Sure. Consider the graph where a = cost and c = capacity:
a=3,c=1
Ao----->oB
\ ^
\ /a=1,c=1
\/
/\
/ \a=1,c=1
/ v
Co----->oD
a=3,c=1
So there are obviously 2 possible max flows. One uses the horizontal edges and the other the diagonals.
For the flow along the horizontals, we have a residual graph:
a=-3,c=1
Ao<-----oB
\ ^
\ /a=1,c=1
\/
/\
/ \a=1,c=1
/ v
Co<-----oD
a=-3,c=1
Notice the cycle B->A->D->C with capacity 1 and cost -3 + 1 -3 + 1 = -4.
The intuitive explanation for this cycle is that every increase in flow of one unit going along the edges in the cycle - or conversely every decrease in flow going along its edges in the opposite direction - we will decrease total cost of flow by 4 because we will be substituting flow along the cheaper diagonal edges for flow along the comparatively expensive horizontal edges.
In the augmenting path algorithm for min cost flow, we'd go ahead and push 1 unit of flow along this cycle and end up with a new, cheaper flow along the diagonals. This would provide the new residual graph:
a=3,c=1
Ao----->oB
^ /
\ /a=-1,c=1
\/
/\
/ \a=-1,c=1
v \
Co----->oD
a=3,c=1
Now the cycle is A->B->C->D and has cost 3 - 1 + 3 - 1 = 4, so the max flow along diagonals is a min cost max flow.