Routing Scheduling Part 1 Transport Service Selection Depends

Routing & Scheduling: Part 1

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

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

Example • • A Birmingham luggage company maintains a finishedgoods 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.

Example Transport Services Available to the Firm 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 20 Air 1. 40 2 40 Transport Service

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

Example Method of Computation Rail Rx. D (. 1)(700, 000) = 70, 000 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 Cost Type Transportation In-transit Inventory Total for Rail $2, 235, 465

Example Cost Type Transportation In-transit Inventory Plant Inventory Warehouse Inventory Method of Computation Piggyback Rx. D (. 15)(700, 000) = 105, 000 ICDT/365 [(. 3)(30)(700, 000)(14)]/365 = 241, 644 ICQ/2 [(. 3)(30)(50, 000)(0. 93)c] = 418. 500 IC”Q/2 [(. 3)(30. 15)(50, 000)(0. 93)c] = 420. 593 Total for Rail C = accounts for improved transport service & number of shipments per year $1, 185, 737

Example Cost Type Transportation In-transit Inventory Plant Inventory Warehouse Inventory Method of Computation Truck Rx. D (. 2)(700, 000) = 140, 000 ICDT/365 [(. 3)(30)(700, 000)(5)]/365 = 86, 301 ICQ/2 [(. 3)(30)(50, 000)(0. 84)c] = 378, 000 IC”Q/2 [(. 3)(30. 2)(50, 000)(0. 84)c] = 380, 520 Total for Rail C = accounts for improved transport service & number of shipments per year $984, 821

Example Cost Type Transportation In-transit Inventory Plant Inventory Warehouse Inventory Method of Computation Air Rx. D (1. 4)(700, 000) = 980, 000 ICDT/365 [(. 3)(30)(700, 000)(2)]/365 = 34, 521 ICQ/2 [(. 3)(30)(25, 000)(0. 80)c] = 378, 000 IC”Q/2 [(. 3)(30. 4)(25, 000)(0. 80)c] = 190, 755 Total for Rail C = accounts for improved transport service & number of shipments per year $1, 387, 526

Example Modal Choice Cost Type Transportation Method of Computation Rail Piggyback Truck Air Rx. D 70, 000 105, 000 140, 000 980, 000 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 In-transit Inventory

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 tradeoffs • 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

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

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.

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

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

Routing & Scheduling: Part 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = d 0, C + d. C, 0 + d. A, B - d. A, C - d. C, B • If stop C were inserted after stop B, the savings value is expressed as • S = d. B, 0 - d. B, C + d 0, C • If stop C were inserted before stop A, the savings value is expressed as • S = d. C, 0 - d. C, A + d 0, A

The Savings Method of Routing Initial routing – Route distance = d 0, A + d. A, 0 + d 0, B + d. B, 0 + d 0, C + d. C, 0 d 0, A Stop A d. A, 0 Depot (0) d. B, 0 Stop B d 0, C d. C, 0 Stop C

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

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

The Savings Method of Routing Combining 3 stops on 1 route – Route distance (additional stop added after 1 st 2 stops) = d 0, A(100) + d. A, B(145) + d. B, C(65) + d. C, 0(25) = 335 miles d 0, A = 100 Atlanta (A) d. A, B = 145 Depot (0) Birmingham (B) d. C, 0 = 25 d. B, C = 65 Gadsden (C) Savings value of S = d. B, 0 (85) - d. B, C(65) + d 0, 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

The Savings Method of Routing Combining 3 stops on 1 route – Route distance (additional stop added between 1 st 2 stops) = d 0, A(100) + d. A, C(125) + d. C, B(65) + d. B, 0(85) = 375 miles d 0, A = 100 Atlanta (A) d. A, C = 125 Depot (0) Gadsden (C) d. B, 0 = 85 d. C, B = 65 Birmingham (B) Savings value of S = d 0, C(25) + d. C, 0(25) + d. A, B(145) - d. A, C(125) - d. C, 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

The Savings Method of Routing Combining 3 stops on 1 route – Route distance (additional stop added before 1 st 2 stops) = d 0, C(25) + d. C, A(125) + d. A, B(145) + d. B, 0(85) = 380 miles d 0, C = 25 Gadsden (C) d. C, A = 125 Depot (0) Atlanta (A) d. B, 0 = 85 d. A, B = 145 Birmingham (B) Savings value of S = d. C, 0(25) - d. C, A(125) + d 0, 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

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