Models Air transport
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Flight connections at a hub
% function Result = flightconnectionsatahubEx(PriLev)
%
% Creates a TOMLAB MIP problem for flight connections at a hub
%
% FLIGHT CONNECTIONS AT A HUB
%
% The airline SafeFlight uses the airport Roissy-Charles-de-Gaulle
% as a hub to minimize the number of flight connections to European
% destinations. Six Fokker 100 airplanes of this airline from
% Bordeaux, Clermont-Ferrand, Marseille, Nantes, Nice, and Toulouse
% are landing between 11am and 12:30pm. These aircraft leave for
% Berlin, Bern, Brussels, London, Rome, and Vienna between 12:30 pm
% and 13:30 pm. The numbers of passengers transferring from the
% incoming flights to one of the outgoing flights are listed in the
% table below.
%
% Numbers of passengers transferring between the different flights
% +----------------+---------------------------------------+
% |Origins | Destinations |
% | +------+----+--------+------+----+------+
% | |Berlin|Bern|Brussels|London|Rome|Vienna|
% +----------------+------+----+--------+------+----+------+
% |Bordeaux | 35 | 12 | 16 | 38 | 5 | 2 |
% |Clermont-Ferrand| 25 | 8 | 9 | 24 | 6 | 8 |
% |Marseille | 12 | 8 | 11 | 27 | 3 | 2 |
% |Nantes | 38 | 15 | 14 | 30 | 2 | 9 |
% |Nice | – | 9 | 8 | 25 | 10 | 5 |
% |Toulouse | – | – | – | 14 | 6 | 7 |
% +----------------+------+----+--------+------+----+------+
%
% For example, if the flight incoming from Bordeaux continues on to
% Berlin, 35 passengers and their luggage may stay on board during
% the stop at Paris. The flight from Nice arrives too late to be
% re-used on the connection to Berlin, the same is true for the
% flight from Toulouse that cannot be used for the destinations
% Berlin, Bern and Brussels (the corresponding entries in the table
% are marked with ‘–’).
%
% How should the arriving planes be re-used for the departing
% flights to minimize the number of passengers who have to change
% planes at Roissy?
%
% VARIABLES
%
% origdest People remaining in plane at transfer
% idximp Impossible combinations
%
% RESULTS
%
% for an interpretation of the results, set PriLev > 1, for example:
% flightconnectionsatahubEx(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) 2005-2006 by Tomlab Optimization Inc., $Release: 5.1.0$
% Written Oct 21, 2005. Last modified Feb 3, 2006.
function Result = flightconnectionsatahubEx(PriLev)
if nargin < 1
PriLev = 1;
end
origdest = [ 35 12 16 38 5 2;...
25 8 9 24 6 8;...
12 8 11 27 3 2;...
38 15 14 30 2 9;...
0 9 8 25 10 5;...
0 0 0 14 6 7];
idximp = find(origdest == 0);
Prob = flightconnectionsatahub(origdest, idximp);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1
p = 6; % the number of planes
in_names = ['Bord';'Cler';'Mars';... % city names in
'Nant'; 'Nice'; 'Toul'];
out_names = ['Berl';'Bern';'Brus';... % city names out
'Lond';'Rome';'Vien'];
transfers = reshape(Result.x_k,p,p); % reshape x_k
for i = 1:p, % in city has number i
for o = 1:p, % out city has number o
if transfers(i,o) == 1,
disp(['plane from ' in_names(i,:) ...
' continues to ' out_names(o,:) ])
end
end
end
end
% MODIFICATION LOG
%
% 051021 med Created.
% 060113 per Added documentation.
% 060125 per Moved disp to end
% 060203 med Print level updatedComposing flight crews
% function Result = composingflightcrewsEx(PriLev)
%
% Creates a TOMLAB MIP problem for composing flight crews
%
% COMPOSING FLIGHT CREWS
%
% 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
%
% RESULTS
%
% Set PriLev to 2 for an interpretation of the results
% Result = composingflightcrewsEx(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) 2005-2006 by Tomlab Optimization Inc., $Release: 5.1.0$
% Written Oct 21, 2005. Last modified Feb 3, 2006.
function Result = composingflightcrewsEx(PriLev)
if nargin < 1
PriLev = 1;
end
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]';
Prob = composingflightcrews(scores, arcs_in, arcs_out, arcs_score);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1
idx = find(Result.x_k);
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 levelScheduling flight landings
% function Result = schedulingflightlandingsEx(PriLev)
%
% Creates a TOMLAB MIP problem for scheduling flight landings
%
% SCHEDULING FLIGHT LANDINGS
%
% The plane movements in large airports are subject to numerous
% security constraints. The problem presented in this section consist
% of calculating a schedule for flight landings on a single runway.
% More general problems have been studied but they are fairly
% complicated (dynamic cases, for instance through planes arriving
% late, instances with several runways, etc.).
%
% Ten planes are due to arrive. Every plane has an earliest arrival
% time (time when the plane arrives above the zone if traveling at
% maximum speed) and a latest arrival time (influenced among other
% things by its fuel supplies). Within this time window the airlines
% choose a target time, communicated to the public as the flight
% arrival time. The early or late arrival of an aircraft with respect
% to its target time leads to disruption of the airport and causes
% costs. To take into account these cost and to compare them more
% easily, a penalty per minute of early arrival and a second penalty
% per minute of late arrival are associated with every plane. The
% time windows (in minutes from the start of the day) and the
% penalties per plane are given in the following table.
%
% Characteristics of flight time windows
% +-----------------+---+---+---+---+---+---+---+---+---+---+
% |Plane | 1| 2| 3| 4| 5| 6| 7| 8| 9| 10|
% +-----------------+---+---+---+---+---+---+---+---+---+---+
% |Earliest arrival |129|195| 89| 96|110|120|124|126|135|160|
% |Target time |155|258| 98|106|123|135|138|140|150|180|
% |Latest Arrival |559|744|510|521|555|576|577|573|591|657|
% |Earliness penalty| 10| 10| 30| 30| 30| 30| 30| 30| 30| 30|
% |Lateness penalty | 10| 10| 30| 30| 30| 30| 30| 30| 30| 30|
% +-----------------+---+---+---+---+---+---+---+---+---+---+
%
% Due to turbulence and the duration of the time during which a plane
% is on the runway, a security interval has to separate any two
% landings. An entry in line p of column q in the following table
% denotes the minimum time interval (in minutes) that has to lie
% between the landings of planes p and q, even if they are not
% consecutive. Which landing schedule minimizes the total penalty
% subject to arrivals within the given time windows and the required
% intervals separating any two landings?
%
% Matrix of minimum intervals separating landings
%
% +--+--+--+--+--+--+--+--+--+--+--+
% | | 1| 2| 3| 4| 5| 6| 7| 8| 9|10|
% +--+--+--+--+--+--+--+--+--+--+--+
% | 1| –| 3|15|15|15|15|15|15|15|15|
% | 2| 3| –|15|15|15|15|15|15|15|15|
% | 3|15|15| –| 8| 8| 8| 8| 8| 8| 8|
% | 4|15|15| 8| –| 8| 8| 8| 8| 8| 8|
% | 5|15|15| 8| 8| –| 8| 8| 8| 8| 8|
% | 6|15|15| 8| 8| 8| –| 8| 8| 8| 8|
% | 7|15|15| 8| 8| 8| 8| –| 8| 8| 8|
% | 8|15|15| 8| 8| 8| 8| 8| –| 8| 8|
% | 9|15|15| 8| 8| 8| 8| 8| 8| –| 8|
% |10|15|15| 8| 8| 8| 8| 8| 8| 8| –|
% +--+--+--+--+--+--+--+--+--+--+--+
%
% VARIABLES
%
% costs Cost or being late
% mintimes Minimal intervals between landings
% var1 30 plus
% var2 30 minus
%
% RESULTS
%
% For an interpretation of the results, set PriLev > 1, for example:
% Result = schedulingflightlandingsEx(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) 2005-2006 by Tomlab Optimization Inc., $Release: 5.1.0$
% Written Oct 21, 2005. Last modified Feb 3, 2006.
function Result = schedulingflightlandingsEx(PriLev)
if nargin < 1
PriLev = 1;
end
costs = [ 129 195 89 96 110 120 124 126 135 160;...
155 258 98 106 123 135 138 140 150 180;...
559 744 510 521 555 576 577 573 591 657;...
10 10 30*ones(1,8);...
10 10 30*ones(1,8)];
mintimes = [ 0 3 15 15 15 15 15 15 15 15;...
3 0 15 15 15 15 15 15 15 15;...
15 15 0 8 8 8 8 8 8 8;...
15 15 8 0 8 8 8 8 8 8;...
15 15 8 8 0 8 8 8 8 8;...
15 15 8 8 8 0 8 8 8 8;...
15 15 8 8 8 8 0 8 8 8;...
15 15 8 8 8 8 8 0 8 8;...
15 15 8 8 8 8 8 8 0 8;...
15 15 8 8 8 8 8 8 8 0];
Prob = schedulingflightlandings(costs, mintimes);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1
planes = 10; % number of planes
[times,idx] = sort(Result.x_k(1:planes));
late = Result.x_k(planes*planes+2*planes+1:planes*planes+3*planes);
early = Result.x_k(planes*planes+1*planes+1:planes*planes+2*planes);
for p = 1:planes,
time = times(p);
plane = idx(p);
if (late(plane) > 0.5),
disp(['plane ' num2str(plane) ' will arrive minute ' ...
num2str(time) ' (' num2str(late(plane)) ' minute(s) late)'])
elseif (early(plane) > 0.5),
disp(['plane ' num2str(plane) ' will arrive minute ' ...
num2str(time) ' (' num2str(early(plane)) ' minute(s) early)'])
else
disp(['plane ' num2str(plane) ' will arrive minute ' ...
num2str(time) ' (on time)' ])
end
end
end
% MODIFICATION LOG
%
% 051021 med Created.
% 060113 per Added documentation
% 060125 per Moved disp to end
% 060203 med Print level updatedAirline hub location
% function Result = airlinehublocationEx(PriLev)
%
% Creates a TOMLAB MIP problem for airline hub location
%
% AIRLINE HUB LOCATION
%
% FAL (the Flying Air Lines) specializes in freight transport. The
% company links the major French cities with cities in the United
% States, namely: Atlanta, Boston, Chicago, Marseille, Nice, and
% Paris. The average quantities in tonnes transported every day by
% this company between these cities are given in the following table.
%
% Average quantity of freight transported between every pair of cities
%
% +---------+-------+------+-------+---------+----+-----+
% | |Atlanta|Boston|Chicago|Marseille|Nice|Paris|
% |Atlanta | 0 | 500 | 1000 | 300 |400 |1500 |
% |Boston |1500 | 0 | 250 | 630 |360 |1140 |
% |Chicago | 400 | 510 | 0 | 460 |320 | 490 |
% |Marseille| 300 | 600 | 810 | 0 |820 | 310 |
% |Nice | 400 | 100 | 420 | 730 | 0 | 970 |
% |Paris | 350 | 1020 | 260 | 580 |380 | 0 |
% +---------+-------+------+-------+---------+----+-----+
%
% We shall assume that the transport cost between two cities i and j
% is proportional to the distance that separates them. The distances
% in miles are given in the next table.
%
% Distances between pairs of cities
%
% +---------+------+-------+---------+----+-----+
% | |Boston|Chicago|Marseille|Nice|Paris|
% +---------+------+-------+---------+----+-----+
% |Atlanta | 945 | 605 | 4667 |4749| 4394|
% |Boston | 866 |3726 | 3806 |3448| |
% |Chicago | 4471 |4541 | 4152 | | |
% |Marseille| 109 | 415 | | | |
% |Nice | 431 | | | | |
% +---------+------+-------+---------+----+-----+
%
% The airline is planning to use two cities as connection platforms
% (hubs) to reduce the transport costs. Every city is then assigned
% to a single hub. The traffic between cities assigned to a given
% hub H1 to the cities assigned to the other hub H2 is all routed
% through the single connection from H1 to H2 which allows the
% airline to reduce the transport cost. We consider that the
% transport cost between the two hubs decreases by 20%. Determine the
% two cities to be chosen as hubs in order to minimize the transport
% cost.
%
% VARIABLES
%
% frights Goods between cities
% distance Distances
% hubs Number of hubs
%
% RESULTS
%
% The result from this example is quite hard to interpret since
% Result.x_k should be split into a tensor of rank 4 with the indices
% i, j, k and l and also into a vector of length c (number of
% cities). The vector indicates the hubs. For all non-zero element
% (i,j,k,l) in the tensor the following interpretation is possible:
% the route from city i to j passes through the hubs k and l.
%
% In the example below temp(3,5,2,6) == 1 since the route from
% 3 (Chicago) to 5 (Nice) pass through 2 (Boston) and 6 (Paris).
% Also temp(4,5,6,6) == 1 since the route from 4 (Marseille) to
% 5 (Nice) pass through only 6 (Paris).
%
% 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) 2005-2006 by Tomlab Optimization Inc., $Release: 5.1.0$
% Written Nov 7, 2005. Last modified Feb 3, 2006.
function Result = airlinehublocationEx(PriLev)
if nargin < 1
PriLev = 1;
end
frights = [ 0 500 1000 300 400 1500;...
1500 0 250 630 360 1140;...
400 510 0 460 320 490;...
300 600 810 0 820 310;...
400 100 420 730 0 970;...
350 1020 260 580 380 0];
distance = [ 0 945 605 4667 4749 4394;...
945 0 866 3726 3806 3448;...
605 866 0 4471 4541 4152;...
4667 3726 4471 0 109 415;...
4749 3806 4541 109 0 431;...
4394 3448 4152 415 431 0];
hubs = 2;
Prob = airlinehublocation(frights, distance, hubs);
Result = tomRun('cplex', Prob, PriLev);
if PriLev > 1
c = 6; % number of cities
c_names = ['Atla';'Bost';'Chic';'Mars';'Nice';'Pari'];
temp = reshape(Result.x_k(1:c^4),c,c,c,c);
hubs = find(Result.x_k(c^4+1:c^4+c));
disp('Put hubs in ')
for h = 1: length(hubs),
disp([' ' num2str(c_names(hubs(h),:)) ])
end
% displaying the routes between cities
for i = 1:c,
for j = i+1:c,
for k = 1:c,
for l = 1:c,
if temp(i,j,k,l) == 1,
% route involving one hub
if k==l & i~=k & j~=k & i~=j,
disp(['the route ' num2str([c_names(i,:),' ', ...
c_names(j,:)]) ' requires hub ' num2str(c_names(k,:))])
% route involving two hubs
elseif i~=j & i~=k & i~=l & j~=k & j~=l & k~=l,
disp(['the route ' num2str([c_names(i,:),' ', ...
c_names(j,:)]) ' requires hubs ' ...
num2str([c_names(k,:),' ', c_names(l,:)])])
% route between hubs
elseif ( (i==k & j==l) | (i==l & j==k) ) & k~=l,
disp(['the route ' num2str([c_names(i,:),' ', ...
c_names(j,:)]) ' is between hubs' ])
end
end
end
end
end
end
end
% MODIFICATION LOG
%
% 051107 med Created.
% 060113 per Added documentation.
% 060116 per Minor changes.
% 060125 per Moved disp to end
% 060203 med Updated print levelPlanning a flight tour
% function Result = planningaflighttourEx(PriLev)
%
% Creates a TOMLAB MIP problem for planning a flight tour
%
% PLANNING A FLIGHT TOUR
%
% A country in south-east Asia is experiencing widespread flooding.
% The government, with international help, decides to establish a
% system of supply by air. Unfortunately, only seven runways are
% still in a usable state, among which is the one in the capital.
%
% The government decides to make the planes leave from the capital,
% have them visit all the other six airports and then come back to
% the capital. The following table lists the distances between the
% airports. Airport A1 is the one in the capital. In which order
% should the airports be visited to minimize the total distance
% covered?
%
% Distance matrix between airports (in km)
%
% +--+---+---+---+---+---+---+
% | | A2| A3| A4| A5| A6| A7|
% +--+---+---+---+---+---+---+
% |A1|786|549|657|331|559|250|
% |A2|668|979|593|224|905| |
% |A3|316|607|472|467| | |
% |A4|890|769|400| | | |
% |A5|386|559| | | | |
% |A6|681| | | | | |
% +--+---+---+---+---+---+---+
%
% VARIABLES
%
% distance The distance matrix
%
% RESULTS
%
% For an interpretation of the results set PriLev > 1, for example:
% Result = planningaflighttourEx(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., $Release: 5.0.0$
% Written Nov 7, 2005. Last modified Nov 7, 2005.
function Result = planningaflighttourEx(PriLev)
if nargin < 1
PriLev = 1;
end
distance = [ 0 786 549 657 331 559 250;...
786 0 668 979 593 224 905;...
549 668 0 316 607 472 467;...
657 979 316 0 890 769 400;...
331 593 607 890 0 386 559;...
559 224 472 769 386 0 681;...
250 905 467 400 559 681 0];
Prob = planningaflighttour(distance);
stop = 0;
n = size(distance,1);
while stop<100
Result = tomRun('cplex', Prob, PriLev);
% Find a subtour
idx = zeros(n,n);
idx(1,1:n) = [1:7];
for i=1:n
for j=2:n
idx(j,i) = find(Result.x_k( (idx(j-1,i)-1)*n+1: idx(j-1,i)*n) ~= 0);
if idx(j,i) == i
break
end
end
end
shortmat = spones(idx);
len = full(sum(shortmat,1));
[val, idx2] = min(len);
idx3 = idx(1:val, idx2);
Aextra = zeros(1,Prob.N);
for k=1:val-1
Aextra(1,idx3(k)+n*(idx3(k+1)-1)) = 1;
end
Prob.A = [Prob.A;Aextra];
Prob.b_L = [Prob.b_L;-inf];
Prob.b_U = [Prob.b_U;1];
Prob.mLin = Prob.mLin+1;
Result.tour = idx(:,1);
if val == n
break
end
stop = stop + 1;
end
if PriLev > 1,
a = 7; % number of airports
temp = reshape(Result.x_k,a,a); % reshape results
not_home_again = true; % help to loop
current_pos = 1; % we start in 1
route = []; % initially empty
route = [route current_pos]; % add position
while not_home_again, % loop if not home
current_pos = find(temp(current_pos,:)); % go to next pos
route = [route current_pos]; % add to route
if current_pos == 1, % if home:
not_home_again = false; % stop loop
end
end
disp(['Shortest route is: ' num2str(route)])
disp([' or: ' num2str(route(end:-1:1))])
end
% MODIFICATION LOG
%
% 051107 med Created.
% 060116 per Added documentation.
% 060125 per Moved disp to end