Transcript Transportation Problem

Transportation problem
Production
capacity
Requirement
for goods
b1
a2
b2
...
Factories
an
ai
i-th factory
bj
delivers to j-th
customer at cost cij
...
a1
bm
Customers
𝑎𝑖 ≥
𝑖
𝑏𝑗
𝑗
necessary condition
Minimum cost of transportation satisfying the demand of customers.
Transportation tableau
amount
transported
cost cij of delivering from
ith factory to jth customer
7
x11
3
x12
4
x21
2
2
shadow
customer’s “price”
2
1
v1
u2
6
5
x33
v2
2
u1
8
x23
x32
4
4
x13
x22
x31
supply ai of ith factory
3
v3
3
u3
shadow
factory “price”
demand bj of jth customer
shadow prices are relative to some baseline
Transportation problem
𝑖 𝑎𝑖
• Overproduction: If produced more then needed (
Dummy customer at zero delivery cost
• Shortage: If required more than produced (
𝑖 𝑎𝑖
<
>
𝑗 𝑏𝑗 )
𝑗 𝑏𝑗 )
Dummy producer at market delivery cost
(what it would cost to deliver it from a third party)
7
x11
3
x12
4
4
x13
u01
x814
2
2
x31
1
x32
v91
x441
33
v82
x242
v1
4
𝑖
𝑎𝑖x33
=
𝑗
𝑏𝑗 x3
34
v3
11
x343
v2
2
5
8
u02
u2
u03
u3
3
x844
b2
a3
b3
a4
b4
u4
0
v4
8
a2
3
v04
v3
b1
u1
x21
x22transportation
x23
x624 problem
6
Balanced
2
a1
Transportation Simplex
• Applying the Simplex method to the problem
• Basic solution
– min-cost method
• Pivoting
– shadow prices
set u1 = 0, then ui + vj=cij
– reduced cost
pivot if ui + vj > cij
• Finding a loop
cost = 3×7 + 3×4 + 2×0
+ 1×2 + 2×1 + 6×0 = 37
7
x311
3
x12
4
x21
2
2
34
0
v1
2
x23
1
x232
02
v2
0
x214
x313
x22
x131
4
0
x624
5
x33
30
680
u2
06
0
v4
02
8
u3
130
x34
v3
u1
z = 37
must mark
takeexactly
the smaller
m + nof
– the
1 = 6two
cells
Transportation Simplex
• Applying the Simplex method to the problem
• Basic solution
– min-cost method
• Pivoting
– shadow prices
set u1 = 0, then ui + vj=cij
– reduced cost
pivot if ui + vj > cij
• Finding a loop
7
x311
3
x12
4
x21
2
2
1
x232
2
8
0
x624
5
x33
v62
v71
4
2
x23
u01
0
x214
x313
x22
x131
4
u02
6
-5
u3
0
x34
v43
3
ui =0 and vj must sum up to cij = 4
3
v04
8
z = 37
vj = 4
Transportation Simplex
• Applying the Simplex method to the problem
• Basic solution
– min-cost method
• Pivoting
– shadow prices
set u1 = 0, then ui + vj=cij
– reduced cost
pivot if ui + vj > cij
• Finding a loop
7 6 >3
x311
x12
4
x313
7>4 6>2 4>2
x21
x22
x23
2
x131
4
x214
8
0
x624
v62
2
v43
3
calculate ui + vj
v04
8
u02
6
1 -1 ≤ 5 -5 ≤ 0
x232
x33
x34
v71
u01
0
-5
u3
3
z = 37
Transportation Simplex
• Applying the Simplex method to the problem
• Basic solution
– min-cost method
• Pivoting
– shadow prices
set u1 = 0, then ui + vj=cij
– reduced cost
pivot if ui + vj > cij
• Finding a loop
• Largest
New basis
feasible Δ = 2
7 6 >3
3x13-Δ
x12
11
4
x313
7>4 6>2 4>2
x21
x222
x23
+Δ
2
1x31+Δ
31
4
2+Δ
2x4
14
8
0
6x4
6-Δ
24
v62
2
v43
3
v04
8
u02
6
1 -1 ≤ 5 -5 ≤ 0
2x2-Δ
x33
x34
32
v71
u01
0
-5
u3
3
z = 37
29