# Models Ground transport

## Car rental

```% function Result = carrentalEx(PriLev)
%
% Creates a TOMLAB MIP problem for car rental
%
% CAR RENTAL
%
% A small car rental company has a fleet of 94 vehicles distributed
% among its 10 agencies. The location of every agency is given by its
% geographical coordinates X and Y in a grid based on kilometers. We
% assume that the road distance between agencies is approximately 1.3
% times the Euclidean distance (as the crow flies). The following
% table indicates the coordinates of all agencies, the number of cars
% required the next morning, and the stock of cars in the evening
% preceding this day.
%
% Description of the vehicle rental agencies
%
% +-------------+--+--+--+--+--+--+--+--+--+--+
% |Agency       | 1| 2| 3| 4| 5| 6| 7| 8| 9|10|
% +-------------+--+--+--+--+--+--+--+--+--+--+
% |X coordinate | 0|20|18|30|35|33| 5| 5|11| 2|
% |Y coordinate | 0|20|10|12| 0|25|27|10| 0|15|
% +-------------+--+--+--+--+--+--+--+--+--+--+
% |Required cars|10| 6| 8|11| 9| 7|15| 7| 9|12|
% |Cars present | 8|13| 4| 8|12| 2|14|11|15| 7|
% +-------------+--+--+--+--+--+--+--+--+--+--+
%
% Supposing the cost for transporting a car is \$ 0.50 per km,
% determine the movements of cars that allow the company to
% re-establish the required numbers of cars at all agencies,
% minimizing the total cost incurred for transport.
%
% VARIABLES
%
% demand                     Required cars per agency
% stock                      Cars present
% cost                       Cost per km to transport a car
% xcord                      The X-coordinate of agencies
% ycord                      The Y-coordinate of agencies
% n                          Number of agencies
% distance                   A matrix of distances
%
% RESULTS
%
% For an interpretation of the results, try this:
% Result          = carrentalEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = carrentalEx(PriLev)

if nargin < 1
PriLev = 1;
end

demand          = [10   6   8  11   9   7  15   7   9  12]';
stock           = [ 8  13   4   8  12   2  14  11  15   7]';
cost            = 0.50;

xcord           = [ 0  20  18  30  35  33   5   5  11   2]';
ycord           = [ 0  20  10  12   0  25  27  10   0  15]';
n               = length(xcord);

distance        = zeros(n,n);

for i=1:n
for j=1:n
distance(i,j) = 1.3*sqrt( (xcord(i) - xcord(j))^2 + (ycord(i) - ycord(j))^2);
end
end

Prob = carrental(cost, distance, stock, demand);
Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
idx_excess      = find(stock-demand > 0);
idx_need        = find(stock-demand < 0);
n_excess        = length(idx_excess);
n_need          = length(idx_need);
temp            = reshape(Result.x_k,n_excess,n_need);

disp('THE SENDING OF CARS')
for i = 1:n_excess,       % scan all positions, disp interpretation
disp(['agency ' num2str(idx_excess(i)) ' sends: ' ])
for j = 1:n_need,
if temp(i,j) ~= 0
disp(['   ' num2str(temp(i,j)) ' car(s) to agency ' ...
num2str(idx_need(j))])
end
end
disp(' ')
end

disp('THE GETTING OF CARS')
for j = 1:n_need,
disp(['agency ' num2str(idx_need(j)) ' gets: ' ])
for i = 1:n_excess,       % scan all positions, disp interpretation
if temp(i,j) ~= 0
disp(['   ' num2str(temp(i,j)) ' car(s) from agency ' ...
num2str(idx_excess(i))])
end
end
disp(' ')
end
end
% MODIFICATION LOG
%
% 051019 med   Created.
% 060125 per   Moved disp to end```

## Choosing the mode of transport

```% function Result = choosingthemodeoftransportEx(PriLev)
%
% Creates a TOMLAB MIP problem for choosing the mode of transport
%
% CHOOSING THE MODE OF TRANSPORT
%
% A company in the south-west of France needs to transport 180 tonnes
% of chemical products stored in depots D1 to D4 to the three
% recycling centers C1, C2, and C3. The depots D1 to D4 contain
% respectively 50, 40, 35, and 65 tonnes, that is 190 tonnes in
% total. Two modes of transport are available: road and rail. Depot
% D1 only delivers to centers C1 and C2 and that by road at a cost of
% \$ 12k/t and \$ 11k/t. Depot D2 only delivers to C2, by rail or road
% at \$ 12k/t and \$ 14k/t respectively. Depot D3 delivers to center C2
% by road (\$ 9k/t) and to C3 by rail or road for \$ 4k/t and \$ 5k/t
% respectively. The depot D4 delivers to center C2 by rail or road at
% a cost of \$ 11k/t and \$ 14k/t, and to C3 by rail or road at \$ 10k/t
% and \$ 14k/t respectively.
%
% Its contract with the railway company for the transport of chemical
% products requires the company to transport at least 10 tonnes and
% at most 50 tonnes for any single delivery. Besides the standard
% security regulations, there are no specific limitations that apply
% to road transport. How should the company transport the 180 tonnes
% of chemicals to minimize the total cost of transport?
%
% Graph of connections.
%
%       +--  2 (D1)        /           \
%      /            \     /             12 (C1)
%     /              \   /             /
%    /              --+--    6 (rail) /         \
%   |    +-  3 (D2)   |     /                    \
%   |   /           --+----/                      \
%   |  /              |                            \
%   | /               |                             \
%                +----+----  8 (rail) \              \
%   1           /      \               \
%     \        /        \               13 (C2) ----- 15
%   |  \      /          \             /
%   |   \                 \  9 (road) /              /
%   |    +-- 5 ------------                         /
%   |     (D4) \     ------                        /
%   |           \   /                             /
%    \           +--+-----  10 (rail) \          /
%     \             /     /            \        /
%      \       ----+     /              14 (C3)
%       +--- 4 ---------+              /
%        (D3) \             11 (road) /
%              +-----------
%
%
% VARIABLES
%
% arcs_out/in                For an arc i arcs_out(i) is its starting node
%                            and arcs_in(i) ts target node
% arcs_min                   Minimal load for an arc
% arcs_max                   Maximal load for an arc
% arcs_cost                  Cost for an arc
% minflow                    Flow from 1 to 15
%
% RESULTS
%
% For an interpretation of the results, try the following
% Result   = choosingthemodeoftransportEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = choosingthemodeoftransportEx(PriLev)

if nargin < 1
PriLev = 1;
end

arcs_out        = [ 1  1  1  1  2  2  3  3  4  4  4  5  5  5  5 ...
6  7  8  9 10 11 12 13 14 ]';
arcs_in         = [ 2  3  4  5  7  9  6  7  9 10 11  8  9 10 11 ...
12 12 13 13 14 14 15 15 15 ]';
arcs_min        = [zeros(1,6) 10 0 0 10 0 10 0 10 0 ...
zeros(1,9)]';
arcs_max        = [50 30 35 65 inf inf 50 inf inf 50 inf 50 inf 50 inf ...
inf*ones(1,9)]';
arcs_cost       = [0 0 0 0 12 11 12 14 9 4 5 11 14 10 14 ...
zeros(1,9) ]';
minflow         = 50+30+35+65;

Prob = choosingthemodeoftransport(arcs_out, arcs_in, arcs_min, arcs_max, arcs_cost, minflow);

Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
arcs     = Result.x_k;

disp('transport as follows:')

for i = 1:length(arcs),
if arcs(i) ~= 0,
disp(['    ' num2str(arcs(i)) ' units from node ' ...
num2str(arcs_out(i)) ' to node ' num2str(arcs_in(i))])
end
end

end

% MODIFICATION LOG
%
% 051019 med   Created.
% 060125 per   Moved disp to end```

## Depot location

```% function Result = depotlocationEx(PriLev)
%
% Creates a TOMLAB MIP problem for depot location
%
% DEPOT LOCATION
%
% A large company wishes to open new depots to deliver to its sales
% centers. Every new set-up of a depot has a fixed cost. Goods are
% delivered from a depot to the sales centers close to the site.
% Every delivery has a cost that depends on the distance covered. The
% two sorts of cost are quite different: set-up costs are capital
% costs which may usually be written off over several years, and
% transport costs are operating costs. A detailed discussion of how
% to combine these two costs is beyond the scope of this text — we
% assume here that they have been put on some comparable basis,
% perhaps by taking the costs over a year.
%
% There are 12 sites available for the construction of new depots and
% 12 sales centers need to receive deliveries from these depots.
%
% The following table gives the costs (in thousand \$) of satisfying
% the entire demand of each customer (sales center) from a depot (not
% the unit costs). So, for instance, the cost per unit of supplying
% customer 9 (who has a total demand of 30 tonnes according to the
% next table) from depot 1 is \$ 60000/30t, i.e. \$ 2000/t. Certain
% deliveries that are impossible are marked with "Inf".
%
% Delivery costs for satisfying entire demand of customers
%
% +-----+-----------------------------------------------+
% |     |               Customer                        |
% |Depot|  1   2   3   4   5   6   7   8   9  10  11  12|
% +-----+---+---+---+---+---+---+---+---+---+---+---+---+
% |   1 |100| 80| 50| 50| 60|100|120| 90| 60| 70| 65|110|
% |   2 |120| 90| 60| 70| 65|110|140|110| 80| 80| 75|130|
% |   3 |140|110| 80| 80| 75|130|160|125|100|100| 80|150|
% |   4 |160|125|100|100| 80|150|190|150|130|Inf|Inf|Inf|
% |   5 |190|150|130|Inf|Inf|Inf|200|180|150|Inf|Inf|Inf|
% |   6 |200|180|150|Inf|Inf|Inf|100| 80| 50| 50| 60|100|
% |   7 |100| 80| 50| 50| 60|100|120| 90| 60| 70| 65|110|
% |   8 |120| 90| 60| 70| 65|110|140|110| 80| 80| 75|130|
% |   9 |140|110| 80| 80| 75|130|160|125|100|100| 80|150|
% |  10 |160|125|100|100| 80|150|190|150|130|Inf|Inf|Inf|
% |  11 |190|150|130|Inf|Inf|Inf|200|180|150|Inf|Inf|Inf|
% |  12 |200|180|150|Inf|Inf|Inf|100| 80| 50| 50| 60|100|
% +-----+---+---+---+---+---+---+---+---+---+---+---+---+
%
% In addition, for every depot we have the following information: the
% fixed cost for constructing the depot that needs to be included
% into the objective function and its capacity limit, all listed in
% the table below.
%
% Fix costs and capacity limits of the depot locations
%
%+------------+----+----+-----+----+----+----+----+----+----+-----+----+----+
%|Depot       |   1|   2|    3|   4|   5|   6|   7|   8|   9|   10|  11|  12|
%+------------+----+----+-----+----+----+----+----+----+----+-----+----+----+
%|Cost (k\$)   |3500|9000|10000|4000|3000|9000|9000|3000|4000|10000|9000|3500|
%|Capacity (t)| 300| 250|  100| 180| 275| 300| 200| 220| 270|  250| 230| 180|
%+------------+----+----+-----+----+----+----+----+----+----+-----+----+----+
%
% The quantities demanded by the sales centers (customers), are
% summarized in the following table.
%
% Demand data
% +----------+---+--+--+---+---+---+--+--+--+---+--+---+
% |Customer  |  1| 2| 3|  4|  5|  6| 7| 8| 9| 10|11| 12|
% +----------+---+--+--+---+---+---+--+--+--+---+--+---+
% |Demand (t)|120|80|75|100|110|100|90|60|30|150|95|120|
% +----------+---+--+--+---+---+---+--+--+--+---+--+---+
%
% In every case, the demand of a customer needs to be satisfied but a
% sales center may be delivered to from several depots. Which depots
% should be opened to minimize the total cost of construction and of
% delivery, whilst satisfying all demands?
%
% VARIABLES
%
% delcosts                   Costs to deliver to centre from depot
% idxinf                     Impossible combination of centre/depot
% delcosts(idxinf)           A large number < Inf
% buildcosts                 Costs to build a depot=
% capacity                   Capacities per potential depot
% demand                     The customers demand in tonnes
%
% RESULTS
%
% To interpret the results run this:
% Result  = depotlocationEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = depotlocationEx(PriLev)

if nargin < 1
PriLev = 1;
end

delcosts     = [ 100  80  50  50  60 100 120  90  60  70  65 110;...
120  90  60  70  65 110 140 110  80  80  75 130;...
140 110  80  80  75 130 160 125 100 100  80 150;...
160 125 100 100  80 150 190 150 130 inf inf inf;...
190 150 130 inf inf inf 200 180 150 inf inf inf;...
200 180 150 inf inf inf 100  80  50  50  60 100;...
100  80  50  50  60 100 120  90  60  70  65 110;...
120  90  60  70  65 110 140 110  80  80  75 130;...
140 110  80  80  75 130 160 125 100 100  80 150;...
160 125 100 100  80 150 190 150 130 inf inf inf;...
190 150 130 inf inf inf 200 180 150 inf inf inf;...
200 180 150 inf inf inf 100  80  50  50  60 100];

idxinf = find(delcosts == inf);
delcosts(idxinf) = 1e6;

buildcosts   = [3500 9000 10000 4000 3000 9000 9000 3000 4000 10000 9000 3500]';
capacity     = [ 300  250   100  180  275  300  200  220  270   250  230  180]';
demand       = [ 120   80    75  100  110  100   90   60   30   150   95  120]';

Prob = depotlocation(delcosts, buildcosts, capacity, demand, idxinf);

Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
c       = length(demand);                % number of customers
d       = length(buildcosts);            % number of possible depots
build   = find(Result.x_k(1:d)');        % the depots to build
deliver = reshape(Result.x_k(d+1:c*d+d),c,d);
deliver = deliver(:,build);
deliver(find(deliver<0.001)) = 0;
disp(['minimal total cost = ' num2str(Result.f_k) ])
disp(['build the depots     ' num2str(build)      ])
[jj,ii] = size(deliver);                 % ii = depots, jj = customers
for j = 1:jj,
disp(['let customer ' num2str(j) ' get '])
for i = 1:ii,
if deliver(j,i) ~= 0,
if deliver(j,i) == 1,
disp(['   all of its demand from depot ' num2str(build(i)) ])
else
disp(['   ' num2str(deliver(j,i)) ...
' of its demand from depot ' num2str(build(i)) ])
end
end
end
end
end

% MODIFICATION LOG
%
% 051019 med   Created.
% 060125 per   Moved disp to end```

## Heating oil delivery

```% function Result = heatingoildeliveryEx(PriLev)
%
% Creates a TOMLAB MIP problem for heating oil delivery
%
% HEATING OIL DELIVERY
%
% A transporter has to deliver heating oil from the refinery at
% Donges to a certain number of clients in the west of France. His
% clients are located at Brain-sur-l’Authion, Craquefou, Guérande, la
% Haie Fouassière, Mésanger and les Ponts-de-Cé. The following table
% lists the demands in liters for the different sites.
%
% Demands by clients (in liters)
%
% +-----------+-----+-----+-----+-----+
% | BslA| Craq| Guér| H Fo| Mésa| P dC|
% +-----+-----+-----+-----+-----+-----+
% |14000| 3000| 6000|16000|15000| 5000|
% +-----+-----+-----+-----+-----+-----+
%
% The next table contains the distance matrix between the clients and
% the refinery.
%
% Distance matrix (in km)
% +-------------------+----+----+----+----+----+----+----+
% |                   |Dong|BslA|Craq|Guér|H Fo|Mésa|P dC|
% +-------------------+----+----+----+----+----+----+----+
% |Donges             |   0| 148|  55|  32|  70| 140|  73|
% |Brain-s.-l’Authion | 148|   0|  93| 180|  99|  12|  72|
% |Craquefou          |  55|  93|   0|  85|  20|  83|  28|
% |Guérande           |  32|  80|  85|   0| 100| 174|  99|
% |Haie Fouassière    |  70|  99|  20| 100|   0|  85|  49|
% |Mésanger           | 140|  12|  83| 174|  85|   0|  73|
% |Ponts-de-Cé        |  73|  72|  28|  99|  49|  73|   0|
% +-------------------+----+----+----+----+----+----+----+
%
% The transport company uses tankers with a capacity of 39000 liters
% for the deliveries. Determine the tours for delivering to all
% clients that minimize the total number of kilometers driven.
%
% VARIABLES
%
% distance                   The distance matrix
% demand                     Demand
% capacity                   Capacity of tanker
%
% RESULTS
%
% Run this for an interpretation of the results
% Result = heatingoildeliveryEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = heatingoildeliveryEx(PriLev)

if nargin < 1
PriLev = 1;
end

distance     = [   0 148  55  32  70 140  73;...
148   0  93 180  99  12  72;...
55  93   0  85  20  83  28;...
32 180  85   0 100 174  99;...
70  99  20 100   0  85  49;...
140  12  83 174  85   0  73;...
73  72  28  99  49  73   0];

demand       = [14000;3000;6000;16000;15000;5000];
capacity     = 39000;

Prob = heatingoildelivery(distance, demand, capacity);

Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
s      = 7;                                         % number of sites
names  = ['Dong'; 'BslA'; 'Craq'; 'Guér'; 'H Fo'; 'Mésa'; 'P dC'];
tour   = reshape(Result.x_k(1:s*s),s,s);            % extract tour
tour(find(tour<0.5)) = 0;                           % remove false zeros
first  = find(tour(1,:));                           % might be an array
disp(['min distance of ' num2str(Result.f_k) ' km by using'])

for init = 1:length(first),                         % there might be many first stops
this_tour = [names(1,:)];                        % start in Dong
site = first(init);
this_tour = [this_tour ' -> ' names(site,:)];    % add city 2
loop_me = 1;
next = find(tour(site,:));                       % what city is next?

if next == 1,                                    % if we are going home:
loop_me = 0;                                  % do not enter while-loop
this_tour = [this_tour  ' -> ' names(1,:)];   % just add home and quit
disp(['Tour ' num2str(init) ': ' num2str(this_tour) ])
end

while loop_me == 1,                              % if more than one stop
this_tour = [this_tour  ' -> ' names(next,:)];% add them one at a time
next = find(tour(next,:));

if next == 1,                                  % when we are going home
loop_me = 0;                                % stop loop
this_tour = [this_tour  ' -> ' names(1,:)]; % finish names and quit
disp(['  tour ' num2str(init) ': ' num2str(this_tour) ])
end
end
end
end

% MODIFICATION LOG
%
% 051020 med   Created.
% 060125 per   Moved disp to end```

## Combining different modes of transport

```% function Result = combiningdiffmodesoftranspEx(PriLev)
%
% Creates a TOMLAB MIP problem for combining different modes of transport
%
% COMBINING DIFFERENT MODES OF TRANSPORT
%
% A load of 20 tonnes needs to be transported on a route passing
% through five cities, with a choice of three different modes of
% transport: rail, road, and air. In any of the three intermediate
% cities it is possible to change the mode of transport but the load
% uses a single mode of transport between two consecutive cities.
% The following table lists the cost of transport in \$ per tonne
% between the pairs of cities.
%
% Transport costs with different modes
%
%
%  Pairs of cities
% +----+---+---+---+---+
% |    |1–2|2–3|3–4|4–5|
% +----+---+---+---+---+
% |Rail| 30| 25| 40| 60|
% |Road| 25| 40| 45| 50|
% |Air | 40| 20| 50| 45|
% +----+---+---+---+---+
%
% The next table summarizes the costs for changing the mode of
% transport in \$ per tonne. The cost is independent of location.
%
%
% Cost for changing the mode of transport
%
% +---------+----+----+---+
% +---------+----+----+---+
% |Rail     |  0 |  5 | 12|
% |Road     |  8 |  0 | 10|
% |Air      | 15 | 10 |  0|
% +---------+----+----+---+
%
% How should we organize the transport of the load at the least cost?
%
% VARIABLES
%
% transpcost                 Transport costs
% changecost                 Cost to change mode of transport
%
% RESULTS
%
% For an interpretation of the results, try this:
% Result = combiningdiffmodesoftranspEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = combiningdiffmodesoftranspEx(PriLev)

if nargin < 1
PriLev = 1;
end

transpcost     = [ 30 25 40 60 ;...
25 40 45 50 ;...
40 20 50 45];

changecost     = [  0  5  12;...
8  0  10;...
15 10   0];

demand         = 30;

Prob = combiningdiffmodesoftransp(transpcost, changecost, demand);

Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
modes  = reshape(Result.x_k(1:3*4),3,4);
disp(['the min cost (' num2str(Result.f_k) ') is found by,'])
for m = 1:size(modes,2),
disp(['   going from town ' num2str(m) ' to town ' num2str(m+1) ...
' by ' means(find(modes(:,m)),:)])
end
end

% MODIFICATION LOG
%
% 051020 med   Created.
% 060125 per   Moved disp to end```

## Fleet planning for vans

```% function Result = fleetplanningforvansEx(PriLev)
%
% Creates a TOMLAB MIP problem for fleet planning for vans
%
% FLEET PLANNING FOR VANS
%
% A chain of department stores uses a fleet of vans rented from
% different rental agencies. For the next six months period it has
% forecast the following needs for vans (table below):
%
% Requirements for vans for six months
%
% +---+---+---+---+---+---+
% |Jan|Feb|Mar|Apr|May|Jun|
% +---+---+---+---+---+---+
% |430|410|440|390|425|450|
% +---+---+---+---+---+---+
%
% At the 1st January, the chain has 200 vans, for which the rental
% period terminates at the end of February.
%
% To satisfy its needs, the chain has a choice among three types of
% contracts that may start the first day of every month: 3-months
% contracts for a total cost of \$1700 per van, 4-months contracts at
% \$2200 per van, and 5-months contracts at \$2600 per van. How many
% contracts of the different types need to be started every month in
% order to satisfy the company’s needs at the least cost and to have
% no remaining vans rented after the end of June?
%
% Running contracts in month 5 (May)    .........
%                                       :       :
%                               +-------:-------:-------+
%                               | Rent34:       :       |
%                       +-------+-------:-------:-------+
%                       |     Rent33    :       :
%                       +---------------:-------:-------+
%                       | Rent43        :       :       |
%               +-------+---------------:-------:-------+
%               |    Rent42             :       :
%               +-----------------------:-------:-------+
%               | Rent52                :       :       |
%       +-------+-----------------------:-------:-------+
%       |    Rent51                     :       :
%       +-------------------------------:-------:
%               .       .       .       :       :       .
%  Month   Jan  :  Feb  :  Mar  :  Apr  :  May  :  Jun  :
%       :.......:.......:.......:.......:       :.......:
%                                       :.......:
%
%
% VARIABLES
%
% demand                     Demand of vans per month
% initialsupply              Vans per month
% contractlength             Contracts available
% contractcost               Cost of contracts
%
% RESULTS
%
% To interpret the result, use PriLev > 1, for example:
% Result = fleetplanningforvansEx(2);
%
% REFERENCES
%
% Applications of optimization... Gueret, Prins, Seveaux
% http://web.univ-ubs.fr/lester/~sevaux/pl/index.php
%
% INPUT PARAMETERS
% PriLev       Print Level
%
% OUTPUT PARAMETERS
% Result       Result structure.

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2005 by Tomlab Optimization Inc., \$Release: 5.0.0\$

function Result = fleetplanningforvansEx(PriLev)

if nargin < 1
PriLev = 1;
end

demand         = [ 430; 410; 440; 390; 425; 450];

initialsupply  = [ 200; 200;   0;   0;   0;   0];

contractlength = [ 3; 4; 5];

contractcost   = [ 1700; 2200; 2600];

Prob = fleetplanningforvans(demand, initialsupply, contractlength, contractcost);

Result = tomRun('cplex', Prob, PriLev);

if PriLev > 1,
months  = length(demand);
c       = length(contractlength);
temp   = reshape(Result.x_k,c,months);
disp(['minimal cost of ' num2str(Result.f_k) ' by '])
for m = 1:months,
if sum(temp(:,m)) > 0,
rent = find(temp(:,m));
disp(['   renting ' num2str(temp(rent,m)) ' vans for ' ...
num2str(contractlength(rent)) ' months in month ' num2str(m)])
end
end

end

% MODIFICATION LOG
%
% 051021 med   Created.