model for solving the k-median problem. More...
Go to the source code of this file.
Functions | |
def | kmedian (I, J, c, k) |
def | distance (x1, y1, x2, y2) |
def | make_data (n, m, same=True) |
Variables | |
int | n = 200 |
m = n | |
I | |
J | |
c | |
x_pos | |
y_pos | |
same | |
int | k = 20 |
model = kmedian(I,J,c,k) | |
int | EPS = 1 |
x | |
y | |
list | edges = [(i,j) for (i,j) in x if model.getVal(x[i,j]) > EPS] |
list | facilities = [j for j in y if model.getVal(y[j]) > EPS] |
G = NX.Graph() | |
other = set(j for j in J if j not in facilities) | |
client = set(i for i in I if i not in facilities and i not in other) | |
dictionary | position = {} |
with_labels | |
False | |
node_color | |
nodelist | |
node_size | |
model for solving the k-median problem.
Definition in file kmedian.py.
def kmedian.distance | ( | x1, | |
y1, | |||
x2, | |||
y2 | |||
) |
return distance of two points
Definition at line 43 of file kmedian.py.
def kmedian.kmedian | ( | I, | |
J, | |||
c, | |||
k | |||
) |
kmedian -- minimize total cost of servicing customers from k facilities Parameters: - I: set of customers - J: set of potential facilities - c[i,j]: cost of servicing customer i from facility j - k: number of facilities to be used Returns a model, ready to be solved.
Definition at line 13 of file kmedian.py.
def kmedian.make_data | ( | n, | |
m, | |||
same = True |
|||
) |
creates example data set
Definition at line 48 of file kmedian.py.