1. Transport Decisions Chapter 7
If you are planning for one year, grow rice. If you are planning for 20 years, grow trees. If you are planning for centuries, grow men.
A Chinese proverb
Chapter 7
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2. Transport Decisions in Transport Strategy
PLANNING
ORGANIZING
CONTROLLING
Transport Strategy •
Transport fundamentals
Transport decisions
Customer
service goals
The product
Logistics service
Ord
. proc. & info. sys.
Inventory Strategy
Forecasting
Inventory decisions
Purchasing and supply
scheduling decisions
Storage fundamentals
Storage decisions
Location Strategy
Location decisions
The network planning process
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3. problems in transportation
Typical Transport Decisions
Mode/Service selection
Private fleet planning -
Carrier routing
Routing from multiple points
Routing from coincident origin
destination
points
Vehicle routing and scheduling
Freight consolidation
Just a few of the many
problems in transportation
CR (2004) Prentice Hall, Inc.

4. Mode/Service Selection
The problem -
Define the available choices
Balance performance effects on inventory against
the cost of transport
Methods for selection
Indirectly through network configuration
Directly through channel simulation
Directly through a spreadsheet approach as f
ollows:
Alternatives
Cost types Air Truck Rail
Transportation In
transit inventory
Source inventory
Destination inventory
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5. Mode/Service Selection (Cont’d)
Example Finished goods are to be shipped from a plant inventory to a warehouse inventory some distance away. The expected volume to be shipped in a year is 1,200,000 lb. The product is worth $25 per lb. and the plant and carrying costs are 30% per year.
Other data are:
Transport
choice
Rate,
$/lb.
Transit
time,
days
Shipment
size, lb.
Rail
0.11 25
100,000
Truck
0.20 13
40,000
Air
0.88 1
16,000
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6. Include transport rate
Transport Selection Analysis
Cost
Compu-
type
tation
Rail
Truck
Air
Trans- RD
.11(1,200,000)
.20(1,200,000)
.88(1,200,000)
portation
= $132,000
= $240,000
= $1,056,000
In-transit
ICDT
365
[.30(25)
[.30(25)
[.30(25)
inventory
1,200,000(25)]/365
1,200,000(13)]/365
1,200,000(1)]/365
= $616,438
= $320,548
= $24,658
Plant
ICQ 2
[.30(25)
[.30(25)
[.30(25)
inventory
100,000]/2
40,000]/2
16,000]/2
= $375,000
= $150,000
= $60,000 IC ' Q 2
Whse
[.30(25.11)
[.30(25.20)
[.30(25.88)
inventory
100,000]/2
40,000]/2
16,000]/2
= $376,650
= $151,200
= $62,112
Include transport rate
Totals
$1,500,088 $
861,748
$1,706,770
Improved service
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7-6

7. Carrier Routing
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8. Can be a weighted index of time and distance
Carrier Routing (Cont’d)
Origin
Amarillo
Oklahoma
City
Destination
Fort Worth A B E I C D G F H J
90 minutes 84
138
348
156 48
132
150
126
120 66 60
Note
: All link times are in minutes 90
Can be a weighted index of time and distance
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9. Shortest Route Method Step Solved Nodes Directly Connected to Unsolved
Its
Closest
Unsolved
Node
Total Cost
Involved
nth
Nearest
Minimu
m Cost
Its Last
Connection a 1 A B 90 AB * 2 C
138 AC
90+66=156 3 D
348 E
90+84=174
174 BE F
138+90=228 4
228 CF I
174+84=258 5
=294
258 EI H
228+60=288 6
228+60=
288 FH J
=384 7
294 CD G
=360
288+48=336 8
=414
384 IJ
Shortest Route Method
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10. MAPQUEST SOLUTION Mapquest at www.mapquest.com
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11. Routing from Multiple Points
This problem is solved by the traditional transportation method of linear programming
Plant 1
Requirements = 600
Plant 2
Requirements = 500
Plant 3
Requirements = 300
Supplier A
Supply  400
Supplier C
Supply  500
Supplier B
Supply  700 4 a 7 6 5 9 8
The transportation rate in $ per ton for an optimal routing between supplier A and plant 1 .
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12. TRANLP problem setup
Solution
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13. Routing with a Coincident Origin/Destination Point
Typical of many single truck routing problems from a
single depot.
Mathematically, a complex problem to solve efficiently.
However, good routes can be found by forming a route
pattern where the paths do not cross
a "tear drop"
pattern.
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14. Single Route Developed by ROUTESEQ in LOGWARE
Y coordinates
Y coordinates 8 8 7 4 9 13 16 7 4 9 13 16 6 10 19 6 10 19 5 6 15 20 5 6 15 20 4 2 8 18 4 2 8 18 3 5 D 12 17 3 5 D 12 17 2 2 3 3 1 7 11 14 1 7 11 14 1 1
X coordinates
X coordinates
(a) Location of beverage accounts
and distribution center (D) with
grid overlay
(b) Suggested routing pattern
7-14
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15. Multi-Vehicle Routing and Scheduling
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16. Practical Guidelines for Good Routing and Scheduling
1. Load trucks with stop volumes that are in closest proximity to each other D
Depot
Stops
Depot
(a) Weak clustering
(b) Better clustering
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17. May need to coordinate with sales to achieve clusters
Guidelines (Cont’d)
2. Stops on different days should be arranged to produce tight clusters F T D
Depot
(a) Weak clustering--
routes cross
(b) Better clustering
Stop
May need to coordinate with sales to achieve clusters
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18. Guidelines (Cont’d)
3. Build routes beginning with the farthest stop from the depot
4. The stop sequence on a route should form a teardrop pattern (without time windows)
5. The most efficient routes are built using the largest vehicles available first
6. Pickups should be mixed into delivery routes rather than assigned to the end of the routes
7. A stop that is greatly removed from a route cluster is a good candidate for an alternate means of delivery
8. Narrow stop time window restrictions should be avoided (relaxed)

19. Application of Guidelines to Casket Distribution
Warehouse
Funeral home
Typical weekly demand and pickups
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20. Application of Guidelines to Casket Distribution (Cont’d)
Territories of
equal size
to minimize
number of trucks
Warehouse
Funeral home
Division of sales territories into days of the week
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21. Application of Guidelines to Casket Distribution (Cont’d)
Warehouse
Funeral home
Route design within territories
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22. “Sweep” Method for VRP
Example A trucking company has 10,000-unit vans for merchandise pickup to be consolidated into larger loads for moving over long distances. A day’s pickups are shown in the figure below. How should the routes be designed for minimal total travel distance?
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23. Stop Volume and Location
Geographical
region
Pickup
points
1,000
4,000
2,000
3,000
2,000
3,000
3,000
2,000
Depot
1,000
2,000
2,000
2,000
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24. “Sweep” Method Solution
Sweep direction
is arbitrary
Route #3
8,000 units
Route #1
10,000 units
1,000
4,000
2,000
3,000
2,000
3,000
3,000
2,000
Depot
1,000
2,000
2,000
2,000
Route #2
9,000 units
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25. The “Savings” Method for VRP
Stop d
A,0 d
0,A A A d
0,A d
A,B d
0,B
Depot
Depot d
B,0 B d
B,0 B
Stop
(a) Initial routing
(b) Combining two stops on a route
Route distance = d +d +d
+ d
Route distance = d +d +d
0,A
A,0
0,B
B,0
0,A
A,B
B,0
“Savings” is better than “Sweep” method—has lower average error
7-25
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26. Savings Method Observation
The points that offer the greatest savings when combined on the same route are those that are farthest from the depot and that are closest to each other.
This is a good principle
for constructing multiple-stop
routes
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27. Route Sequencing in VRPAM PM 8 9 10 11 12 1 2 3 4 5 6
Route #1
Route #10
Route #6
Truck #1
Route #9
Route #4
Truck #2
Route #5
Route #8
Truck #3
Route #2
Route #7
Truck #4
Route #3
Truck #5
Minimize number of trucks
by maximizing number of routes
handled by a single truck
7-27
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28. Freight Consolidation
Combine small shipments into larger ones
A problem of balancing cost savings against customer service reductions
An important area for cost reduction in many firms
Based on the rate-shipment size relationship for for-hire carriers
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29. Freight Consolidation Analysis
Suppose we have the following orders for the next three days.
Consider shipping these orders each day or consolidating them into one shipment. Suppose that we know the transport rates.
Note: Rates from an interstate tariff
From:
Ft Worth
Day 1
Day 2
Day 3
To:
Topeka
5,000 lb.
25,000 lb.
18,000 lb.
Kansas City
7,000
12,000
21,000
Wichita
42,000
38,000
61,000
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30. Freight Consolidation Analysis (Cont’d)
Separate shipments
Day 1
Day 2
Rate x volume = cost
Topeka
3.42 x 50 = $171.00
1.14 x 250 = $285.00
Kansas City
3.60 x 70 =
1.44 x 120 =
Wichita
0.68 x =
285.60
0.68 x 400 a =
272.00
Total $708.60
Total $729.80
a Ship 380 cwt., as if full truckload of 400 cwt.
Day 3
Totals
1.36 x 180 = $244.80
$700.80
1.20 x 210 =
676.80
0.68 x 610 =
414.80
972.40
Total $911.60
$2,350.00
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7-30

31. Freight Consolidation Analysis (Cont’d)
Consolidated shipment
Computing transport cost for one combined, three-day shipment
Day 3
Rate x volume = cost a
Topeka
0.82 x 480
= $393.60
Kansas City
0.86 x =
Wichita
0.68 x =
958.80
Total
$1,696.40
a 480 =
Cheaper, but what about
the service effects of holding
early orders for a longer time
to accumulate larger shipment
sizes?
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7-31