ENM 212 Integer Programming and Network Models-HW 1 Solutions
1)
a) NWC Final Allocation Table is :
S1
S2
S3
S4
S5
Demand
D1
13 (3)
14
3
18
30
3
D2
10 (2)
13 (3)
0
9
24
5
D3
22
16 (3)
M(1)
19
34
4
D4
29
21
11 (5)
23
36
5
D5
18
M
6 (1)
11 (4)
28 (1)
6
Ddummy
0
0
0
0
0 (2)
2
Supply
5
6
7
4
3
M=10000
The minimum total transportation cost
𝑧𝑁𝑊𝐶 =13×3+10×2+13×3+16×3+10000×1+11×5+6×1+11×4+28×1+0×2=10279=13×3+10×2+13×3+16×3
+10000×1+11×5+6×1+11×4+28×1+0×2=279+M
Here, the number of allocated cells = 10 is equal to m + n - 1 = 5 + 6 - 1 = 10
b) VAM
Final Allocation Table is
D1
D2
D3
D4
D5
Ddumm
y
Suppl
y
Row Penalty
S1
13
10(5
)
22
29
18
0
5
10 | 3 | 8 | 8 | -- | --| -- | -|
S2
14
13
16(4
)
21(1
)
M(1)
0
6
13 | 1 | 3 | 3 | 5 | 5 | 5
| 5 |
S3
3(3
)
0
M
11(4
)
6
0
7
0 | 3 | 6 | -- | -- | -- | -- |-- |
S4
18
9
19
23
11(4
)
0
4
9 | 2 | 2 | 2 | 8 | -- | -- | -|
S5
30
24
34
36
28(1
)
0(2)
3
24 | 4 | 4 | 4 | 6 | 6 | -- |
-- |
Demand
3
5
4
5
6
2
Col
Penalty
10
10
------
9
9
9
1
----
3
3
3
3
3
18
16
10
10
10
2
2
15
21
5
5
5
7
17
M
M
0
-------
--
--
16
21
--
--
The minimum total transportation cost
𝑧𝑝𝑒𝑛𝑎𝑙𝑡𝑦 =10×5+16×4+21×1+M×1+3×3+11×4+11×4+28×1+0×2=260+M
Here, the number of allocated cells = 9, which is one less than to m + n - 1 = 5 + 6 - 1 = 10This
solution is degenerate.
OR
Final Table
D1
D2
D3
D4
D5
Ddum
my
Suppl
y
S1
13
10(3
)
22
29
18(2
)
0
5
10 | 3 | 3 | 3 | 3| 3 | 3 |
3|
S2
14(1
)
13(1
)
16(4)
21
M
0
6
13 | 1 | 1 | 1 | 1 | 1 | 1
| 1|
S3
3(2)
0
1000
0
11(5
)
6
0
7
0 | 3 | 3 | -- | -- | -- | -- |
-- |
S4
18
9
19
23
11(4
)
0
4
9 | 2 | 2 | 2 | -- | -- | -| -- |
S5
30
24(1
)
34
36
28
0(2)
3
24 | 4 | 4 | 4 | 4 | 6 | 6
| -- |
Demand
3
5
4
5
6
2
10
10
10
------
9
9
9
1
3
3
3
3
3
3
3
3
6
6
---
10
10
-------
5
5
5
7
10
----
0
--------
Col
Penalty
Row Penalty
Here, the number of allocated cells = 10, which is equal to m + n - 1 = 5 + 6 - 1 = 10.
𝑧𝑝𝑒𝑛𝑎𝑙𝑡𝑦 =3*10+18*2+14*1+13*1+16*4+3*2+11*5+11*4+24*1+0*2=286
Interpretition:
𝑧𝑁𝑊𝐶 > 𝑧𝑝𝑒𝑛𝑎𝑙𝑡𝑦
VAM(penalty) gave better solution than NWC.
2) MODI: Start with a BFS. ( Be careful : m + n – 1 cells must be allocated before each step. )
Final optimal solution :
D1
D2
D3
D4
D5
Ddummy
Supply
S1
13 (0) 10 (5) 22
29
18
0
5
S2
14
13
16 (4)
21 (2) 10000 0
6
S3
3 (3)
0
10000 11 (3) 6 (1)
0
7
S4
18
9
19
23
11 (4)
0
4
S5
30
24
34
36
28 (1)
0 (2)
3
Demand
3
5
4
5
6
2
D1
D2
D3
D4
D5
Ddummy
Supply
S1
13 (3) 10 (2) 22
29
18
0
5
S2
14
13
16 (4)
21 (2) 10000 0
6
S3
3
0(3)
10000 11 (3) 6 (1)
0
7
S4
18
9
19
23
11 (4)
0
4
S5
30
24
34
36
28 (1)
0 (2)
3
Demand
3
5
4
5
6
2
Or
𝑧 ∗ = 276
3) GAMS CODE
$title GENERAL MODEL OF TRANSPORTATION PROBLEM
OR