Exercises of Chapter 2
2. Consider the transshipment problem in a network with seven vertices labelled as
1,2,…,7, with b T = [-6 -4 -8 0 0 11 7], and nine arcs (1, 2), (1, 4), (2, 3), (2, 5), (3,
5), (4, 6)，(5,4), (5, 7) and (6, 7). Obtain an FTS by enlarging the network with the
construction of two additional arcs (1, 5) and (5, 6) so that the entire supply is sent
to vertex 5 from each source and the entire demand is sent from vertex 5 to each
sink. After obtaining an FTS, solve the problem with c = [5 9 3 8 2 7 6 3 8].
Solution:
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
x  [ x12 , x14 , x 23 , x 25 , x35 , x 46 , x54 , x57 , x67 ]T
5, 0
b  [6,4,8,0,0,11,7]T
Iteration 1.Introduce arcs a10  (1,5) and a11  (5,6) Initial spanning tree T (1)
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
Let c '  0 0 0 0 0 0 0 0 0 1 1

Initial FTS
x (1)  [0 0 0 4 8 0 0 7 0
6 11]T
y 7  0,
'
y 5  y 7  c57
 0,
'
y 6  y 5  c56
 1,
'
y 4  y 5  c54
 0,
'
y 3  y 5  c35
 0,
'
y 2  y 5  c 25
 0,
y1  y 5  c15'  1
y (1)  [1,0,0,0,0,1,0]
Since y2-y1 = 0 – (-1) = 1 > 0 (cost of transportation from 1 to 2),
arc a1  (1,2) is profitable.
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
The fundamental cycle of a1 in T(1) is (1, 2, 5, 1).
The only backward arc is a10  (1,5) , which is the leaving arc. t=6
The new FTS and the corresponding ST are:
x ( 2)  [6 0 0 10 8 0 0 7 0
0 11]T
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
2nd iteration: Find the dual solution y for T(2).
y7  0,
'
y5  y7  c57
 0,
'
y6  y5  c56
 1,
'
y 4  y5  c54
 0,
'
y3  y5  c35
 0,
'
y 2  y5  c25
 0,
y1  y 2  c12'  0
y ( 2 )  [0,0,0,0,0,1,0]
For arc (4,6), y6 – y4 = 1 > c46’ = 0.
So, (4,6) is profitable.
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
Only (4,6) is backward arc. So it is also the leaving arc.
The new FTS and the corresponding ST are:
x (3)  [6 0 0 10 8 11 11 7 0
0 0]T
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
3rd iteration (Phase II): The cost = cx(3) = 290
Find the dual solution corresponding to T(3)
y7 = 0
y7 – y5 = c57 = 3. So, y5 = -3
y4 – y5 = c54 = 6. So, y4 = 3,
Continuing this way …, you will find the dual solution
y = (-16, -11, -5, 3, -3, 10, 0)
For arc (1, 4), y4 – y1 = 19 > 9 = c14.
So, arc (1, 4) is profitable.
The fundamental cycle of (1,4) is (1, 2, 5, 4, 1)
(1,2), (2,5) and (5,4) are all backward arcs.
Since the flow x12 is the smallest about these 3 lines, (1, 2) is the leaving arc.
The new FTS and the corresponding ST are:
x(4) = [0, 6, 0, 4, 8, 11, 5, 7, 0]
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
5, 0
The 4th iteration: The cost = cx(4) = 230
Find the dual solution corresponding to T(3)
y = (-6, -11, -5, 3, -3, 10, 0)
(2,3) is a profitable arc.
The fundamental cycle of (2,3) is (2, 3, 5, 2)
Only (2,5) is a backward arc. Thus (2,5) is the leaving arc.
The new FTS and the corresponding ST are:
x(5) = [0, 6, 4, 0, 12, 11, 5, 7, 0]
1, -6
2,-4
3, -8
7, 7
4, 0
6, 11
The 5th iteration: The cost = cx(5) = 218
Find the dual solution corresponding to T(3)
y = (-6, -8, -5, 3, -3, 10, 0)
There is no more profitable arc.
5, 0
So, the optimal solution is x(5) = [0, 6, 4, 0, 12, 11, 5, 7, 0].
The optimal value is 218.