Efficiency of Hospitals

Contents

Problem description

The administration of the hospitals in Paris decides to measure the efficiency of the surgery departments in four major hospitals with a desire to improve the service to the public. To keep this study anonymous, the hospitals are named H1 to H4. The method suggested to measure the efficiency is DEA (Data Envelopment Analysis). This method compares the performance of a fictitious hospital with the performances of the four hospitals.

Three initial indicators (resources) are taken into account: the number of non-medical personnel, the general expenses, and the available number of beds. In addition, four final indicators (services) are analyzed: the number of hospital admissions per day, the number of consultations in the outpatients’ clinic, the number of nurses on duty every day, and the number of interns and doctors on duty every day. The corresponding data have been analyzed over a period of two years and the numbers representing a day of average activity in every hospital are given in the following two tables.

Resource indicators

+---------------------+-----+------+-----+-----+
|                     | H1  | H2   | H3  | H4  |
+---------------------+-----+------+-----+-----+
|Non-medical personnel| 90  | 87   | 51  | 66  |
|General expenses (k$)|38.89|109.48|40.43|48.41|
|Number of beds       | 34  | 33   | 20  | 33  |
+---------------------+-----+------+-----+-----+

Service indicators

+-------------------+-----+-----+-----+-----+
|                   | H1  | H2  | H3  | H4  |
+-------------------+-----+-----+-----+-----+
|Admissions         |30.12|18.54|20.88|10.42|
|Consultations      |13.54|14.45| 8.52|17.74|
|Interns and doctors| 13  |  7  |   8 | 26  |
|Nurses on duty     | 79  | 55  |  47 | 50  |
+-------------------+-----+-----+-----+-----+

Justify through the DEA method how hospital H2 is performing compared to the others.

Variables

resources                  A matrix describing the resources
services                   A matrix describing the services

Reference

Applications of optimization... Gueret, Prins, Seveaux

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2009 by Tomlab Optimization Inc., $Release: 7.2.0$
% Written Oct 7, 2005.   Last modified Apr 8, 2009.

Problem setup

resources  = [90 87 51 66;...
    38.89 109.48 40.43 48.41;...
    34 33 20 33];
services   = [30.12 18.54 20.88 10.42;...
    13.54 14.45 8.52 17.74;...
    13 7 8 26;...
    79 55 47 50];

h = size(resources,2);    %coef
s = size(services,1);     %services
r = size(resources,1);    %resources

coef = tom('coef',h,1);
fserv = tom('fserv',s,1);
fres = tom('fres',r,1);
eff = tom('eff',1,1);

% No variables are binary
bnds = {coef >= 0, fserv >= 0, fres >= 0, eff >= 0};

% Coef constraints
con1 = {sum(coef) == 1};

% Service constraint
con2 = {services*coef == fserv};

% Resource constraint
con3 = {resources*coef == fres};

% Objective
objective = eff;

indices = zeros(size(resources,2),1);
for i=1:size(resources,2)
    % Service indicators, greater than ficticious ones
    con4 = {fserv >= services(:,i)};

    % Efficiency relationship
    con5 = {fres <= resources(:,i)*eff};

    constraints = {bnds, con1, con2, con3, con4, con5};
    options = struct;
    options.solver = 'cplex';
    options.name   = 'Efficiency of Hospitals';
    sol = ezsolve(objective,constraints,[],options);

    indices(i,1) = sol.eff;
end

PriLev = 1;
if PriLev > 0
    for i = 1:length(indices),
        disp(['H' num2str(i) ' is ' num2str(100*indices(i)) '% efficient'])
    end
end

% MODIFICATION LOG
%
% 051206 med   Created
% 060118 per   Added documentation
% 060125 per   Moved disp to end
% 090325 med   Converted to tomSym
Problem type appears to be: lp
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Efficiency of Hospitals        f_k       0.999999999999999890
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv    2 Iter    2 
Problem type appears to be: lp
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Efficiency of Hospitals        f_k       0.921049539123350750
                                       sum(|constr|)      0.000000000000007105
                              f(x_k) + sum(|constr|)      0.921049539123357850
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv    5 Iter    5 
Problem type appears to be: lp
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Efficiency of Hospitals        f_k       1.000000000000000000
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv    2 Iter    2 
Problem type appears to be: lp
===== * * * =================================================================== * * *
TOMLAB - Tomlab Optimization Inc. Development license  999001. Valid to 2010-02-05
=====================================================================================
Problem: ---  1: Efficiency of Hospitals        f_k       1.000000000000000000
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv    3 Iter    3 
H1 is 100% efficient
H2 is 92.105% efficient
H3 is 100% efficient
H4 is 100% efficient