The following special-case modifications to the linear programming formulation can be made: – Minimum shipping guarantees from i to j:

• Outline:
• – Transportation: starting basic feasible solution
• References: – Frederick Hillier and Gerald J. Lieberman.
• – Hamdy A. Taha. Operations Research: An Introduction. 8th Edition. Prentice-Hall, Inc , 2007.
• LP Formulation

  11 c

  > 0 for all i and j

  >= d j for each destination j i x ij

  Sx ij < s i for each source i j Sx ij

  , can be written as: Min SSc ij x ij i j s.t.

  The linear programming formulation in terms of the amounts shipped from the origins to the destinations, x ij

  1 s

  3 s

  2 d

  1 d

  23 d

  22 c

  21 c

  13 c

  12 c

  2 c

  PENELITIAN OPERASIONAL I (TIN 4109)

  2

  1

  1

  3

  3

  2

  2

  1

  1

  A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points.

  Transportation Description

  Introduction to Operations Research. 7th ed. The McGraw-Hill Companies, Inc, 2001.

  Lecture 12

  Lecture 12 TRANSPORTATION

2 SOURCES DESTINATIONS Transportation Problem

  Transportation Problem Example: BBC Building Brick Company (BBC) has orders for 80 tons of

• LP Formulation Special Cases

  bricks at three suburban locations as follows: Northwood -- 25 The following special-case modifications to tons, Westwood -- 45 tons, and Eastwood -- 10 tons. BBC has the linear programming formulation can be two plants, each of which can produce 50 tons per week. made:

  How should end of week shipments be made to fill the

• – Minimum shipping guarantees from i to j:

  above orders given the following delivery cost per ton: x > L ij ij

• – Maximum route capacity from i to j: Northwood Westwood Eastwood

  x < L ij ij

  Plant 1

  24

  30

  40

• – Unacceptable routes: Plant 2 30 40

  42 delete the variable Example: BBC Example: BBC

• LP Formulation • LP Formulation
• – Objective Function – Decision Variables Defined

  Minimize total shipping cost per week: x = amount shipped from plant i to suburb j ij

  Min 24x + 30x + 40x + 30x + 40x + 42x

  11

  12

  13

  21

  22

  23

  where i = 1 (Plant 1) and 2 (Plant 2)

• – Constraints

  j = 1 (Northwood), 2 (Westwood), s.t. x + x + x < 50 (Plant 1 capacity)

  11

  12

  13

  and 3 (Eastwood) + x + x < 50 (Plant 2 capacity) x

  21

  22

  23

  x + x >= 25 (Northwood demand)

  11

  21

  x + x >= 45 (Westwood demand)

  12

  22

  x + x >= 10 (Eastwood demand)

  13

  23

  all x > 0 (Non-negativity) ij

  1. Decision Variable: Powerco has three electric power plants that supply the electric Since we have to determine how much electricity is sent from each needs of four cities. plant to each city;

  Solution Exercise….

• The associated supply of each plant and demand of each city is

  X = Amount of electricity produced at plant i and sent to city j ij given in the table as follows:

  X = Amount of electricity produced at plant 1 and sent to city 4

  14

• The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.

  2. Objective Function: From To

  Since we want to minimize the total cost of shipping from plants to City 1 City 2 City 3 City 4 Supply cities;

  (Million kwh) Minimize Z = 8X +6X +10X +9X +9X +12X +13X +7X

  11

  12

  13

  14

  21

  22

  23

  24 $8 $6 $10 $9

  35 Plant 1

• 14X +9X +16X +5X

  31

  32

  33

  34 $9 $12 $13 $7

  50 Plant 2 $14 $9 $16 $5

  40 Plant 3

  45

  20

  30

  30 Demand (Million kwh)

• 6X
• 10X
• 9X
• 9X
• 12X
• 13X
• 7X
• X
• X
• X
• 14X
• 9X
• 16X
• 5X
• X
• X
• X

  31

  23

  11

  12

  13

  14

  21

  22

  23

  24

  X

  24 <= 50

  21

  22

  Since each supply point has a limited production capacity; Since each supply point has a limited production capacity;

  X

  14 <= 35

  13

  12

  11

  X

  3. Supply Constraints

  32

  33

  34 S.T.:

  Min Z = 8X

• X
• X
• X

• X
• X
• X

• X
• X
• X
• X
• X
• X
• X
• X
• X
• X
• X
• X

  12

  12

  13

  X

  32 >= 20

  22

  X

  33 >= 30

  31 >= 45 (Demand Constraints)

  21

  11

  X

  34 <= 40

  33

  23

  X

  31

  Methods to find the BFS for a balanced Transportation Problem There are three basic methods:

  11 value can not be greater than minimum of this 2 values).

  11 to a larger number, will be the demand of demand point 1 and the supply of supply point 1. Your x

  11 as large as possible (here the limitations for setting x

  Vogel’s Method To find the bfs by the NWC method: Begin in the upper left (northwest) corner of the transportation tableau and set x

  2. Minimum Cost Method 3.

  1. North West Corner Method

  The reason for that is, if a set of decision variables (x ij ’s) satisfy all but one constraint, the values for x ij ’s will satisfy that remaining constraint automatically.

  14

  Transportation Problem Unlike other Linear Programming problems, a balanced TP with m supply points and n demand points is easier to solve, although it has m + n equality constraints.

  1

     n j j j m i i i d s

     

  Balanced Transportation Problem If Total supply equals to total demand, the problem is said to be a balanced transportation problem:

  34 >= 30 Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

  24

  32

  X

  22

  X

  32 >= 20

  X

  13

  23

  33 >= 30

  X

  14

  24

  34 >= 30

  X

  Xij >= 0 (i= 1,2,3; j= 1,2,3,4) LP Formulation of Powerco’s Problem

  31 >= 45

  21

  11

  4. Demand Constraints Solution

  24 <= 50

  34 <= 40

  23

  22

  21

  X

  14 <= 35 (Supply Constraints)

  13

  12

  11

  X

  31

  32

  33

• X
• X
• X
• X

• X
• X
• X
• X
• X
• X

5. Sign Constraints

1 Finding Basic Feasible Solution for

  After we check the east and south cells, we saw that we can go east

  According to the explanations in the previous slide we can (meaning supply point 1 still has capacity to fulfill some demand). set x =3 (meaning demand of demand point 1 is satisfied

  11 by supply point 1).

  5

  3

  2 X

  6

  6

  2

  2

  3

  5

  2

  3 X

  3

  2

  3

  3

  2

  3

  2 X

  6

  3

  3

  2

  2 X

  5

  2

  3 X

  X

  2

  3 After applying the same procedure, we saw that we can go

  Finally, we will have the following bfs, which is: south this time (meaning demand point 2 needs more supply x =3, x =2, x =3, x =2, x =1, x =2 by supply point 2).

  11

  12

  22

  23

  24

  34

  3

  2 X

  3

  2

  1

  3

  2 X

  2

  3

  2

  1 X

  X X

  X

  3

  3

  2 X

  2 X

  X X

  X X

  3

  2

  1 X

  2 X

  X X

  2

2. Minimum Cost Method The Steps: The Northwest Corner Method dos not utilize shipping costs.

  It can yield an initial BFS easily but the total shipping cost may

  1. First, We look for the cell with the minimum cost of shipping in the overall transportation tableau. be very high.

  2. We should cross out row i and column j and reduce the supply The minimum cost method uses shipping costs in order come or demand of the noncrossed-out row or column by the value up with a BFS that has a lower cost. of Xij.

  3. Then we will choose the cell with the minimum cost of shipping from the cells that do not lie in a crossed-out row or column and we will repeat the procedure. An example for Minimum Cost Method Step 1: Select the cell with minimum cost.

  2

  6 Step 5: Find the new cell with minimum shipping cost and cross-

  8

  2

  5

  3

  1

  2

  5

  6

  5

  3

  2

  out column 1

  4

  8

  5 X

  15

  X

  6 X

  4

  8

  3

  8

  2

  5

  3

  1

  2

  3

  4

  6

  3

  X X

  6 X

  X

  4 X

  5

  6

  4

  8

  3

  8

  2

  5

  1

  6

  2

  5

  6

  5

  3

  2

  out column 3

  6 Step 6: Find the new cell with minimum shipping cost and cross-

  4

  X

  10 X

  X

  5 X

  5

  5

  3

  8

  3

  8

  5

  3

  1

  2

  6

  5

  3

  2

  6 Step 2: Cross-out column 2

  4

  12

  4

  15

  10

  5

  6

  4

  8

  3

  5

  3

  1

  2

  6

  5

  8

  6

  3

  2

  2

  out row 1

  6 Step 4: Find the new cell with minimum shipping cost and cross-

  4

  10 X

  15

  5 X

  6

  4

  8

  3

  8

  5

  12 X

  3

  1

  2

  6

  5

  3

  2

  out row 2

  15 Step 3: Find the new cell with minimum shipping cost and cross-

  2

  5

  6

  4

  6 Step 7: Finally assign 6 to last cell. The bfs is found as: X =5,

  11

  3. Vogel’s Method X =2, X =8, X =5, X =4 and X =6

  21

  22

  31

  33

  34

  1. Begin with computing each row and column a penalty. The penalty will be equal to the difference between the two smallest

  2

  3

  5

  6 shipping costs in the row or column.

  X

  5 2. Identify the row or column with the largest penalty.

  2

  1

  3

  5

  3. Find the first basic variable which has the smallest shipping cost

  X

  2

  8 in that row or column.

  3

  8

  4

  6

  4. Then assign the highest possible value to that variable, and cross-

  X

  out the row or column as in the previous methods. Compute new

  5

  4

  6 penalties and use the same procedure.

  X X

  X X

  Step 2: Identify the largest penalty and assign the highest An example for Vogel’s Method Step 1: Compute the penalties. possible value to the variable.

  Supply Row Penalty Supply Row Penalty

  6

  7

  8

  6

  7

  8 10 7-6=1 5 8-6=2

  5

  15

  80

  78

  15

  80

  78 15 78-15=63 15 78-15=63

  Demand

  15

  5

  5 Demand

  15 X

  5 Column Penalty 15-6=9 80-7=73 78-8=70 Column Penalty 15-6=9 _ 78-8=70

  Step 3: Identify the largest penalty and assign the highest Step 4: Identify the largest penalty and assign the highest possible value to the variable. possible value to the variable.

  Supply Row Penalty Supply Row Penalty

  6

  7

  8

  6

  7

  8 _

  X _

  5

  5

  5

  5

  15

  80

  78

  15

  80

  78 15 _ 15 _

  Demand

  15 X

  X Demand

  15 X

  X Column Penalty 15-6=9 _ _ Column Penalty _ _ _ Solve the following transportation problem: Lecture 13

  15 c ij matrix

  2

  7

  5

  7 d j

  2

  5

  3 Origin Destination c ij matrix

  Exercise Bazaraa Chapter 10:

  1

  2

  3 4 s i

  1

  7 2 -1

  10

  4

  2

  3

  2

  3

  30

  3

  2

  1

  3

  4

  25 d j

  10

  15

  25

  1

  3

  Step 5: Finally the bfs is found as X

  5

  11

  =0, X

  12

  =5, X

  13

  =5, and X

  21

  =15

  Supply Row Penalty

  6

  7

  8

  5

  15

  1

  80

  78

  15 Demand Column Penalty _ _ _

  _ _

  X X

  X X

  X Exercise

  Bazaraa Chapter 10: Solve the following transportation problem:

  1

  2

  3 s i

  1

  5

  4

• – Preparation
• Materi:
• – Transportasi: Optimum Solution