Management Science Chapter 5 Transportation, Assignment, and Transshipment Problems - A11.4801i Chapter 5 Transportation Problem

1 Slide

Management Science Management Science

Chapter 5 Chapter 5 Transportation, Assignment, and Transportation, Assignment, and Transshipment Problems Transshipment Problems

2 Slide

and functions (e.g. costs, supplies, demands, etc.) associated with the arcs and/or nodes. etc.) associated with the arcs and/or nodes.

and PERT/CPM problems (Chapter 10) are all

tree, and maximal flow problems (Chapter 9)

well as the shortest route, minimal spanning

transshipment problems of this chapter, as

Transportation, assignment, and



and functions (e.g. costs, supplies, demands,

Transportation, Assignment, and

represented by a set of nodes, a set of arcs,

is one which can be

network model



Transshipment Problems Transshipment Problems

and PERT/CPM problems (Chapter 10) are all examples of network problems. examples of network problems.

3 Slide

Transportation, Assignment, and

Transshipment Problems Transshipment Problems

 Each of the three models of this chapter

Each of the three models of this chapter (transportation, assignment, and transshipment

(transportation, assignment, and transshipment

models) can be formulated as linear programs

and solved by general purpose linear and solved by general purpose linear programming Algorithms (simplex method). programming Algorithms (simplex method).



For each of the three models, if the right-hand

side of the linear programming formulations are

all integers, the optimal solution will be in terms all integers, the optimal solution will be in terms of integer values for the decision variables. of integer values for the decision variables.

 However, there are many computer packages

However, there are many computer packages (including

(including The Management Scientist, DS, QSB

The Management Scientist, DS, QSB )

) which contain separate computer codes for which contain separate computer codes for these models which take advantage of their these models which take advantage of their network structure. network structure.

4 Slide

Transportation Problem

 The

The transportation problem transportation problem seeks to seeks to minimize the total shipping costs of minimize the total shipping costs of transporting goods from transporting goods from m m origins or origins or sources (each with a supply sources (each with a supply s s i i ) to

) to n n destinations (each with a demand destinations (each with a demand d d j j ),

), when the unit shipping cost from source, when the unit shipping cost from source, i i

, to a destination, , to a destination, j j

, is , is c c ij ij .

 The

The network representation network representation for a for a transportation problem with two sources transportation problem with two sources and three destinations is given on the and three destinations is given on the next slide. next slide.

5 Slide

DESTINATIONS

2 c c ₁₁ ₁₁ c c ₁₂ ₁₂ c c ₁₃ ₁₃ c c ₂₁ ₂₁ c c ₂₂ ₂₂ c c ₂₃ ₂₃ d d ₁ ₁ d d ₂ ₂ d d ₃ ₃ s s ₁ ₁ s ₂ SOURCES SOURCES

Transportation Problem

Network Representation



DESTINATIONS

6 Slide

Transportation Problem

 LP Formulation

LP Formulation The linear programming formulation in terms of the

The linear programming formulation in terms of the amounts shipped from the sources to the destinations, amounts shipped from the sources to the destinations, x x _{ij ij ,} , can be written as: can be written as:

Min Min

  c c _ij _{ij x} x _ij _{ij (total transportation cost)} (total transportation cost) i j i j s.t. s.t.

  x x _{ij ij <} < s s _{i i for each source} for each source i i

(supply constraints) (supply constraints) j j

  x x _ij _{ij =} = d d _j _{j for each destination} for each destination j (demand j (demand constraints) constraints) i i x x _{ij ij >} > for all for all i i and and j (nonnegativity constraints) j (nonnegativity constraints)

7 Slide

Transportation Problem Transportation Problem



To solve the transportation problem by its special

purpose algorithm, it is required that the sum of purpose algorithm, it is required that the sum of

the supplies at the sources equal the sum of the

the supplies at the sources equal the sum of the demands at the destinations. If the total supply is

demands at the destinations. If the total supply is

greater than the total demand, a dummy greater than the total demand, a dummy destination is added with demand equal to the destination is added with demand equal to the excess supply, and shipping costs from all sources excess supply, and shipping costs from all sources are zero. Similarly, if total supply is less than are zero. Similarly, if total supply is less than total demand, a dummy source is added. total demand, a dummy source is added.

 When solving a transportation problem by its

When solving a transportation problem by its

special purpose algorithm, unacceptable shipping

routes are given a cost of + routes are given a cost of +

M M (a large number).

(a large number).

8 Slide

Demand

25 S1 S1 S2 S2 D3 D3 D2 D2 D1 D1

Transportation Problem Transportation Problem

cell represents a shipping route (which is an



transportation tableau

is given below. Each

cell represents a shipping route (which is an

arc on the network and a decision variable in

the LP formulation), and the unit shipping

costs are given in an upper right hand box in

costs are given in an upper right hand box in the cell. the cell.

Supply Supply

30 X
30 X

₁₂

^{12<20 X
20 X
₁₃ ^{13<30 X
30 X
₂₁ ^{21<40X
40X
₂₂ ^{22<35X
35X
₂₃ ²³ S.t.X
X
₂₁ ²¹
9 Slide

Problem formulation
Problem formulation

The LP model for this problem is as follows: The LP model for this problem is as follows:
Min Z = 15 X Min Z = 15 X ₁₁ ¹¹
X X ₁₁ _{11 + X} + X ₁₂ _{12 + X} + X ₁₃ _{13 ≤ 50} ≤ 50 Supply constraints Supply constraints
X X ₂₁ _{21 + X} + X ₂₂ _{22 + X} + X ₂₃ _{23 ≤ 30} ≤ 30
X X ₁₁ ¹¹
= 25 = 25
X X ₁₂ _{12 + X} + X ₂₂ _{22 = 45} = 45
X X ₁₃ _{13 + X} + X ₂₃ _{23 = 10 demand constraints} = 10 demand constraints
X X ₁₁ ¹¹ , …, X
, …, X ₂₃ ²³ 

Phase I -- Obtaining an initial feasible solution
Phase II -- Moving toward optimality
10 Slide
Transportation Problem Transportation Problem
 The transportation problem is solved in two
The transportation problem is solved in two phases: phases:
Phase I -- Obtaining an initial feasible solution
Phase II -- Moving toward optimality 
In Phase I, the Minimum-Cost Procedure can be In Phase I, the Minimum-Cost Procedure can be
used to establish an initial basic feasible solution
used to establish an initial basic feasible solution
without doing numerous iterations of the simplex
without doing numerous iterations of the simplex
method. method.
 In Phase II,
In Phase II, the Stepping Stone, by using the the Stepping Stone, by using the
MODI method for evaluating the reduced costs MODI method for evaluating the reduced costs may be used to move from the initial feasible may be used to move from the initial feasible solution to the optimal one. solution to the optimal one.
11 Slide

Initial Tableau
Initial Tableau

There are many method for finding the initial

There are many method for finding the initial
tableau for the transportation problem which

tableau for the transportation problem which
are:
are: 1.
1. Northwest corner Northwest corner 2.
2. Minimum cost of the row Minimum cost of the row 3.
3. Minimum cost of the column Minimum cost of the column 4.
4. Least cost Least cost 5.
5. Vogle’s approximation method Vogle’s approximation method 6.
6. Russell’s approximation method Russell’s approximation method
12 Slide
30
20
20
10
10
35
35
20
20
40
40
30
25
30
30 Demand
Demand
10
10
45
45
25
25 S1 S1 S2 S2 D3 D3 D2 D2 D1 D1
15
25
25
Northwest corner Northwest corner
(that is,

Northwest corner: Begin by selecting X
Northwest corner: Begin by selecting X ₁₁ _{11 (that is,} (that is,
start in the northwest corner of the transportation

start in the northwest corner of the transportation
tableau). Therefore, if X
tableau). Therefore, if X _ij _{ij was the last basic variable} was the last basic variable
(occupied cell) selected, then select X
(occupied cell) selected, then select X _ij+1 _ij+1
(that is,
move one column to the right) if source I has any
25

move one column to the right) if source I has any
supply remaining. Otherwise, next select X
supply remaining. Otherwise, next select X _{i+1 j} _{i+1 j}
(that
(that is, move one row down). is, move one row down).
Supply Supply
30
30
50
50
15 Total cost is $2275
Step 1:
Step 2:
30
30
35
35
20
20
40
40
30
10
30

30
Demand
10
45
25 S1 S1 S2 S2 D3 D3 D2 D2 D1 D1
15

15
Total cost is
30
10
13 Slide
Decrease the row and column availabilities by Decrease the row and column availabilities by this amount and remove from consideration all other this amount and remove from consideration all other cells in the row or column with zero availability/ cells in the row or column with zero availability/ demand. (If both are simultaneously reduced to 0, demand. (If both are simultaneously reduced to 0, assign an allocation of 0 to any other unoccupied cell in assign an allocation of 0 to any other unoccupied cell in the row or column before deleting both.) GO TO STEP 1. the row or column before deleting both.) GO TO STEP 1.
Transportation Algorithm Transportation Algorithm

Phase I - Minimum-Cost Method
Phase I - Minimum-Cost Method
Step 1:
Select the cell with the least cost. Assign to Select the cell with the least cost. Assign to this cell the minimum of its remaining row supply or this cell the minimum of its remaining row supply or remaining column demand. remaining column demand.
Step 2:
Supply Supply
15
30
30
50
50
25
25
15
$2225
Step 1:
Step 2:
14 Slide

Transportation Algorithm
Transportation Algorithm
 Phase II - Stepping Stone Method
Phase II - Stepping Stone Method
Step 1: For each unoccupied cell, calculate the
For each unoccupied cell, calculate the reduced cost by the MODI method described below. reduced cost by the MODI method described below.
Select the unoccupied cell with the most negative Select the unoccupied cell with the most negative reduced cost. (For maximization problems select reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) the unoccupied cell with the largest reduced cost.) If none, STOP.
If none, STOP.
Step 2: For this unoccupied cell generate a
For this unoccupied cell generate a stepping stone path by forming a closed loop with stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between
alternating horizontal and vertical lines between
them. them.
Determine the minimum allocation where a Determine the minimum allocation where a subtraction is to be made along this path. subtraction is to be made along this path.
Step 3:
allocation to all cells where subtractions are to be made along the stepping stone path. be made along the stepping stone path.
others occupied with 0's.)
0, make only one unoccupied; make the
0, make only one unoccupied; make the
(unoccupied). If more than one cell becomes
(unoccupied). If more than one cell becomes
stepping stone path now becomes 0
stepping stone path now becomes 0
(Note: An occupied cell on the
(Note: An occupied cell on the
allocation to all cells where subtractions are to
15 Slide
additions are to be made, and subtract this
additions are to be made, and subtract this
Add this allocation to all cells where
Add this allocation to all cells where
Step 3:
Phase II - Stepping Stone Method (continued)
Phase II - Stepping Stone Method (continued)


Transportation Algorithm
Transportation Algorithm
others occupied with 0's.) GO TO STEP 1.
Step 1:
Step 2:
by solving the relationship c c _ij _ij
cost =
), the reduced
), the reduced
, j j
,
For unoccupied cells ( i i
For unoccupied cells (
Step 3:
for occupied cells. occupied cells.
for
v v _j _j
= u u _i _i
=
by solving the relationship
Calculate the remaining u u _i _{i 's and} 's and v v _j _{j 's} 's
Calculate the remaining
Step 2:
Set u u ₁ _{1 = 0.} = 0.
Set
Step 1:
Associate a number, u u _{i i , with each row and} , with each row and v v _{j j} with each column. with each column.
Associate a number,
MODI Method (for obtaining reduced costs)
MODI Method (for obtaining reduced costs)


Transportation Algorithm
Transportation Algorithm
Step 3:
16 Slide
cost = c c _ij _{ij -} - u u _i _{i -} - v v _j _{j .} .
17 Slide
40
42
40
40
30
30
40 Plant 2 Plant 2
30
Example: BBC Example: BBC
30
Plant 1 24 Plant 1 24
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
How should end of week shipments be made to fill How should end of week shipments be made to fill the above orders given the following delivery cost the above orders given the following delivery cost per ton: per ton:
Building Brick Company (BBC) has orders for 80 Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as tons of bricks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 follows: Northwood -- 25 tons, Westwood -- 45
tons, and Eastwood -- 10 tons. BBC has two plants,
tons, and Eastwood -- 10 tons. BBC has two plants,
each of which can produce 50 tons per week. each of which can produce 50 tons per week.
42
18 Slide
45
Demand Supply Supply
50
50
50
50
20
20
10
10
45
30
25
25 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
24
24
24
30
30

Example: BBC
Example: BBC
42

Initial Transportation Tableau
Initial Transportation Tableau
Since total supply = 100 and total demand =
Since total supply = 100 and total demand =
80, a dummy destination is created with demand
80, a dummy destination is created with demand of 20 and 0 unit costs. of 20 and 0 unit costs.
42
42
42
40
30
40
40
40
40
40
40
40
30
30
30
24
Iteration 1:
Iteration 2:
Iteration 3:
Iteration 4:
19 Slide

Example: BBC
Example: BBC
 Least Cost Starting Procedure
Least Cost Starting Procedure
Iteration 1: Tie for least cost (0), arbitrarily select
Tie for least cost (0), arbitrarily select x x ₁₄ _{14 . Allocate 20. Reduce} . Allocate 20. Reduce s s ₁ _{1 by 20 to 30 and} by 20 to 30 and delete the Dummy column. delete the Dummy column.
Iteration 2: Of the remaining cells the least cost
Of the remaining cells the least cost is 24 for is 24 for x x ₁₁ _{11 . Allocate 25. Reduce}
. Allocate 25. Reduce
s s ₁ _{1 by 25 to 5} by 25 to 5 and eliminate the Northwood column. and eliminate the Northwood column.
Iteration 3: Of the remaining cells the least cost
Of the remaining cells the least cost is 30 for is 30 for x x ₁₂ _{12 . Allocate 5. Reduce the Westwood} . Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. column to 40 and eliminate the Plant 1 row.
Iteration 4: Since there is only one row with two
Since there is only one row with two cells left, make the final allocations of 40 and 10 cells left, make the final allocations of 40 and 10 to to x x ₂₂ ₂₂ and and x x ₂₃ ₂₃ , respectively.
, respectively.
20 Slide
10
30
30
30
30 Demand
Demand Supply Supply
50
50
50
50
20
20
10
30
45
45
25
25 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
24
24
24
30
30

Example: BBC
Example: BBC
10

Initial tableau
Initial tableau
25
25
5
5
20
20
40
40
10
30
42
42
42
42
40
40
40
40
40
40
40
40
24 Total transportation cost is $2770
MODI Method
3. Since
= = c c ₂ _{2 j j} for occupied cells in row 2, for occupied cells in row 2, then then
v v _{j j}
4. Since u u ₂ ₂
4. Since
2, 2, then then u u ₂ _{2 + 30 = 40, hence} + 30 = 40, hence u u ₂ _{2 = 10.} = 10.
= = c c _{i i} ₂ ₂ for occupied cells in column for occupied cells in column
v v ₂ ₂
3. Since u u _{i i}
2. Since u u ₁ _{1 +} + v v _j _{j =} = c c ₁ ₁ _j _{j for occupied cells in row 1,} for occupied cells in row 1, then then v v ₁ _{1 = 24,} = 24, v v ₂ _{2 = 30,} = 30,
v
v
₄ _{4 = 0.} = 0.
2. Since
= 0 = 0
1. Set u u ₁ ₁
1. Set
MODI Method
Iteration 1
 Iteration 1

Example: BBC
Example: BBC
21 Slide
10 + 10 + v v ₃ _{3 = 42, hence} = 42, hence v v ₃ _{3 = 32.} = 32.
Example: BBC Example: BBC
 Iteration 1
Iteration 1
MODI Method (continued)
MODI Method (continued)
Calculate the reduced costs (circled numbers
Calculate the reduced costs (circled numbers _{ij i j} + . - on the next slide) by c u v on the next slide) by c ij u i v j .
Unoccupied Cell Reduced Cost
Unoccupied Cell Reduced Cost
(1,3) 40 - 0 - 32 = 8
(1,3) 40 - 0 - 32 = 8
(2,1) 30 - 24 -10 = -4
(2,1) 30 - 24 -10 = -4
(2,4) 0 - 10 - 0 = -10
(2,4) 0 - 10 - 0 = -10
Since some of the reduced cost are negative, the current solution is not optimal.
Cell (2,4) has the most negative; therefore, it will be the basic variable that must be occupied in the next iteration.
22 Slide
24
10
40
40
40
40
40
30
30
30
30
30
30
30
30 v v _j _j u u _i _i
10
40
32
32
30
30
24
24 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2 Eastwood
Eastwood Westwood
Westwood Northwood
Northwood
24
24
24
40
40
8
>>8 42
23 Slide

Example: BBC
Example: BBC

Iteration 1 Tableau
Iteration 1 Tableau
25
25
25
25
5
5
5
5
20
20
20
20
42
42
42
4
10
>4 -4 10
40
10
10
40
40
10>410
40
10
Stepping Stone Method
24 Slide

Example: BBC
Example: BBC
 Iteration 1
Iteration 1
Stepping Stone Method The stepping stone path for cell (2,4) is (2,4), (1,4),
The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells are
(1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and
20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next hence reallocate 20 along this path. Thus for the next tableau: tableau: x x ₂₄ ₂₄
= 0 + 20 = 20 (0 is its current allocation) = 0 + 20 = 20 (0 is its current allocation) x x ₁₄ ₁₄
= 20 - 20 = 0 (blank for the next = 20 - 20 = 0 (blank for the next tableau) tableau) x x ₁₂ ₁₂
= 5 + 20 = 25 = 5 + 20 = 25 x x ₂₂ ₂₂
= 40 - 20 = 20 = 40 - 20 = 20 The other occupied cells remain the same.
The other occupied cells remain the same.
25 Slide
45
30
30 Demand
Demand Supply Supply
50
50
50
50
20
20
10
10
45
30
25
25 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
24
24
24
24 Total transportation cost is $2570 = 2770 – 10 (20) Reduced cost of cell (2,4)
30
30
Example: BBC Example: BBC
42
25
25
25
25
20
20

10

10
20
20
42
42
30
42
40
40
40
40
40
40
40
40
30
30
New quantity
MODI Method
= 0.
10 + 10 + v v ₃ _{3 = 42 or} = 42 or v v ₃ _{3 = 32; and, 10 +} = 32; and, 10 + v v ₄ _{4 =} =
4. Since u u ₂ _{2 +} + v v _j _{j =} = c c ₂ ₂ _j _{j for occupied cells in row 2, then} for occupied cells in row 2, then
4. Since
3. Since u u _{i i +} + v v ₂ _{2 =} = c c _{i i} ₂ _{2 for occupied cells in column 2,} for occupied cells in column 2, then then u u ₂ _{2 + 30 = 40, or} + 30 = 40, or
u
u
₂ _{2 = 10.} = 10.
3. Since
2. Since u u ₁ _{1 +} + v v _{j j =} = c c _{ij ij for occupied cells in row 1, then} for occupied cells in row 1, then v v ₁ _{1 = 24,} = 24, v v ₂ _{2 = 30.} = 30.
2. Since
1. Set u u ₁ ₁ = 0.
26 Slide
1. Set
's for this tableau.
's and 's and v v _{j j} 's for this tableau.
The reduced costs are found by calculating the the u u _{i i}
MODI Method The reduced costs are found by calculating
Iteration 2
 Iteration 2

Example: BBC
Example: BBC
0 or 0 or v v ₄ _{4 = -10.} = -10.
MODI Method (continued)
u _i _i
>
Reduced Cost
30 - 10 - 24 = -4
(2,1)
(2,1)
= 10
= 10
0 - 0 - (-10)
0 - 0 - (-10)
(1,4)
(1,4)
40 - 0 - 32 = 8
40 - 0 - 32 = 8
(1,3)
(1,3)
Reduced Cost
Unoccupied Cell
27 Slide
Unoccupied Cell
.
v v _j _j .
next slide) by c c _ij _ij
next slide) by
on the
on the
Calculate the reduced costs (circled numbers
Calculate the reduced costs (circled numbers
MODI Method (continued)
Iteration 2
Iteration 2


Example: BBC
Example: BBC
30 - 10 - 24 = -4 Since there is still negative reduced cost for cell (2,1), the solution is not optimal. Cell (2,1) must be occupied
24
10
40

40
40
40
40

30
30
30

30
30
30
30
30 v v _{j j} u u _i _i
10
40
36
36
30
30
24
24 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
24
24
24
40
40
25
Iteration 2 Tableau
25
25
25
42
25
25
20
25
Iteration 2 Tableau
25
10


Example: BBC
Example: BBC
8
10>8 <>10 8 20
42
42
42
4
>4
20
20
10
10
10
20
20
6
6
28 Slide
20
20
Stepping Stone Method
29 Slide

Example: BBC
Example: BBC
 Iteration 2
Iteration 2
Stepping Stone Method
The most negative reduced cost is = -4 determined

The most negative reduced cost is = -4 determined
by by x x ₂₁ _{21 . The stepping stone path for this cell is (2,1),} . The stepping stone path for this cell is (2,1),
(1,1),(1,2),(2,2). The allocations in the subtraction (1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus the new cells are 25 and 20 respectively. Thus the new solution is obtained by reallocating 20 on the stepping solution is obtained by reallocating 20 on the stepping stone path. Thus for the next tableau: stone path. Thus for the next tableau: x x ₂₁ ₂₁
= 0 + 20 = 20 (0 is its current allocation) = 0 + 20 = 20 (0 is its current allocation) x x ₁₁ ₁₁
= 25 - 20 = 5 = 25 - 20 = 5 x x ₁₂ ₁₂
= 25 + 20 = 45 = 25 + 20 = 45 x x ₂₂ ₂₂
= 20 - 20 = 0 (blank for the next tableau) = 20 - 20 = 0 (blank for the next tableau) The other occupied cells remain the same.
The other occupied cells remain the same.
30 Slide
10
30
30 Demand
Demand Supply Supply
50
50
50
50
20
20
10
45
30
45
25
25 Dummy Dummy
Plant 1 Plant 1
Plant 2 Plant 2
Eastwood Eastwood
Westwood Westwood
Northwood Northwood
24
24
24
30
30
Example: BBC Example: BBC
42
5
5
45
45
20
20
10
10
20
20
42
42
30
42
40
40
40
40
40
40
40
40
30
30
24 Total cost is $2490 = 2570 - 4(20)
MODI Method
3. Since u u _i _{i +} + v v ₁ _{1 =} = c c _i _i ₁ _{1 for occupied cells in column 2,} for occupied cells in column 2, then then u u ₂ ₂
= 0 or = 0 or v v ₄ ₄
= 36, and 6 + = 36, and 6 + v v ₄ ₄
= 42 or = 42 or v v ₃ ₃
6 + 6 + v v ₃ ₃
= = c c ₂ _{2 j j} for occupied cells in row 2, then for occupied cells in row 2, then
v v _{j j}
4. Since u u ₂ ₂
4. Since
= 6.
u u
₂
₂
= 6.
3. Since
2. Since u u ₁ _{1 +} + v v _j _{j =} = c c ₁ ₁ _j _{j for occupied cells in row 1, then} for occupied cells in row 1, then v v ₁ _{1 = 24 and} = 24 and v v ₂ _{2 = 30.} = 30.
2. Since
1. Set u u ₁ _{1 = 0} = 0
1. Set
's 's and and v v _j _{j 's for this tableau.} 's for this tableau.
The reduced costs are found by calculating the the u u _{i i}
MODI Method The reduced costs are found by calculating
Iteration 3
 Iteration 3

Example: BBC
Example: BBC
24 = 30 or
24 = 30 or
6.
31 Slide
= = -6.
MODI Method (continued)
Reduced Cost
40 - 6 - 30 = 4
(2,2)
(2,2)
6
6
0 - 0 - (-6) =
0 - 0 - (-6) =
(1,4)
(1,4)
40 - 0 - 36 = 4
40 - 0 - 36 = 4
(1,3)
(1,3)
Reduced Cost
Unoccupied Cell
32 Slide
Calculate the reduced costs (circled numbers

Example: BBC
Example: BBC

Iteration 3
Iteration 3
MODI Method (continued)
Calculate the reduced costs (circled numbers
on the
Unoccupied Cell
on the
next slide) by
next slide) by c c _ij _ij
u u _i _i
v v _j _j .
.
40 - 6 - 30 = 4 Since all the reduced cost are nonnegative, the current solution is optimal
4
<>6 <>>>4 6
40
40

40
40
40
40

30
30
30
30
30
30
30}}}}

Management Science Chapter 5 Transportation, Assignment, and Transshipment Problems - A11.4801i Chapter 5 Transportation Problem

Dokumen yang terkait

Summary Chapter 5

Chapter 7: Project Cost Management

4 Chapter 5 – Arena Basics.pdf

Chapter 5 Lets Visit Niagara Falls

RPP no 5.4 Chapter 5 binatang, bangunan publik dan evaluasi

RPP no 5.3 Chapter 5 benda di sekolah

Slides for Chapter 13: Transactions and Concurrency Control - Chapter 13 slides

5. Flexural Analysis and Design of Beams 5.1. Reading Assignment - Chapter 5

Chapter 5 Introduction to Microsoft Excel

A11.4801i Chapter 1 Linear Programming

Dukungan

Links

Management Science Chapter 5 Transportation, Assignment, and Transshipment Problems - A11.4801i Chapter 5 Transportation Problem

Dokumen yang terkait

Summary Chapter 5

Chapter 7: Project Cost Management

4 Chapter 5 – Arena Basics.pdf

Chapter 5 Lets Visit Niagara Falls

RPP no 5.4 Chapter 5 binatang, bangunan publik dan evaluasi

RPP no 5.3 Chapter 5 benda di sekolah

Slides for Chapter 13: Transactions and Concurrency Control - Chapter 13 slides

5. Flexural Analysis and Design of Beams 5.1. Reading Assignment - Chapter 5

Chapter 5 Introduction to Microsoft Excel

A11.4801i Chapter 1 Linear Programming

Dokumen yang Anda mencari sudah siap untuk unduhkan