1
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows}
\begin{document}
\tikzstyle{block} = [draw, fill=white, rectangle, 
    minimum height=3em, minimum width=6em]
\tikzstyle{sum} = [draw, fill=white, rectangle, node distance=1cm]
\tikzstyle{input} = [coordinate]
\tikzstyle{output} = [coordinate]

\begin{tikzpicture}[auto, node distance=2cm,>=latex']
\node [input, name=input] {};
\node [sum, right of=input] (sum) {S};
\node [block, right of=S] (E) {E};
\node [block, right of=E] (I) {I};
\node [block, above of=I] (J) {J};
\node [block, below of=J] (T) {T};

\node [output, below of=S] (output) {};
\node [output, below of=E] (output) {};
\node [output, below of=I] (output) {};
\node [output, right of=J] (output) {};
\node [output, right of=T] (output) {};

\draw [draw,->] (input) -- node {$A$} (S);
\draw [draw,->] (input) -- node {$\Lambda$} (S);
\draw [->] (S) -- node[name=$m \beta \left( \frac{I+l J}{N}  \right)$] {$m \beta \left( \frac{I+l J}{N}  \right)$} (E);
\draw [->] (S) -- node[name=$(1-m) \beta \left( \frac{I+l J}{N}  \right)$] {$m \beta \left( \frac{I+l J}{N}  \right)$} (I);
\draw [->] (E) -- node[name=k] {$k$} (I);
\draw [->] (I) -- node[name=n] {$k$} (J);
\draw [->] (I) -- node[name=$r_1$] {$r_1$} (T);
\draw [->] (J) -- node[name=$r_2$] {$r_2$} (T);
\draw [->] (T) -- node[name=$q\delta$] {$q\delta$} (E);
\draw [->] (T) -- node[name=$(1-q)\delta$] {$q\delta$} (S);

\draw [->] (s) -- node [name=$\mu$] {$\mu$}(output);
\draw [->] (E) -- node [name=$\mu$] {$\mu$}(output);
\draw [->] (I) -- node [name=$\mu+d_1$] {$\mu$}(output);
\draw [->] (J) -- node [name=$\mu+d_2$] {$\mu$}(output);
\draw [->] (T) -- node [name=$\mu$] {$\mu$}(output);

\end{tikzpicture}
\end{document}

enter image description here

Sandy G
  • 42,558

1 Answers1

3

After some weeding I arrive at

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{positioning}
\begin{document}

\begin{tikzpicture}[auto, node distance=2cm,>=latex,block/.style={draw, fill=white, rectangle, 
    minimum height=3em, minimum width=6em}]
 \node[block] (S) {S};
 \node[block, right=of S] (E) {E};
 \node[block, right=of E] (I) {I};
 \node[block, above right=of I] (J) {J};
 \node[block, below right=of I] (T) {T};
 %
 \draw[->] (S) -- ++ (0,2) -| node[pos=0.25,above]{$(1-m) \beta \left( \frac{I+l J}{N}  \right)$} (I); 
 \draw[->] (S) -- node[pos=0.5,below]{$m \beta \left( \frac{I+l J}{N}\right)$}(E);
 \draw[->] (E) -- node[pos=0.5,above]{$k$} (I);
 \draw[->] (E.90) -- ++ (0,0.5) node[right] {$\mu$};
 \draw[->] (I.20) -- node[pos=0.5,above,sloped] {$n$} (J.180);
 \draw[->] (I.-20) -- node[pos=0.5,below,sloped] {$r_1$} (T.180);
 \draw[->] (J) -- node[pos=0.5,right] {$r_2$} (T);
 \draw[->] (T.-135) -- ++ (0,-1) -| node[pos=0.25] {$q\delta$} (E);
 \draw[->] (T.-45) -- ++ (0,-2) -| node[pos=0.25] {$(1-q)\delta$} (S.-45);
 \draw[->] (S.-135) -- ++ (0,-1) node[below]{$\mu$};
 \draw[<-] (S.160) -- ++ (-1,0) node[left] {$\Lambda$};
 \draw[<-] (S.200) -- ++ (-1,0) node[left] {$A$};
 \draw[->] (T.0) -- ++ (1,0) node[right]{$\mu$};
 \draw[->] (J.0) -- ++ (1,0) node[right]{$\mu+d_2$};
\end{tikzpicture}
\end{document}

enter image description here

Note:

  1. \tikzstyle is deprecated.
  2. So is the arrows library.
  3. And the positioning syntax you were using.
  4. You cannot give nodes a name that is a formula.
  5. It is not particularly useful to give different nodes the same names.
  6. This code can be optimized further. It is an attempted compromise between elegance and being very explicit and thus more understandable.
  • Darn you :-) I should have known you were awake, I got only one percent of it untangled in the last 30 mins, but am learning some TikZ at last :-) I see you 3.14 and raise you tau https://www.scientificamerican.com/article/let-s-use-tau-it-s-easier-than-pi/ –  Mar 20 '19 at 01:55
  • @KJO Sorry, I didn't know you were looking at it. (After looking at it I figured that the quickest way is to copy the formulae, delete the rest and draw it more or less from scratch.) –  Mar 20 '19 at 01:57