Transportation Problems MHA 6350. Medical Supply Transportation Problem A Medical Supply company...

Transportation Problems

MHA 6350

Medical Supply Transportation Problem

• A Medical Supply company produces catheters in packs at three productions facilities.

• The company ships the packs from the production facilities to four warehouses.

• The packs are distributed directly to hospitals from the warehouses.

• The table on the next slide shows the costs per pack to ship to the four warehouses.

Medical Supply

Seattle New York Phoenix MiamiFROMPLANT

Juarez $19 $ 7 $ 3 $21Seoul 15 21 18 6Tel Aviv 11 14 15 22

TO WAREHOUSE

CapacityJuarez 100Seoul 300Tel Aviv 200

DemandSeattle 150New York 100Phoenix 200Miami 150

Source: Adapted from Lapin, 1994

J Xjs Xjn Xjp Xjm 100

S N P M

Xss Xsn Xsp Xsm

Xts Xtn Xtp Xtm

150 100 200 150

S 300

T 200

WarehouseDemand 600

TO WAREHOUSEPlant

CapacityFrom Plant

Number of constraints = number of rows + number of columns

Total plant capacity must equal total warehouse demand.Although this may seem unrealistic in real world application, it is possible to construct any transportation problem using this model.


S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

Northwest Corner Method

Begin with a blank shipment schedule. Note the shipping costs in the upper right hand corner of each cell.


S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11


100

Start in the upper left-hand corner, “northwest corner” of the schedule and place the largest amount of capacity and demand available in that cell. Seattle demands 150 and Jaurez has a capacity of 100.


S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11


100

Since Juarez capacity is depleted move down to repeat the process for the Seoul to Seattle cell. Seoul has sufficient capacity but Seattle can only take another 50 packs of demand.

50


S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11


100

50

Now move to the next cells to the right and assign capacity for Seoul to warehouse demand until depleted. Then move down to the Tel Aviv row and repeat the process.

100 150

50 150


S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11


The previous slides show the process of satisfying all constraints and allows us to begin with a starting feasible solution. Multiply the quantity in each cell by the cost.

1900

2100750 2700

750 3300

1900

750

2100

2700

750

3300

C =11,500


rj = 0

ks = 19

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

a

b

For non empty cells: cij = ri+ kj

Assign zero as the row number for the first row.

19 = (0) + ks


rj = 0

rs = -4

ks = 19

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

b



15 = rs + 19rs = -15 + 19 = -4 Note: Always use the newest r value to compute the next k.


rj = 0

rs = -4

ks = 19 kp = 22

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

*c

a

b d



18 =-4 + kp

18 + 4 = kp

= 22

Skip cell SN, mark it * for later and move on to cell SP .


rj = 0

rs = -4

rt = -7

ks = 19 kp = 22

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

*c

a

e

b d

For non empty cells: ctp = rt+ kp

Assign zero as the row number for the first row thenuse the newest r value to compute the next k.

15 = rt + 2215 - 22 = rt

= -7


rj = 0

rs = -4

rt = -7

ks = 19 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

*c

a

e

b d f



22 = -7 + km

22 + 7 = km

= 29


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

*c

a

e

b g d f



21= -4 + kn

21 + 4 = kn

= 25



rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f

Next calculate empty cells using: cij - ri - kj

JN = 7 – 0 – 25 = -18

-18

Improvement Difference >>


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


JP = 3 – 0 – 22 = -19

-18 -19



rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


JM = 21 – 0 – 29 = -8

-18 -19 -8



rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


SM = 6 – (-4) – 29 = -19

-18 -19 -8

-19



rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


TS = 11 – (-7) – 19 = -1

-18

-19

-8

-1


-19


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


TN = 14 – (-7) – 25 = -4

-18

-19

-8

-1 -4


-19


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

c

a

e

b g d f


-18

-19

-8

-1 -4


-19


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

100

50 100 150

50 150

Next calculate the entering cell by finding the empty cell with the greatest absolute negative improvement difference.

-18 -19 -8

-1 -4

-19

Cells JP and SM are tied for the greatest improvement at $19 per pack. Break the tie and arbitrarily choose JP. JP becomes the entering cell. Place a + sign in cell JP

(+)(-)

(+) (-)

Note: Except for the entering cell all changes must involve nonempty cells.

100


rj = 0

rs = -4

rt = -7

ks = 19 kn = 25 kp = 22 km = 29

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

50 100

50 150

-18 -19 -8

-1 -4

-19

Continue around the closed loop until all tradeoffs are completed.

(+)(-)

(+) (-)

Note: Except for the entering cell all changes must involve nonempty cells.

100

150 50

Previous cost was $11,500 and the new is:

300 22502100

900750

3300C = $9,600


rj = 0

rs = 15

rt = 12

ks = 0 kn = 6 kp = 3 km = 10

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

50 100

50 150

119 11

-1 -4

-19(+)

(-)(+)

(-)

Note: The r and k values and the improvement difference values have changed.

100

150 50

Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference.>>>>>SM


rj = 0

rs = 15

rt = 12

ks = 0 kn = 6 kp = 3 km = 10

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

50 100

50 150

119 11

-1 -4

-19(+)

(-)(+)

(-)


100

150 50

Previous cost was $9,600, now the new is:

300 22502100

30015002200

C = $8,650

Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference.SM

50

100100


rj = 0

rs = -4

rt = 12

ks = 19 kn = 25 kp = 3 km = 10

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

50 100

-180 11

-20 -23

-19(+)

(-)(+)

(-)


100

150

Previous cost was $8,650, now the new is:

300 2250

90014001500

C = $6,350


50

100100100

150

0

kn = 2


rj = 0

rs = 16

rt = 12

ks = -1 kp = 3 km = 10

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

520 31

20

3 -1(+)

(-)(+)

(-)


100

150

$6,350

300 2250

90014001500

C = $6,350


100100

150

0

kn = 2


rj = 0

rs = 15

rt = 11

ks = 0 kp = 3 km = -9

S N P M

150 100 200 150

J 100

S 300

T 200

Demand 600

CapacityFrom

To

19 7 3 21

6

22

18

15

21

14

15

11

419 30

20

3

1

(+)

(-)(+)

(-)


100

50

$6,250

300 750

1800900

11001400

C = $6,250

Optimal Solution

100

100

150

100

kn = 3

In five iterations the shipping cost has moved from $11,500 to $6,250. There are no remaining empty cells with a negative value.

Transportation Problems MHA 6350. Medical Supply Transportation Problem A Medical Supply company...

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