1. Routing & Scheduling: Part 1

2. Transport Service Selection
Depends on variety of service characteristics
Not all service characteristics are of equal importance
Most common bases used for modal choice:
Cost of service
Average transit time (speed)
Transit-time variability (dependability)
Other bases used
Capability
Availability & adequacy of equipment
Availability of service
Frequency of service
Security
Claims handling
Shipment tracing
Problem-solving assistance

3. Basic Cost Trade-Offs
When alternative modes are available, the one chosen should be the one that offers the lowest total cost consistent with customer service goals.
Often, cost trade-offs must be used.
Speed & dependability affect both the seller’s & buyer’s inventory level, as well as the inventory that is in transit.
Slower, less reliable modes require more inventory in the distribution channel

4. Example
A Birmingham luggage company maintains a finished-goods inventory at its plant
Currently, rail is used to ship between Birmingham and the firm’s West Coast warehouse
Average transit time is T = 21 days
100,000 units are kept at each stocking point with the luggage having an average value of C = $30 per unit
Inventory carrying costs are I = 30 percent per year
There are D = 700,000 units sold per year out of the West Coast warehouse
Average inventory levels can be reduced by 1 percent for each day of transit time that is eliminated.

5. Example Transport Services Available to the Firm Transport Service
Rate ($/unit)
Door-to-Door Transit Time (days)
No. of Shipments per year
Rail
0.10 21 10
Piggyback
0.15 14 20
Truck
0.20 5
Air
1.40 2 40

6. Example Different modes affect the time inventory is in transit
Annual demand (D) will be in transit by the fraction of the year represented by T/365 days, where T is the average transit time
Annual cost of carrying this in-transit inventory is ICDT/365
Average inventory at both ends of the channel can be approximated as Q/2, where Q is the shipment size
Holding cost per unit is I x C
Note that C must reflect where the inventory is in the channel
Value of C at the plant is the price ($30 per unit)
Value of C at the WC warehouse is C + transportation rate
Total annual transportation cost is R x D

7. Example Cost Type Method of Computation Rail Transportation R x D
(.1)(700,000) = 70,000
In-transit Inventory
ICDT/365
[(.3)(30)(700,000)(21)]/365 = 363,465
Plant Inventory
ICQ/2
[(.3)(30)(100,000)] = 900,000
Warehouse Inventory
IC”Q/2
[(.3)(30.1)(100,000)] = 903,000
Total for Rail
$2,235,465

8. Example Cost Type Method of Computation Piggyback Transportation R x D
(.15)(700,000) = 105,000
In-transit Inventory
ICDT/365
[(.3)(30)(700,000)(14)]/365 = 241,644
Plant Inventory
ICQ/2
[(.3)(30)(50,000)(0.93)c] =
Warehouse Inventory
IC”Q/2
[(.3)(30.15)(50,000)(0.93)c] =
Total for Rail
$1,185,737
C = accounts for improved transport service & number of shipments per year

9. Example Cost Type Method of Computation Truck Transportation R x D
(.2)(700,000) = 140,000
In-transit Inventory
ICDT/365
[(.3)(30)(700,000)(5)]/365 = 86,301
Plant Inventory
ICQ/2
[(.3)(30)(50,000)(0.84)c] = 378,000
Warehouse Inventory
IC”Q/2
[(.3)(30.2)(50,000)(0.84)c] = 380,520
Total for Rail
$984,821
C = accounts for improved transport service & number of shipments per year

10. Example Cost Type Method of Computation Air Transportation R x D
(1.4)(700,000) = 980,000
In-transit Inventory
ICDT/365
[(.3)(30)(700,000)(2)]/365 = 34,521
Plant Inventory
ICQ/2
[(.3)(30)(25,000)(0.80)c] = 378,000
Warehouse Inventory
IC”Q/2
[(.3)(30.4)(25,000)(0.80)c] = 190,755
Total for Rail
$1,387,526
C = accounts for improved transport service & number of shipments per year

11. Example Modal Choice Cost Type Method of Computation Rail Piggyback
Truck
Air
Transportation
R x D
70,000
105,000
140,000
980,000
In-transit Inventory
ICDT/365
363,465
241,644
86,301
34,521
Plant Inventory
ICQ/2
900,000
418,500
378,000
182,250
Warehouse Inventory
IC”Q/2
903,000
420,593
380,520
190,755
Totals
$2,235465
$1,185,737
$984,821
$1,387,526

12. Factors Other Than Transportation Cost that Affect Modal Choices
Effective buyer/seller cooperation is encouraged if a reasonable knowledge of the other party’s costs is available
If there are competing suppliers, buyer & supplier should act rationally to gain optimum cost-transport service trade-offs
Offering higher-quality transportation services than the competition may allow the seller to charge a higher price for the product
Elements in the mix change frequently
Transport rate fees, product mix changes, inventory cost changes, & transport service retaliation by competitors
If buyer makes the transport choice, seller’s inventories are impacted as well, which may impact price charged for the product

13. Vehicle Routing & Scheduling
Selecting the best paths for the transport mode to follow to minimize travel time or distance reduces transportation costs and improves customer service
Start with determining shortest possible routes based on
Transit time
Distance
Cost
Incorporate restrictions

14. Restrictions on Vehicle Routing & Scheduling
Each stop on the route may have volume to be picked up as well as delivered
Multiple vehicles may be used using different capacity limits to both weight and cube
Maximum total driving time allowed before a rest period must be taken is 8 hours
Stops may permit pickups/deliveries only at certain times of day (time windows)
Pickups may be permitted on a route only after deliveries are made
Drivers may be allowed to take short rests or lunch breaks at certain times of the day.

15. 8 Principles for Good Routing & Scheduling
Load trucks with stop volumes that are in the closest proximity to each other – minimizes interstop travel between them
Stops on different days should be arranged to produce tight clusters – develop overall route, plus daily routes
Build routes beginning with the farthest stop from the depot
Sequence of stops on a truck route should form a teardrop pattern – try to keep route paths from crossing

16. 8 Principles for Good Routing & Scheduling
The most efficient routes are built using the largest vehicles available – allocate largest vehicles first, then smaller
Pickups should be mixed into delivery routes rather than assigned to the end of routes
A stop that is greatly removed from the other stops in a route cluster is a good candidate for an alternative means of delivery
Narrow stop time window restrictions should be avoided – see if you can renegotiate the time window restrictions

17. Routing & Scheduling: Part 2

18. The Sweep Method of Routing
Simple method to use
Fairly accurate with a projected error rate of about 10%
Good to use when results must be obtained in short order or
Good to use when a good solution is needed as opposed to an optimum solution

19. The Sweep Method of Routing
Locate all stops including the depot on a map or grid
Extend a straight line from the depot in any direction
Rotate the line (clockwise or counterclockwise) until it intersects a stop
Will the inserted stop exceed the vehicle’s capacity?
If not, continue rotating the line until the next stop is intersected
Will the cumulative volume exceed the vehicle’s capacity?
Continue process until vehicle’s capacity would be exceeded
Sequence the stops to minimize distance

20. Sweep Method Example
Gofast Trucking uses vans to pickup merchandise from outlying customers
Merchandise is returned to a depot where it is consolidated into large loads for intercity transport
Firm’s vans can haul 10,000 units
Completing a route typically takes a full day
Firm wants to know
How many routes (trucks) are needed
Which stops should be on the routes
And sequence of stops for each truck

21. Sweep Method Example Pickup Stop Data: quantities shown in units E
1000 H
4000 A
2000 F
3000 D
3000
Depot B
3000 G
2000 I
1000 C
2000 L
2000 K
2000 J
2000

22. Sweep Method Example “Sweep” method solution E 1000 H 4000 A 2000 F
3000 D
3000
Depot B
3000 G
2000 I
1000 C
2000 L
2000 K
2000 J
2000

23. Sweep Method Example Route 1 10,000 units “Sweep” method solution E
1000 H
4000 A
2000 F
3000
Route 1
10,000 units D
3000
Depot B
3000 G
2000 I
1000 C
2000 L
2000 K
2000 J
2000

24. Sweep Method Example Route 3 8,000 units Route 1 10,000 units Route 2
“Sweep” method solution E
1000 H
4000
Route 3
8,000 units A
2000 F
3000
Route 1
10,000 units D
3000
Depot B
3000 G
2000 I
1000
Route 2
9,000 units C
2000 L
2000 K
2000 J
2000

25. The Savings Method of Routing
Developed by Clarke & Wright (1963)
Objective is to minimize the total distance traveled by all vehicles and
To minimize (indirectly) the number of vehicles needed to serve all stops
Has been proven to be
Flexible enough to handle wide range of practical constraints (forms routes & sequences of stops on routes simultaneously)
Relatively fast for problems with a moderate number of stops
Capable of generating near optimum solutions

26. The Savings Method of Routing
Begin with a dummy vehicle serving each stop and returning to the depot.
Gives the maximum distance to be experienced in the routing problem
Two stops are then combined together on the same route
Eliminates one vehicle and travel distance is reduced
To determine which stops to combine on a route, the distance saved is calculated before and after each combination
This calculation is repeated for all stop pairs
The stop pair with the largest savings value is selected to be combined

27. The Savings Method of Routing
Initial routing – Route distance
= d0,A + dA,0 + d0,B + dB,0
d0,A
Stop A
dA,0
Depot (0)
dB,0
Stop B
d0,B

28. The Savings Method of Routing
Combing 2 stops on 1 route – Route distance
= d0,A + dA,B + dB,0
d0,A
Stop A
Depot (0)
dA,B
Stop B
dB,0
Savings value of S
= d0,A + dB,0 - dA,B

29. The Savings Method of Routing
Initial routing – Route distance
= d0,A + dA,0 + d0,B + dB,0
d0,A = 100
Atlanta (A)
dA,0 = 100
Jacksonville (0)
dB,0 = 85
Birmingham (B)
d0,B = 85
Initial routing – Route distance
= d0,A(100) + dA,0(100) + d0,B(85) + dB,0(85) = 370 miles total

30. The Savings Method of Routing
Combining 2 stops on 1 route – Route distance
= d0,A(100) + dA,B(145) + dB,0(85) = 330 total miles
Atlanta (A)
d0,A = 100
Jacksonville (0)
dA,B = 145
Birmingham (B)
dB,0 = 85
Savings value of S
= d0,A(100) + dB,0(85) - dA,B(145) = 40 miles saved

31. The Savings Method of Routing
If a third stop (C) is to be inserted between stops A and B, where A and B are on the same route, the savings value is expressed as
S = d0,C + dC,0 + dA,B - dA,C - dC,B
If stop C were inserted after stop B, the savings value is expressed as
S = dB,0 - dB,C + d0,C
If stop C were inserted before stop A, the savings value is expressed as
S = dC,0 - dC,A + d0,A

32. The Savings Method of Routing
Initial routing – Route distance
= d0,A + dA,0 + d0,B + dB,0 + d0,C + dC,0
Stop A
d0,A
dA,0
Depot (0)
dB,0
Stop B
d0,B
Stop C
d0,C
dC,0

33. The Savings Method of Routing
Combining 3 stops on 1 route – Route distance (additional stop added after 1st 2 stops)
= d0,A + dA,B + dB,C + dC,0
Stop A
d0,A
dA,B
Depot (0)
Stop B
dB,C
dC,0
Stop C
Savings value of S = d0,C + dC,0 + dA,B - dA,C - dC,B

34. The Savings Method of Routing
Initial routing – Route distance = d0,A(100) + dA,0 (100) + d0,B(85) + dB,0 (85) + d0,C (25) + dC,0 (25) =
420 total miles
d0,A = 100
Atlanta (A)
dA,0 = 100
Jacksonville (0)
dB,0 = 85
Birmingham (B)
d0,B = 85
d0,C = 25
Gadsden (C)
dC,0 = 25

35. The Savings Method of Routing
Combining 3 stops on 1 route – Route distance (additional stop added after 1st 2 stops)
= d0,A(100) + dA,B(145) + dB,C(65) + dC,0(25) = 335 miles
Atlanta (A)
d0,A = 100
dA,B = 145
Depot (0)
Birmingham (B)
dB,C = 65
dC,0 = 25
Gadsden (C)
Savings value of S = dB,0 (85) - dB,C(65) + d0,C(25) = 45 miles saved
Earlier O to A to B to O route (330 miles) plus 0 to C to 0 route (50 miles) = 380 miles

36. The Savings Method of Routing
Combining 3 stops on 1 route – Route distance (additional stop added between 1st 2 stops)
= d0,A(100) + dA,C(125) + dC,B(65) + dB,0(85) = 375 miles
Atlanta (A)
d0,A = 100
dA,C = 125
Depot (0)
Gadsden (C)
dC,B = 65
dB,0 = 85
Birmingham (B)
Savings value of S = d0,C(25) + dC,0(25) + dA,B(145) - dA,C(125) - dC,B(65) = 5 miles saved
Earlier O to A to B to O route (330 miles) plus 0 to C to 0 route (50 miles) = 380 miles

37. The Savings Method of Routing
Combining 3 stops on 1 route – Route distance (additional stop added before 1st 2 stops)
= d0,C(25) + dC,A(125) + dA,B(145) + dB,0(85) = 380 miles
Gadsden (C)
d0,C = 25
dC,A = 125
Depot (0)
Atlanta (A)
dA,B = 145
dB,0 = 85
Birmingham (B)
Savings value of S = dC,0(25) - dC,A(125) + d0,A(100) = 0 miles saved
Earlier O to A to B to O route (330 miles) plus 0 to C to 0 route (50 miles) = 380 miles

38. The Savings Method of Routing
When 3rd Stop Added
Routes Added
Routes Eliminated
Miles Saved
After 1st 2
B > C; C > 0
B > 0 45
Between 1st 2
A > C; C > B
0 > C; C > 0; A > B 5
Before 1st 2
C > A
0 > A; C > 0