Problem (directed chinese postman problem): Consider the following directed graph:
(a) Does the graph shown above have a Eulerian tour? If yes, show it; if no, explain why.
(b) Determine shortest Chinese Postman tour in the above graph. Show all steps.
Solution: (a) The graph shown above is not Eulerian, as the outdegrees of the nodes are 3, 2, 1, 2, 4, and
1, while the indegrees are 1, 3, 2, 2, 2, and 3. Thus we have the sets N– = {n 1 , n 5 } and N+ = {n 2 , n 3 , n 6 }
with “supplies” 1, 1, and 2 and “demands” of 2 and 2. The distance matrix of the resulting transportation
problem is then
11 9 
C = 14 8  ,
 2 4 
so that the matematical formulation of the problem is as follows.
P: Min z = 11x 21 + 9x 25 + + 8x 35 + 2x 61 + 4x 65
s.t. x 21 + x 25 = 1
x 31 + x 35 = 1
x 61 + x 65 = 2
x 21 + x 31 + x 61 = 2
x 25 + x 35 + x 65 = 2
x 21 , x 25, x 31 , x 35 , x 61 , x 65 ≥ 0.
The optimal solution of this transportation plan is x25 = 1 (thus including the arcs a 23 , a 34 , and a 45 in the
solution), x35 = 1 (hence including the arcs a 34 and a 45 in the augmented graph), and x61 = 2 (thus
including two copies of the arc a 61 ), while the optimal values of all other variables are zero. The resulting
augmented graph is shown below:
It can be seen that the augmented graph is Eulerian (i.e., indegrees and outdegrees are equal for all nodes;
here they are 3, 3, 3, 4, 4, and 3). A possible Chinese Postman tour is n 5 , n 4 , n 2 , n 6 , n 1 , n 2 , n 3 , n 4 , n 5 , n 6 ,
n1, n6, n1, n5, n2, n3, n4, n5, n3, n4, n5.
The total length of the tour is 58 (the lengths of all arcs in the original graph that have to be traversed
once each) plus 21 (the value of the objective function of the transportation problem, i.e., the sum of all
arcs that have to be traversed more than once).