TomSym Composing Flight Crews

This page is part of the TomSym Manual. See TomSym Manual.

Problem description

During the Second World War the Royal Air Force (RAF) had many foreign pilots speaking different languages and more or less trained on the different types of aircraft. The RAF had to form pilot/co-pilot pairs (‘crews’) for every plane with a compatible language and a sufficiently good knowledge of the aircraft type. In our example there are eight pilots. In the following table every pilot is characterized by a mark between 0 (worst) and 20 (best) for his language skills (in English, French, Dutch, and Norwegian), and for his experience with different two-seater aircraft (reconnaissance, transport, bomber, fighterbomber, and supply planes).

Ratings of pilots

+----------+-------------------------+--+--+--+--+--+--+--+--+
|          |Pilot                    | 1| 2| 3| 4| 5| 6| 7| 8|
+----------+-------------------------+--+--+--+--+--+--+--+--+
|Language  |English                  |20|14| 0|13| 0| 0| 8| 8|
|          |French                   |12| 0| 0|10|15|20| 8| 9|
|          |Dutch                    | 0|20|12| 0| 8|11|14|12|
|          |Norwegian                | 0| 0| 0| 0|17| 0| 0|16|
+----------+-------------------------+--+--+--+--+--+--+--+--+
|Plane type|Reconnaissance           |18|12|15| 0| 0| 0| 8| 0|
|          |Transport                |10| 0| 9|14|15| 8|12|13|
|          |Bomber                   | 0|17| 0|11|13|10| 0| 0|
|          |Fighter-bomber           | 0| 0|14| 0| 0|12|16| 0|
|          |Supply plane             | 0| 0| 0| 0|12|18| 0|18|
+----------+-------------------------+--+--+--+--+--+--+--+--+

A valid flight crew consists of two pilots that both have each at least 10/20 for the same language and 10/20 on the same aircraft type.

Question 1: Is it possible to have all pilots fly?

Subsequently, we calculate for every valid flight crew the sum of their scores for every aircraft type for which both pilots are rated at least 10/20. This allows us to define for every crew the maximum score among these marks. For example, pilots 5 and 6 have marks 13 and 10 on bombers and 12 and 18 on supply planes. The score for this crew is therefore max(13 + 10, 12 + 18) = 30.

Question 2: Which is the set of crews with maximum total score?

Compatibility graph for pilots Nodes = pilots Arcs = possible combinations (with scores)

1 ----30--- 2 ----27--- 3

| \       / |         / |
 |  24   28  |        /  |
 |   \   /   27     26   |
 |           |      /    |
 |     4     |     /    30
25           |    /      |
 |   /   \   |   /       |
 |  29    21 |  /        |
 | /       \ | /         |

5 ---30---- 6 -----28-- 7

\         |         /
    \        |        /
     \       |       /
      \      36     /
       \     |     /
       30    |   25
         \   |   /
          \  |  /
           \ | /

8

Variables

scores                     Scores for language and piloting skills
arcs_out/in                For an arc i arcs_out(i) is its starting node
                           and arcs_in(i) ts target node
arcs_score                 Strength of an arc

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

scores = [ 20 14  0 13  0  0  8  8;...
    12  0  0 10 15 20  8  9;...
    0 20 12  0  8 11 14 12;...
    0  0  0  0 17  0  0 16;...
    18 12 15  0  0  0  8  0;...
    10  0  9 14 15  8 12 13;...
    0 17  0 11 13 10  0  0;...
    0  0 14  0  0 12 16  0;...
    0  0  0  0 12 18  0 18];

arcs_in    = [1 1 1 2 2 2 3 3 4 4 5 5 6 6 7]';
arcs_out   = [5 4 2 4 6 3 6 7 5 6 6 8 8 7 8]';

arcs_score = [25 24 30 28 27 27 26 30 29 21 30 30 ...
    36 28 25]';

n = length(arcs_in);
n1 = size(scores,2);

fly = tom('fly',n,1,'int');

% All variables are integers.
bnds = {0 <= fly <= 1};

% Orig constr.
con = cell(n1,1);
for i=1:n1
    idx_in  = find(arcs_in  == i);
    idx_out = find(arcs_out == i);
    idx = unique([idx_in;idx_out]);
    con{i} = {sum(fly(idx)) <= 1};
end

% Objective
objective = -arcs_score'*fly;

constraints = {bnds, con};
options = struct;
options.solver = 'cplex';
options.name   = 'Composing Flight Crews';
sol = ezsolve(objective,constraints,[],options);

PriLev = 1;
if PriLev > 0
    idx = find(sol.fly);
    for crew = 1:idx,
        id = idx(crew);
        disp(['crew ' num2str(crew) ' has pilots ' ...
            num2str(arcs_in(id)) ' and ' num2str(arcs_out(id))])
    end
end

% MODIFICATION LOG
%
% 051021 med   Created
% 060113 per   Added documentation
% 060125 per   Moved disp to end
% 060203 med   Updated print level
% 090308 med   Converted to tomSym

Problem type appears to be: mip
Time for symbolic processing: 0.03719 seconds
Starting numeric solver
===== * * * =================================================================== * * *
TOMLAB - TOMLAB Development license  999007. Valid to 2011-12-31
=====================================================================================
Problem: ---  1: Composing Flight Crews         f_k    -125.000000000000000000
                                              f(x_0)      0.000000000000000000

Solver: CPLEX.  EXIT=0.  INFORM=101.
CPLEX Branch-and-Cut MIP solver
Optimal integer solution found

FuncEv    9 
Elapsed time: 0.003000 sec. 
crew 1 has pilots 1 and 2
crew 2 has pilots 3 and 7
crew 3 has pilots 4 and 5