Question

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.

Table 10.1: Description of the vehical 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 BC 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. 10.1.1 Model formula.

this question is required to solve it as linear simple form ,and code it by lingo.

subject: opreation research

name of the book is Applications of optimization with Xpress-MP, problem chapter(10) first question (car rental).

Answer 1

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

idx_excess = find(stock-demand > 0);
n_excess   = length(idx_excess);

idx_need   = find(stock-demand < 0);
n_need     = length(idx_need);

move = tom('move',n_excess,n_need,'int');

% Bounds
bnds = {0 <= move};

% Excess constraint
con1 = {sum(move,2) == stock(idx_excess) - demand(idx_excess)};

% Need constraint
con2 = {sum(move,1)' == demand(idx_need) - stock(idx_need)};

% Objective
objective = sum(sum(move.*cost.*distance(idx_excess,idx_need)));

constraints = {bnds, con1, con2};
options = struct;
options.solver = 'cplex';
options.name   = 'Car Rental';
sol = ezsolve(objective,constraints,[],options);

PriLev = 1;
if PriLev > 0
    temp = sol.move;
    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