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 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.

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

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 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

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] = 418.500 Warehouse Inventory IC”Q/2 [(.3)(30.15)(50,000)(0.93)c] = 420.593 Total for Rail $1,185,737 C = accounts for improved transport service & number of shipments per year

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

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

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

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

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 1000 H 4000 A 2000 F 3000 D 3000 Depot B 3000 G 2000 I 1000 C 2000 L 2000 K 2000 J 2000

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

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

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

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 = d0,A + dA,0 + d0,B + dB,0 d0,A Stop A dA,0 Depot (0) dB,0 Stop B d0,B

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

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

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

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

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

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

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

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

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

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

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