Chapter 08S – The Transportation Model
8S–157
Example
TO
FROM
Chicago
South Bend
Indianapolis
Fort Wayne
Total
Cleveland
80
60
70
50
150
Columbus
75
65
55
40
175
Bowling Green
70
85
55
35
275
Cincinnati
90
55
48
60
100
Total
200
100
300
100
700
Step 1
In establishing the penalty cost for row 1 (Cleveland), we subtract the lowest cost in row 1 from the
second lowest cost in row 1. For Cleveland, the lowest cost is $50 (unit shipping cost from Cleveland
to Ft. Wayne). The second lowest cost is $60 (unit shipping cost from Cleveland to South Bend).
Chicago = 5 South Bend = 5 Indianapolis = 0 Ft. Wayne = 5
Step 2
Since row three (Bowling Green) has the largest penalty cost, it is selected. In row three, the shipping
route from Bowling Green to Ft. Wayne has the lowest shipping cost per unit ($35). Thus, we allocate
as many units as possible (100 units) to it.
Step 3
Since the demand in Ft. Wayne is reduced to zero, we eliminate Ft. Wayne from further consideration.
Chapter 08S – The Transportation Model
8S–158
Step 2
Since Bowling Green has the largest penalty cost, it is selected again. Allocate 175 units from
Bowling Green to Indianapolis.
Step 3
Eliminate Bowling Green from further consideration.
Step 3
Eliminate Indianapolis from further consideration.
Continuing in this fashion gives the following completed transportation table.
To
FROM
Chicago
South Bend
Indianapolis
Fort Wayne
Total
Cleveland
80
60
70
50
150
Columbus
75
65
55
40
175
Bowling Green
70
85
55
35
275
Cincinnati
90
55
48
60
100
Total
200
100
300
100
700
150
50
125
175
100
100
Chapter 08S – The Transportation Model
8S–159
Exercise 1
For the following transportation tableau determine the initial feasible solution using Vogel’s
approximation.
To:
A
B
C
D
Supply
From
1
18
12
14
16
40
2
23
24
27
33
80
3
42
34
31
26
130
Demand
90
80
30
50
Exercise 2
For the following transportation tableau, determine the initial feasible solution using Vogel’s
approximation method.
To:
Milwaukee,
WI
St, Louis,
MO
Dayton,
OH
Supply
From
Wichita, KS
6
9
10
150
Omaha, NE
8
7
11
175
Ames, IO
4
5
12
275
Demand
200
100
300
Chapter 08S – The Transportation Model
8S–160
Solution to Exercise 1 To
From
A
B
C
D
Supply
Iteration 1
Penalty
Cost
Iteration 2
Penalty
Cost
Iteration 3
Penalty
Cost
Iteration 4
Penalty
Cost
1
18
12
14
16
40
2
4
–
–
2
23
24
27
33
80
1
1
1
–
3
42
34
31
26
Demand
90
80
30
50
250
Iteration 1
Penalty
Cost
Iteration 2
Penalty
Cost
Iteration 3
Penalty
Cost
TC = 6,480
Iteration 4
Penalty
Cost
–
–
–
7
Solution to Exercise 2 To
MILW
SL
DAY
SUPPLY
PC1
PC2
PC3
PC4
Wichita
6
9
10
8
7
11
Ames
4
5
12
Demand
200
100
300
600
*PC1
2
2
1
PC3
–
2
1
*Penalty costs in iteration 1.
TC = 6(150) + 11(175) + 4(50) +5(100) + 12(125) = 5025
80
10
70
10
30
50
50
150
100
175
125
175
275
150
12
13
8
3
7
1
4
19
–
–
2
1
PC2
PC4
Omaha
–
–
–
1
1
1
1
4
–
–
7
–
10
10
–
10
12
5
5
130
5
8
8
Chapter 08S – The Transportation Model
8S–161
Supplemental Problems
1. Refer to supplement Chapter 8, Problem 1 on the text web site and formulate it as a linear
programming problem with an objective function and a set of constraints.
2. Refer to supplement Chapter 8, Problem 2 on the text web site and formulate it as a linear
programming problem with an objective function and a set of constraints.
3. Refer to supplement Chapter 8, Problem 3 on the text web site and formulate it as a linear
programming problem with an objective function and a set of constraints.
Solutions to Supplemental Problems
1. x11 = quantity shipped from 1 to A, x12 = quantity shipped from 1 to B, etc.
Minimize Z = 3x11 + 4x12 + 2x13 + 5x21 + 1x22 + 7x23 + 8x31 + 7x32 + 4x33
s.t.
Supply x11 + x12 + x13 = 40
All variables 0
2. x11 = quantity shipped from source 1 to destination 1, etc.
Minimize = 3x11 + 6x12 + 2x13 + 3x21 + x22 + 3x23 + 7x31 + 6x32 + 4x33
s.t.
Supply x11 + x12 + x13 = 40
All variables 0
3. x11 = quantity shipped from 1 to A, x12 = quantity shipped from 1 to B, etc.
Minimize Z = 18x11 + 12x12 + 14x13 + 16x14 + 23x21 + … + 26x34
s.t.
x13 + x23 + x33 = 30
x14 + x24 + x34 = 50
All variables 0