Transportation Problems - PowerPoint PPT Presentation

1 / 30
About This Presentation
Title:

Transportation Problems

Description:

Transportation Problems MHA 6350 Medical Supply Transportation Problem A Medical Supply company produces catheters in packs at three productions facilities. – PowerPoint PPT presentation

Number of Views:88
Avg rating:3.0/5.0
Slides: 31
Provided by: facultyWe5
Category:

less

Transcript and Presenter's Notes

Title: Transportation Problems


1
Transportation Problems
  • MHA 6350

2
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.

3
Medical Supply
Source Adapted from Lapin, 1994
TO WAREHOUSE
FROM PLANT
Seattle New York Phoenix Miami
Juarez 19 7 3
21 Seoul 15 21 18 6 Tel
Aviv 11 14 15 22
Capacity Juarez 100 Seoul 300 Tel Aviv 200
Demand Seattle 150 New York 100 Phoenix 200 Miam
i 150
4
Source Adapted from Lapin, 1994
TO WAREHOUSE
Plant Capacity
From Plant
S
N
P
M
J
Xjs
Xjn
Xjp
Xjm
100
Xss
Xsn
Xsp
Xsm
S
300
Xts
Xtn
Xtp
Xtm
T
200
150
100
200
150
Warehouse Demand
600
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.
5
Source Adapted from Lapin, 1994
Northwest Corner Method
S
N
P
M
Capacity
To
From
19
7
3
21
6
18
21
15
22
15
14
11
150
100
200
150
Demand
600
Begin with a blank shipment schedule. Note the
shipping costs in the upper right hand corner of
each cell.
6
Source Adapted from Lapin, 1994
Northwest Corner Method
S
N
P
M
Capacity
To
From
19
7
3
21
100
6
18
21
15
22
15
14
11
150
100
200
150
Demand
600
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.
7
Source Adapted from Lapin, 1994
Northwest Corner Method
S
N
P
M
Capacity
To
From
19
7
3
21
100
6
18
21
15
50
22
15
14
11
150
100
200
150
Demand
600
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.
8
Source Adapted from Lapin, 1994
Northwest Corner Method
S
N
P
M
Capacity
To
From
19
7
3
21
100
6
18
21
15
50
100
150
22
15
14
11
50
150
150
100
200
150
Demand
600
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.
9
Source Adapted from Lapin, 1994
Northwest Corner Method
S
N
P
M
Capacity
To
From
1900 750 2100 2700 750 3300 C 11,500
19
7
3
21
1900
6
18
21
15
2100
750
2700
22
15
14
11
750
3300
150
100
200
150
Demand
600
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.
10
Source Adapted from Lapin, 1994
For non empty cells cij ri kj
Assign zero as the row number for the first row.
rj 0
a
ks 19
19 (0) ks
b
11
Source Adapted from Lapin, 1994
For non empty cells cij ri kj
Assign zero as the row number for the first row.
rj 0
a
c
rs -4
ks 19
15 rs 19 rs -15 19 -4
b
Note Always use the newest r value to compute
the next k.
12
Source Adapted from Lapin, 1994
For non empty cells cij ri kj
Assign zero as the row number for the first row.
rj 0
a
c
rs -4
ks 19
kp 22
18 -4 kp 18 4 kp 22
b
d
Skip cell SN, mark it for later and move on to
cell SP .
13
Source Adapted from Lapin, 1994
For non empty cells ctp rt kp
Assign zero as the row number for the first row
then use the newest r value to compute the next k.
rj 0
a
c
rs -4
rt -7
e
ks 19
kp 22
15 rt 22 15 - 22 rt -7
b
d
14
Source Adapted from Lapin, 1994
For non empty cells cij ri kj
Assign zero as the row number for the first row
then use the newest r value to compute the next k.
rj 0
a
c
rs -4
rt -7
e
ks 19
kp 22
km 29
22 -7 km 22 7 km 29
b
d
f
15
Source Adapted from Lapin, 1994
For non empty cells cij ri kj
Assign zero as the row number for the first row
then use the newest r value to compute the next k.
rj 0
a
c
rs -4
rt -7
e
ks 19
kn 25
kp 22
km 29
21 -4 kn 21 4 kn 25
b
g
d
f
16
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
JN 7 0 25 -18
Improvement Difference gtgt
rj 0
a
-18
c
rs -4
rt -7
e
ks 19
kn 25
kp 22
km 29
b
g
d
f
17
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
JP 3 0 22 -19
Improvement Difference gtgt
rj 0
a
-18
-19
c
rs -4
rt -7
e
ks 19
kn 25
kp 22
km 29
b
g
d
f
18
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
JM 21 0 29 -8
Improvement Difference gtgt
rj 0
a
-18
-19
-8
c
rs -4
rt -7
e
ks 19
kn 25
kp 22
km 29
b
g
d
f
19
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
SM 6 (-4) 29 -19
Improvement Difference gtgt
rj 0
a
-18
-19
-8
c
rs -4
-19
rt -7
e
ks 19
kn 25
kp 22
km 29
b
g
d
f
20
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
TS 11 (-7) 19 -1
Improvement Difference gtgt
rj 0
a
-18
-8
-19
c
rs -4
-19
rt -7
e
-1
ks 19
kn 25
kp 22
km 29
b
g
d
f
21
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
TN 14 (-7) 25 -4
Improvement Difference gtgt
rj 0
a
-18
-8
-19
c
rs -4
-19
rt -7
e
-1
-4
ks 19
kn 25
kp 22
km 29
b
g
d
f
22
Source Adapted from Lapin, 1994
Next calculate empty cells using cij - ri -
kj
Improvement Difference gtgt
rj 0
a
-18
-8
-19
c
rs -4
-19
rt -7
e
-1
-4
ks 19
kn 25
kp 22
km 29
b
g
d
f
23
Source Adapted from Lapin, 1994
Next calculate the entering cell by finding the
empty cell with the greatest absolute negative
improvement difference.
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
S
N
P
M
Capacity
To
From
19
7
3
21
rj 0
100
100
(-)
()
-18
-19
-8
6
18
21
15
50
100
150
rs -4
()
(-)
-19
22
15
14
11
50
150
rt -7
-1
-4
150
100
200
150
Demand
600
ks 19
kn 25
kp 22
km 29
Note Except for the entering cell all changes
must involve nonempty cells.
24
Source Adapted from Lapin, 1994
Continue around the closed loop until all
tradeoffs are completed.
Previous cost was 11,500 and the new is
S
N
P
M
Capacity
To
300 2250 2100 900 750 3300 C 9,600
From
19
7
3
21
rj 0
100
(-)
()
-18
-19
-8
6
18
21
15
50
50
150
100
rs -4
()
(-)
-19
22
15
14
11
50
150
rt -7
-1
-4
150
100
200
150
Demand
600
ks 19
kn 25
kp 22
km 29
Note Except for the entering cell all changes
must involve nonempty cells.
25
Source Adapted from Lapin, 1994
Begin another iteration choosing the empty cell
with the greatest absolute negative improvement
difference. gtgtgtgtgtSM

S
N
P
M
Capacity
To
From
19
7
3
21
rj 0
100
1
19
11
6
18
21
15
50
50
150
100
()
(-)
rs 15
-19
22
15
14
11
150
50
rt 12
(-)
()
-1
-4
150
100
200
150
Demand
600
ks 0
kn 6
kp 3
km 10
Note The r and k values and the improvement
difference values have changed.
26
Source Adapted from Lapin, 1994
Begin another iteration choosing the empty cell
with the greatest absolute negative improvement
difference. SM

Previous cost was 9,600, now the new is
S
N
P
M
Capacity
To
300 2250 2100 300 1500 2200 C 8,650
From
19
7
3
21
rj 0
100
1
19
11
6
18
21
15
50
50
50
150
100
()
(-)
rs 15
-19
22
15
14
11
150
100
100
50
rt 12
(-)
()
-1
-4
150
100
200
150
Demand
600
ks 0
kn 6
kp 3
km 10
Note The r and k values and the improvement
difference values have changed.
27
Source Adapted from Lapin, 1994
Begin another iteration choosing the empty cell
with the greatest absolute negative improvement
difference. SM

Previous cost was 8,650, now the new is
S
N
P
M
Capacity
To
300 2250 900 1400 1500 C 6,350
From
19
7
3
21
rj 0
100
-18
0
11
6
18
21
15
50
100
150
50
150
()
(-)
rs -4
-19
22
15
14
11
0
100
100
100
(-)
()
rt 12
-20
-23
150
100
200
150
Demand
600
ks 19
kn 25
kp 3
km 10
kn 2
Note The r and k values and the improvement
difference values have changed.
28
Source Adapted from Lapin, 1994
Begin another iteration choosing the empty cell
with the greatest absolute negative improvement
difference. SM

6,350
S
N
P
M
Capacity
To
300 2250 900 1400 1500 C 6,350
From
19
7
3
21
rj 0
100
5
20
31
6
18
21
15
150
150
()
(-)
rs 16
3
-1
22
15
14
11
0
100
100
rt 12
(-)
()
20
150
100
200
150
Demand
600
ks -1
kp 3
km 10
kn 2
Note The r and k values and the improvement
difference values have changed.
29
Source Adapted from Lapin, 1994
Optimal Solution

In five iterations the shipping cost has moved
from 11,500 to 6,250. There are no remaining
empty cells with a negative value.
6,250
S
N
P
M
Capacity
To
300 750 1800 900 1100 1400 C 6,250
From
19
7
3
21
rj 0
100
4
19
30
6
18
21
15
50
100
150
()
(-)
rs 15
3
22
15
14
11
100
100
rt 11
(-)
()
20
1
150
100
200
150
Demand
600
ks 0
kp 3
km -9
kn 3
Note The r and k values and the improvement
difference values have changed.
30
Adios!
Write a Comment
User Comments (0)
About PowerShow.com