinetic parameters
(1) Half-life($ t_{1/2} $ ): Integrate $ \frac{dc}{dt}=-kc^n $ when following primary elimination kinetics. $ C=C_0e^{-kt} $ , When $ C=\frac{C0}2 $, $ t=\frac{\ln\left(2\right)}k=\frac{0.693}k $ ,The half - life was found to be constant when eliminated according to primary kinetics.
(2) Apparent volume of distribution: Vd is the plasma volume occupied by an intravenous infusion of an amount of(X) drug that is then equilibrated and the blood concentration C is measured $ Vd=\frac XC $ , which reflects the plasma volume occupied by the drug.
(3) Plasma clearance Cl: the volume of drug cleared per unit of time, measured in $ L*h^{-1} $
(4) Curve and area under the curve AUC at the time of dosing.Using time as the horizontal coordinate and blood concentration as the vertical coordinate, this curve is the drug - time curve.The area under the curve AUC reflects the total amount of drug absorbed after a single dose and reflects the degree of absorption of the drug.
(5) Css : When the same dose is administered several times at regular intervals, the concentration of the drug may be increased several times until it remains at a certain level or fluctuates at a certain level, which is the steady - state blood concentration.
(6) Bioavailability(F): $ F=\frac AD $ A is the amount of drug in the body and D is the oral dose, the amount of drug A that can enter circulation in the body is only a fraction of D due to the presence of first elimination and not 100% absorption.
2. Pharmacokinetic model selection
In this experiment, mice were given oral tetracycline feed, so the oral chamber model should be selected and bioavailability calculated to fully understand the questions to be answered. For For the oral atrial model, due to the absorption phase from the intestine into the blood circulation, the 1 - chamber model or the 2 - chamber model is usually used, but in order to avoid special circumstances, an additional 3 - chamber model is calculated in order to make the results reliable.
3. Pharmacokinetic model calculation
3.1 Oral one - compartment model
$ X_a $ is the absorption site, which is transferred to the central chamber with a rate constant $ k_a $ and the central chamber is eliminated with a rate constant $ k_{10} $ The kinetic equations are $ \left\{\begin{array}{l}\frac{dX_a}{dt}=X_a\left(-k_a\right)\\\frac{dX_c}{dt}=X_ak_a-k_{10}X_c\end{array}\right. $ ，Taking the Laplace transform on both sides of the equation. $ \left\{\begin{array}{l}SX_a'-X_a'\left(0\right)=-k_aX_a'\\SX_c'-X_c'\left(0\right)=k_aX_a'-k_{10}X_c'\end{array}\right. $ , here $ X_a'\left(0\right)=FD $ , $ X_c'\left(0\right)=0 $ ,Solve the equation for $ X_a' $ and $ X_c' $, $ \left\{\begin{array}{l}X_a'=\frac{FD}{k_a+S}\\X_c'=\frac{FDk_a}{\left(k_{10}+S\right)\left(k_a+S\right)}\end{array}\right. $ ,taking inverse Laplace transform, $ X_c=\frac{\left(FDk_a\right)\left(e^{-k_at}-e^{-k_{10}t}\right)}{k_{10}-k_a} $ ,concentration is $ \frac{X_c}{Vd} $ ,simple form is $ M\left(e^{-k_at}-e^{-k_{10}t}\right) $
3.2 Oral two - room model
$ X_a $ is the absorption site, transferred to the central chamber $ X_c $ at a rate constant, the central chamber is eliminated at a rate constant $ k_{10} $, $ X_p $ is the peripheral chamber, $ X_c $ flows to $ X_p $ at a constant rate $ k_{12} $ , $ X_p $ flows back to $ X_c $ at a constant rate $ k_{21} $ The kinetic equations are $ \left\{\begin{array}{l}\frac{dX_c}{dt}=X_ak_a-k_{10}X_c-k_{12}X_c+k_{21}X_p\\\frac{dX_p}{dt}=k_{12}X_c-k_{21}X_p\\\frac{dX_a}{dt}=X_a\left(-k_a\right)\end{array}\right. $ . Taking the Laplace transform on both sides of the equation.$ \left\{\begin{array}{l}SX_c'-X_c'\left(0\right)=k_aX_a'-k_{10}X_c'-k_{12}X_c'+k_{21}X_p'\\SX_a′-\;X_a′\left(0\right)=-k_aX_a′\\SX_p′-\;X_p′(0)=k_{12}X_c′-k_{21}X_p\end{array}\right. $ , here $ X_a'(0)=\;FD,\;X_c'(0)=\;0，X_p'(0)=\;0 $ . Solve the equation $ \left\{\begin{array}{l}X_a'=\frac{FD}{k_a+S}\\X_c'=\frac{\left(k_{21}+S\right)\left(FDk_a\right)}{\left(k_a+S\right)\left(\left(k_{10}+S\right)\left(k_{21}+S\right)+Sk_{12}\right)}\\X_p'=\frac{k_{12}(k_{21}+S)\left(FDk_a\right)}{\left(k_{21}+S\right)\left(k_a+S\right)\left(\left(k_{10}+S\right)\left(k_{21}+S\right)+Sk_{12}\right)}\end{array}\right. $, taking inverse Laplace transform, $ \mathrm{X} \mathrm{c}=\frac{\left(\mathrm{FD} \mathrm{k}_{\mathrm{a}}\right)\left(\mathrm{e}^{-\mathrm{k}_{\mathrm{a}} \mathrm{t}}-\mathrm{e}^{-\mathrm{k}_{10} \mathrm{t}}\right)}{\left(\mathrm{k}_{10}-\mathrm{k}_{\mathrm{a}}\right)} $ ,concentration is $ \frac{X_c}{Vd} $ , simple form is $ M\left(e^{-k_at}-e^{-k_{10}t}\right) $ . Inverse conversion using partial fractional method, Let $ \frac{\left(k_{21}+S\right)\left(FDk_a\right)}{\left(k_a+S\right)\left(\left(k_{10}+S\right)\left(k_{21}+S\right)+Sk_{12}\right)}=\frac{\left(k_{21}+S\right)\left(FDk_a\right)}{\left(k_a+S\right)\left(\alpha+S\right)\left(\beta+S\right)}=\frac A{k_a+S}+\frac B{\alpha+S}+\frac D{\beta+S} $ ,here $ \alpha+\beta=\;k_{10}+\;k_{12}+\;k_{21}；\alpha\beta=\;k_{10}k_{21} $ . Then $ B\left(\beta+S\right)\left(k_a+S\right)+D\left(\alpha+S\right)\left(k_a+S\right)+A\left(\alpha+S\right)\left(\beta+S\right) \\=\;\left(k_{21}+S\right)\left(FD\;k_a\right)\frac{\left(k_{21}+S\right)\left(FDk_a\right)}{\left(k_a+S\right)\left(\left(k_{10}+S\right)\left(k_{21}+S\right)+Sk_{12}\right)}\\=\frac{\left(k_{21}+S\right)\left(FDk_a\right)}{\left(k_a+S\right)\left(\alpha+S\right)\left(\beta+S\right)}=\frac A{k_a+S}+\frac B{\alpha+S}+\frac D{\beta+S} $ , Let $ S=-k_a $ , $ A=\frac{\left(k_{21}-ka\right)\left(FDk_a\right)}{\left(\beta-k_a\right)\left(\alpha-k_a\right)} $ ,Let $ S=-\alpha $ , $ B=\frac{\left(k_{21}-\alpha\right)\left(\mathrm{FD} k_{a}\right)}{(\beta-\alpha)\left(k_{a}-\alpha\right)} $ ; Let $ S=-\beta $ , $ D=\frac{\left(k_{21}-\beta\right)\left(\mathrm{FD} k_{a}\right)}{\left(k_{a}-\beta\right)(\alpha-\beta)} $ ; $ \mathrm{Xc}=\mathrm{FDk}_{a}\left(A e^{t\left(-k_{a}\right)}+B e^{-\alpha \mathrm{t}}+D e^{-\beta \mathrm{t}}\right) $ Then $ c_{t}=\frac{X_{c}}{V d} $ , simple form is $ A^{\prime} e^{t\left(-k_{a}\right)}+B^{\prime} e^{-\alpha t}+D^{\prime} e^{-\beta t} $
3.3 Oral three - room chamber model
$ X_a $
is the absorption site,
which is transferred to the central chamber $ X_c $
with a rate constant $ k_a $
the central chamber is eliminated with a rate constant $ k_{10} $
,and $ X_p $
and $ X_{p2} $
are the peripheral chambers.Calculations
are performed in the same way and are no longer performed.
4. Pharmacokinetic parameter calculation
4.1 Choice of rooms
(1) Direct observation method: If there is one obvious inflection point, it is the two - compartment model；if there are two obvious inflection points, it is the three - compartment model.
(2) The minimum residual sum method : substitute the measured blood concentration C into each model for calculation, calculate the corresponding parameters, the corresponding equation, the corresponding time substitution, to find the theoretical blood concentration Ce, calculate the residual sum of multiple groups of C and Ce, and take the model with the smallest residual sum.
(3) AIC discriminant : $ AIC=N \ln(R e)+2P $ , N is the number of experimental data points; $ P=2n $ n is the number of rooms: Re is the weighted sum of residuals squared,the smaller the AIC value,the better.
4.2 Model Fitting
1. Residual method
Since most drugs fit the two - chamber model, here is an example of the two - chamber model.two - room model $ C_t=A e^{t\left(-k_{a}\right)}+B e^{-\alpha t}+D e^{-\beta t} $ usually $ \mathbf{k}_{\alpha}>>\beta $ ,and $ \boldsymbol{\alpha}>>\boldsymbol{\beta} $ ,when $ t \rightarrow \infty $ , $ C_{t}=D e^{-\beta t} $ Take the logarithm of both sides at the same time, $ \ln C=\ln D-\beta t $ D and β can be obtained from the slope of the curve at the end of the drug - time curve and the extended intercept.
(2) Extrapolation of the last curve to the first curve at different times, and subtracting the extrapolated concentration of C1 from the measured concentration of Ct to obtain the first residual concentration of Cr1 $ \mathrm{Cr} 1=A e^{t\left(-k_{a}\right)}+B e^{-\alpha t} $ when t is much larger,because of ka>α,we can get $ \mathrm{Cr} 1=B e^{-\alpha t} $ ,then $ \ln \mathrm{Cr} 1=\ln \mathrm{B}-\alpha t $ B and α is obtained;
(3) Similarly extrapolating Cr1 to obtain the second residue concentration Cr2 $ \mathrm{Cr} 2=A e^{t\left(-k_{a}\right)} $ ,then $ \ln \mathrm{Cr}2=\ln \mathrm{A}-t k_{\mathrm{a}} $ and α is obtained;
2. Software fitting
Mathematica and Matlab implementations
4.3 Model parameter calculations(two - chamber model)
1. Calculation of kinetic constants
$ k_{21}: \frac{B}{D}=-\frac{\left(k_{21}-\alpha\right)\left(k_{a}-\beta\right)}{\left(k_{21}-\beta\right)\left(k_{a}-\alpha\right)}$ , then $ \mathrm{k}_{21}=\frac{\beta B\left(k_{a}-\alpha\right)+\alpha D\left(k_{a}-\beta\right)}{B\left(k_{a}-\alpha\right)+D\left(k_{a}-\beta\right)} $ $ k_{10}=\frac{\alpha \beta}{k_{21}} $ , $ k_{12}=\alpha+\beta-k_{10}-k_{21} $
2. AUC $ AUC=\int_{0}^{\infty} C d t=\int_{0}^{\infty} A^{\prime} e^{t\left(-k_{a}\right)}+B^{\prime} e^{-\alpha t}+D^{\prime} e^{-\beta t} d t=\frac{A^{\prime}}{k_{a}}+\frac{B^{\prime}}{\alpha}+\frac{D^{\prime}}{\beta} $
3. Half-life Absorbtion: $ t_{\frac{1}{2}(a)}=\frac{0.693}{k_{a}} $ Distribution: $ t_{\frac{1}{2}(a)}=\frac{0.693}{k_{a}} $ Elimination: $ t_{\frac{1}{2}}(\beta)=\frac{0.693}{\beta} $
4. Vd Center Room Vc: $ v_{c}=\frac{FD k_{21}}{\alpha \beta AUC}$ Total Vd: $ t_{\frac{1}{2}(\alpha)}=\frac{0.693}{\alpha} $
5. Css $ t_{\frac{1}{2}}(\beta)=\frac{0.693}{\beta} $
6. Overall clearance rate $ v_{c}=\frac{F D k_{21}}{\alpha \beta A U C} $
5. Modeling ideas and code implementation
Model ideas
first through the mathematica software, non - linear regression fitting to a room model, two room model, three room model were fitted to obtain the respective room model parameters, drawing images; and this parameter integer as the initial value, the use of matlab, respectively, function fitting and residual method of fitting, respectively, to draw three images.From nine images, choose the best fit to an image,
Mathemactica
One - Room:
data = {{0.25, 82.6}, {0.5, 113.4}, {1, 115.4}, {1.5, 96.6},
{2, 77.5}, {4, 34.5}, {6, 20.3}, {8, 15.1}, {12, 11.1}, {24, 6}}
ClearAll[A, m, n, t]
modle1=A*(E^((-m))t)-E^((-n))t));
FindFit[data, modle1, {A, m, n}, t]
equa = modle1 /. %
Show[ ListPlot[data], Plot[ Evaluate[equa], {t, 0, 24}]]
Two - Room:
ClearAll[A, B, P, m, n, z]
modle2 = A*(E-m t) + B*(E-n t) + P*(E-z t);
FindFit[data, modle2, {A, B, P, m, n, z}, t]
equa2 = modle2 /. %
Show[ ListPlot[data], Plot[ Evaluate[equa2], {t, 0, 24}]]
Three - Room:
ClearAll[A, B, P, Q, m, n, z, y]
modle3 = A( E^(-m t)) + B( E^(-n t)) + P( E^(-z t)) + Q( E^(-y t));
FindFit[data, modle3, {A, B, P, Q, m, n, z, y}, t]
equa3 = modle3 /. %
Show[ ListPlot[data], Plot[ Evaluate[equa3], {t, 0, 24}]]
Matlab
X = input(' Please enter the drug concentration at each time point：');
Y = input(' Please enter each time point：');
= length(X); % Return the number of data points
myfun1 = inline(' K(1)*(exp(K(2)*t) - exp(K(3)*t))', ' K', ' t');
% One - room model formula
K1 = lsqcurvefit(myfun1, [A m n], X, Y);
% Fitting，A m n is the parameters from Mathematica
S1 = 0; % Initialize the squared residuals
for i = 1 : s % Caculate the squared residuals
Ct = K1(1)*(exp(K1(2)*Y(i) - exp(K1(3)*Y(i))));
S1 =(X(i) - Ct)^2 + S1;
end
AIC1 = s*log(S1) + 4; % AIC
myfun2 = inline
(' K(1)*exp(K(2)*t) + K(3)*exp(K(4)*t) + K(5)*exp(K(6)*t)', ' K', ' t');
% Two - room model formula
K2 = lsqcurvefit(myfun2, [A B P m n z], X, Y);
% Fitting，A B P m n z is the parameters from Mathematica
S2 = 0;
for i = 1 : s
Ct = K2(1)*exp(K2(2)*Y(i)) + K2(3)*exp(K2(4)*Y(i)) + K2(5)*exp(K2(6)*Y(i));
S2 =(X(i) - Ct)^2 + S2;
end
AIC2 = s*log(S2) + 8;
myfun3 = inline(' K(1)*exp(K(2)*t) + K(3)*exp(K(4)*t) +
K(5)*exp(K(6)*t) + K(7)*exp(K(8)*t)', ' K', ' t');
% Three - room model formula
K3 = lsqcurvefit(myfun3, [A B P Q m n z y], X, Y);
% Fitting，A B P Q m n z y is the parameters from Mathematica
S3 = 0
for i = 1 : s
Ct = K3(1)*exp(K3(2)*Y(i)) + K3(3)*exp(K3(4)*Y(i)) +
K3(5)*exp(K3(6)*Y(i)) + K3(7)*exp(K3(8)*Y(i));
S3 =(X(i) - Ct)^2 + S3;
end
AIC3 = s*log(S3) + 12;
A = [S1 S2 S3];
best1 = max (A);
n1 = find (A ≥ max (A));
fprintf (' The model with the smallest squared residuals is number % d \r', n1);
B = [AIC1 AIC2 AIC3];
best2 = max (B);
n2 = find (B ≥ max (B));
fprintf (' The model with the smallest AIC is number % d', n2);
Residual method (Matlab implementation)：
X = input (' Please enter the drug concentration at each time point：');
Y = input (' Please enter each time point：');
= length (X); % Return the number of data points
n = find (X ⩵ max (X));
l1 = length (n + 1 : s);
X1 = log (X); % Logarithm of blood levels
A = X1 (n + 1 : end);
B = Y (n + 1 : end);
L11 = polyfit (B, A, 1);
% Return the coefficient of the polynomial fitted to the last segment
fprintf (' The model with the smallest squared residuals is number % d \r', n1);
for i = 1 : n
X1 (i) = exp (L11 (2)) * exp (L11 (1) * Y (i)) - X (i);
end
A = X1 (1 : n);
B = Y (1 : n);
L12 = real (polyfit (B, A, 1));
% Return the coefficient of the initial fitted polynomial
for i = 1 : s
X1 (i) = exp (L11 (2)) * exp (L11 (1) * Y (i)) - exp (L12 (2)) * exp (L12 (1) * Y (i));
% Calculation of theoretical blood levels at each time point
end
S1 = 0;
for i = 1 : s % Calculation of the squared residuals
S1 = S1 + (X (i) - X1 (i))^2;
end
n = find (X ⩵ max (X));
l2 = length (n + 1 : s);
X2 = log (X); % Logarithm of blood levels
A = X2 (n + ceil (l2 / 2) + 1 : end);
B = Y (n + ceil (l2 / 2) + 1 : end);
L21 = polyfit (B, A, 1);
% Return the coefficient of the polynomial fitted to the last segment
for i = 1 : (n + 1) % Calculation of second residue concentration
X2 (i) = exp (L22 (2)) * exp (L22 (1) * Y (i)) + exp (L21 (2)) * exp (L21 (1) * Y (i)) - X (i);
X2 (i) = log (X2 (i));
end
% Return the coefficient of the polynomial fitted to the last segment
A = X2 (1 : n);
B = Y (1 : n);
L23 = real (polyfit (B, A, 1));
% Return the first segment fitting polynomial coefficient
for i = 1 : s
X2 (i) = exp (L21 (2)) * exp (L21 (1) * Y (i)) +
exp (L22 (2)) * exp (L22 (1) * Y (i)) - exp (L23 (2)) * exp (L23 (1) * Y (i));
end
S2 = 0;
for i = 1 : s % Squared residuals
S2 = S2 + (X (i) - X2 (i))^2;
end
F1 = @(t) exp (L11 (2)) * exp (L11 (1) * t) - exp (L12 (2)) * exp (L12 (1) * t);
% One - chamber model fit curve
F2 = @(t) exp (L21 (2)) * exp (L21 (1) * t) +
exp (L22 (2)) * exp (L22 (1) * t) - exp (L23 (2)) * exp (L23 (1) * t);
% Two - chamber model fit curve
subplot (1, 2, 1) % Plotting Curve One
fplot (F1, [0, 24], ' b')
hold on
plot (Y, X, ' or') % Scatter plotting of measured data points
subplot (1, 2, 2) % Plotting Curve Two
fplot (F2, [0, 24], ' b')
hold on
plot (Y, X, ' or') % Scatter plotting of measured data points
if S1 < S2
fprintf (' It is preferable to choose a one -
room room model with a dynamics formula of % f * exp (% f * t) - % f * exp (% f * t)',
exp (L11 (2)), L11 (1), exp (L12 (2)), L12 (1))
else
fprintf (' It is preferable to choose a two - room room model with a
dynamics formula of % f * exp (% f * t) + % f * exp (% f * t) - % f * exp (% f * t)',
exp (L21 (2)), L21 (1), exp (L22 (2)), L22 (1), exp (L23 (2)), L23 (1))
end
Catastrophic medical costs
Background
In most countries, cancer accounts for a significant share of household health spending. Non-small cell lung cancer is one of the most common malignant tumors. Non-small cell lung cancer treatment includes surgery, chemotherapy, drugs, targeted therapy and other methods. Here we consider the cost of surgery and chemotherapy combined with the distribution of household disposable income and Engel's coefficient to estimate the proportion of households unable to afford non-small cell lung cancer treatment.
Modeling
1. Catastrophic medical expenditures
There is a universal international concept of catastrophic household medical expenditures. If the ratio is greater than or equal to 40%, it means that a catastrophic medical expenditure has occurred in this family.
We selected the United Kingdom, the United States and China to predict the proportion of families have difficulty to afford treatment for NSCLC. The choice of these three countries is actually due to language constraints, as official household income data from each government are published in each country's official language.
2. Cost of non-small cell lung cancer treatment
2.1 Open lobectomy
2.2 Chemotherapy
2.3 Medical insurance
In Britain, thanks to the National Health Service (NHS), health care is largely free. However, while the NHS provides cancer treatment, including chemotherapy, free of charge to all who are settled and legally resident in the UK, people may incur other medical costs when they are treated. Here we assume that other medical costs account for 10% of the total cost of treatment. Health care facilities in the United States are largely owned and operated by private sector businesses. After the Affordable Care Act (ACA) mandate, the percentage of adult Americans without health insurance dropped significantly. In the fourth quarter of 2018, it was 13.7%. Here we assume that since people with health insurance are not going to be in trouble because of the high cost of treatment, people whose income level is unaffordable times the percentage of people without health insurance are the people who are really going to be in trouble because of NSCLC. In China, non-small cell lung cancer (NSCLC) is a major disease regulated by the state, and residents with NSCLC can receive a certain percentage (generally 50% to 60%) of reimbursement. However, the annual limit of compensation for hemodialysis, radiotherapy and chemotherapy in the outpatient department of uremia in the township cooperative medical care and the outpatient department of uremia is 11,000 yuan. So we subtracted 11,000 from the cost of treatment to represent the actual expenditure. So the actual health expenditure of each country is
2.4 the minimum annual income
We use the following equation to calculate the minimum annual income that would guarantee that medical expenses would not exceed 40% of the ability to pay: $$ \frac{ME}{D_i-E_n*D_i}=40 \% $$
Where ME refers to actual medical expenditure, Di refers to disposable income and EN refers to Engel's coefficient. The value of Engel coefficient is[10]:
Based on the above rules, we came up with the minimum annual disposable income that would guarantee that medical expenses would not exceed 40% of the ability to pay:
3. The distribution of disposable income across countries
The United Kingdom:
In the UK, less than 10 per cent of people struggle financially to get treatment for the open lobectomy and chemotherapy of NSCLC.
The United States:
From this graph and the average disposable income we can figure out the disposable income of the five quintile in 2018：
Considering the health insurance, at least 8.22 percent of Americans can't afford Open lobectomy and 10.96 percent can't afford chemotherapy.
China
As the National Bureau of Statistics of China only updates data for 2012 until now on its website, while the rest of the data is for 2018, the estimate is subject to some error. At least 80 percent of families in China struggle with the cost of an open lobectomy, and 60 percent of families in China struggle with the cost of chemotherapy.
References
1. Xu K, Evans D B, Kawabata K, et al. Household catastrophic health expenditure: a multicountry analysis[J]. The lancet, 2003, 362(9378): 111-117. 2. Droghetti A, Marulli G, Vannucci J, et al. Cost analysis of pulmonary lobectomy procedure: comparison of stapler versus precision dissection and sealant[J]. ClinicoEconomics and outcomes research: CEOR, 2017, 9: 201. 3. http://ask.39.net/question/61425053.html 4. https://uk.gofundme.com/c/blog/cost-of-chemotherapy-uk 5. Vergnenègre A, Chouaïd C. Review of economic analyses of treatment for non-small–cell lung cancer (NSCLC)[J]. Expert review of pharmacoeconomics & outcomes research, 2018, 18(5): 519-528. 6. 于启林, 朱士俊, 戴广海. 晚期非小细胞肺癌化疗效果与费用分析[J]. 中国医院管理, 2001, 21(012):9-10. 7. https://uk.gofundme.com/c/blog/cost-of-chemotherapy-uk 8. https://en.wikipedia.org/wiki/Health_care_in_the_United_States 9. https://baike.baidu.com/item/%E5%8C%BB%E7%96%97%E4%BF%9D%E9%99%A9/637613?fr=aladdin 10. https://www.usda.gov/ 11. https://www.ons.gov.uk/peoplepopulationandcommunity/personalandhouseholdfinances/incomeandwealth/bulletins/householddisposableincomeandinequality/financialyearending2019 12. https://www.census.gov/content/dam/Census/library/visualizations/2019/demo/p60-266/figure3.pdf 13. https://data.stats.gov.cn/easyquery.htm?cn=C01&zb=A0A07&sj=2019