Download Shortest Route Problem: Finding the Optimal Route with Minimum Cost - Prof. Yanqiu Wang and more Study notes Mathematics in PDF only on Docsity!
Shortest route problem
Find the shortest route from the starting point (p1) to the ending point (p6).
Start
End
15
13
11
9
16
12
4
17
14
p
P
P
P
P
P
1. Set optimization variables
For each "link" in the graph, we set one variable. For example, there’s a "link" from P1 to P2, so we set a variable x12. Notice that
there are two variables associated with points P2 and P3, they are x23 and x32. If the value of a variable is 1, it means the "route"
will pass through this link. Value 0 means this link will not be taken by the route.
In[1]:= vars = 8 x12, x13, x23, x32, x24, x25, x35, x54, x46, x56<;
2. Set the objective function
We would like to minimize the total "distance" (cost) of the chosen route :
In[2]:= f = 15 * x12 + 13 * x13 + 9 * x23 + 9 * x32 +
11 * x24 + 12 * x25 + 16 * x35 + 4 * x54 + 17 * x46 + 14 * x56;
3. Set the constraints
To make sure the route is a connected path from the starting point to the ending point, we need to following constrains:
ü (1) for each point other than the starting and the ending points, the total entering links (inflow)
should be equal to the total leaving links (outflow).
ü
(1) for each point other than the starting and the ending points, the total entering links (inflow)
should be equal to the total leaving links (outflow).
In[3]:= g2 = x12 + x32 ä x24 + x25 + x23;
g3 = x13 + x23 ä x32 + x35;
g4 = x24 + x54 ä x46;
g5 = x35 + x25 ä x54 + x56;
ü (2) for the starting point, out flow - inflow = 1
In[7]:= g1 = x12 + x13 ä 1;
ü (3) for the ending point, inflow - outflow = 1
In[8]:= g6 = x46 + x56 ä 1;
ü (4) we also need to set all variables greater than or equal to 0
In[9]:= NonNegativeness = And ûû Thread@vars ≥ 0 D
Out[9]= x12 ≥ 0 && x13 ≥ 0 && x23 ≥ 0 && x32 ≥ 0 && x24 ≥ 0 && x25 ≥ 0 && x35 ≥ 0 && x54 ≥ 0 && x46 ≥ 0 && x56 ≥ 0
4. Solve the problem
ü (1) First, we can use the command "Minimize", which accepts equations/inequalities as inputs
In[10]:= Minimize@8f, g1 && g2 && g3 && g4 && g5 && g6 && NonNegativeness<, varsD
Out[10]= 8 41, 8 x12 Æ 1, x13 Æ 0, x23 Æ 0, x32 Æ 0, x24 Æ 0, x25 Æ 1, x35 Æ 0, x54 Æ 0, x46 Æ 0, x56 Æ 1 <<
The aboce solution tells us that the shortest route is P1 -> P2 -> P5 -> P6, and the total cost is 41.
ü (2) Mathematica also provides a function "LinearProgramming", which is more efficient.
However, it only accepts matrices/vectors in the standard LP form as inputs. Therefore, we first
need to derive the coefficients in the standard form minimizing f = c
t
x , subject to constraints
Ax >= b, x>=0. Notice that Mathematica uses Ax >= b, different from the Ax=b in the text book.
Hence when we feed the command with inputs, we have to specify for each RHS value b
i
that it
is an exact "equal". This can be down by setting a matrix of the form 88 b
1
, 0}, 8 b
2
, 0}, ... 8 b
k
In[11]:= c = Normal@CoefficientArrays@f, varsD D @@ 2 DD
Out[11]= 8 15, 13, 9, 9, 11, 12, 16, 4, 17, 14<
2 shortestroute.nb
