Линейное программирование. Практикум. Учебное пособие для бакалавриата

7. Tranportation problem

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7. Tranportation problem

Finally we get the initial plan.

z(X) = 80 2 +70 1+60 4 +40 5 +140 7 +10 10 =1750 <2970.

Note. The resulting initial plan has a lower transportation cost than plan obtained by the North West Corner Rule.

7.3. The Distribution Method (U-V Method)

7.3.1. Testing for Optimality

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7. Tranportation problem

subject to

ui +vj ≤ cij,

i =1, ,m,

j =1, ,n.

Theorem. If a basic plan is optimal then there are the numbers ui (i =1, ,m) and vj (j =1, ,n) such that:

Definition. The numbers ui and vj are called potentials of i-th producer and j-th consumer respectively and

∆ij = ui +vj −cij ,

are called net evaluations (net opportunity costs) of empty cells.

Note. The Theorem above states: if a basic plan is optimal then all net evaluations are non-positive.

Example. Let us test for optimality the initial basic plan obtained by the North West Corner Rule.

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7. Tranportation problem

Solution.

Let u1 = 0 . So we get

A1B1 u1 +v1 =11 0 +v1 =11 v1 =11 ,

A1B2 u1 +v2 =5 0 +v2 =5 v2 =5,

A2B1 u2 +v2 = 4 u2 +5 = 4 u2 = −1,

A2B3 u2 +v3 =5 −1+v3 =5 v3 = 6,

A3B3 u3 +v3 =7 u3 +6 =7 u3 =1,

A3B4 u3 +v4 =10 1+v4 =10 v4 = 9.

Net evaluations:

∆13 = u1 +v3 −c13 = 0 +6 −4 =2 > 0 , ∆14 = u1 +v4 −c14 = 0 +9 −2 =7 > 0, ∆21 = u2 +v1 −c21 = −1+11−1 = 9 > 0, ∆24 = u2 +v4 −c24 = −1+9 −9 = −1 < 0, ∆31 = u3 +v1 −c31 =1+11−9 = 3 > 0, ∆32 = u3 +v2 −c32 =1+5 −8 = −2 < 0.

The obtained results we record to the table (where the net evaluations are recorded in the square in the free cells).

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7. Tranportation problem

There are positive net evaluations so the basic plan is not optimal.

Example. Test for optimality the basic plan obtained by the the Least Cost Rule.

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is optimal and zmin = z(X ) =1750.

7.3.2. Improving of a basic plan

Definition. In a transportation table, an ordered set of four or more cells is said to form a loop if:

a) Any two and only two adjacent cells in the ordered set lie in the same row or in the same column.

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7. Tranportation problem

Remark. Any three or more adjacent cells in the loop do not lie in the same row or in the same column.

The loop is usually represented in the form of a closed broken line connecting the vertex of the loop located in the cells of the table.

Example. Let us solve the transportation problem by the distribution method using the initial basic plan obtained by the NWCR.

Solution.

To build new basic plan we select empty cell with the largest positive net evaluation ( A2B1 ) and form a loop connecting the selected cell and the occupied (basic) cells.

We get: A2B1 , A1B1 , A1B2 , A2B2 .

At each vertex of the loop we write the transportation value and alternating signs «+» or «–» beginning with «+» in the empty cell:

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7. Tranportation problem

Let us perform the net cost change. First we identify the minimum transportation value marked with symbol «–»:

∆ = min(50,70) =50

Then we subtract ∆ from the vertices marked with «–»

and add ∆ to all vertices with the sign «+»:

The cell A2B2 gets empty and the cell gets occupied.

A2B1 :

z(X) =20 11+60 5 +50 1+120 5 +60 7 +90 10 =2490 <2940. u1 = 0 v1 =11,v2 =5 ; v1 =11 u2 = −10 v3 =15 u2 = −10 v3 =15 u3 = −8 v4 =18 .

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7. Tranportation problem

Net evaluations:

∆13 =11 > 0 , ∆14 =16 > 0, ∆22 = −9 < 0, ∆24 = −1 < 0, ∆31 = −6 < 0, ∆32 = −11 < 0.

We select empty cell ( A1B4 ) with the largest positive net evaluation and form a loop: A1B4 , A3B4 , A3B3 , A2B3 ,

A2B1 , A1B1 .

∆ = min{20,120,90}=20

A1B4 ↔ A1B1 .

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7. Tranportation problem

z(X) =2170 <2490.

u1 = 0 v2 =5,v4 =2 ; v4 =2 u3 = 8 v3 = −1 u2 = 6 v1 = −5 .

Net evaluations:

∆11 = −16 < 0 , ∆13 = −5 < 0, ∆22 =7 > 0, ∆24 = −1 < 0, ∆31 = −6 < 0, ∆32 =2 > 0 .

We select A2B2 A2B2 , A1B2 , A1B4 , A3B4 , A3B3 ,

A2B3 :

∆ = min{60,70,100}= 60

A1B2 ↔ A2B2

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