Logistics Management Transportation Decisions. Özgür Kabak, Ph.D.

Transcription

1 Logistics Management Transportation Decisions Özgür Kabak, Ph.D.

2 Typical Transport Decisions Mode/Service selection Freight consolidation Trade-off between transportation cost and customer responsiveness Vehicle Routing Seperate and Single Origin and Destination Points Multiple Origin and Destination points Routing with a Coincident Origin/Destination Point

3 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 follows Alternatives: Air / Truck / Rail Cost types Transportation In-transit inventory Source inventory Destination inventory

4 Mode/Service Selection Example The CarryAll Luggage Company produces a line of luggage goods. The typical distribution plan is to produce finished goods inventories to be kept at the plant site. Goods are then shipped to company-owned field warehouses by way of common carriers. Rail is currently used to ship between the East Coast plant and a West Coast warehouse. The average transit time for rail shipments is T = 21 days. At each stocking point, there is an average of 100,000 units of luggage having an average cost of C = $30 per unit. Inventory carrying cost is i = 30 percent per unit cost per year. The company wishes to select the mode of transportation that will minimize the total costs. It is estimated that for every day that transit time can be reduced from the current 21 days, average inventory levels can be reduced by 1 percent.

5 Mode/Service Selection Example The demand is D = 700,000 units sold per year out of the West Coast warehouse. The company can use the following transportation services: Transport Service Rate($/unit)) Door-to-Door Transit Time (days) Number of shipment per year Rail Piggyback (TOFC) Truck Air

6 Mode/Service Selection Example The four factors to be considered are: Transport costs: If R denotes unit transport rate and D denotes annual demand, then RD gives an estimate of the annual transport cost In-transit inventory: Each unit, valued at $C, spends T days in transit. If i is annual holding rate, each item costs ICT/365 in holding charges during transport time. Since D is annual demand, total in-transit inventory cost equals [ICT/365]*D. Letting Q = shipment size, and assuming production occurs instantaneously at the plant, average annual inventory cost at the plant equals ICQ/2.(for rail Q/2=100 both at the plant and at the warehouse) Letting C = C + R, i.e., product value at the field warehouse, then average inventory cost at the field warehouse equals IC Q/2, assuming constant lead time (All of these inventory cost estimates assume constant and deterministic demand rate).

7 Mode/Service Selection Example Cost type Formule Rail TOFC Truck Air Transp n RXD (0.1)( ) = (0.15)( ) = (0.2)( ) = (1.4)( ) = In-transit Inv. ICDT/365 ((0.3)(30) X(700 00)X(21))/365 = Plant Inventory Warehouse Inventory ICQ/2 ((0.3)(30) X( )) = IC Q/2 ((0.3)(30.1) X( )) = ((0.3)(30) X( ) X(14))/365 = ((0.3)(30) X(50.000)(0.93) = ((0.3)(30.15) X(50.000)(0.93) = ((0.3)(30) X( )X(5))/36 5 = ((0.3)(30) (50.000) (0.84)) = ((0.3)(30.2) ( X(0.84)) = ((0.3)(30) X( )X(2))/3 65 = (0.3)(30) X(25.000) X(0.8) = ((0.3)(30.4) X(25000) X(0.8)) = Total

8 Mode/Service Selection Example

9 Characteristics of Transport Mode Selection Based on Total Cost Although assumptions in these back of the envelope calculations seem restrictive, they generally lead to good decisions in transport mode (since costs of different modes can be significantly different) given that we do not invest in transportation equipment. Calculations do not consider costs of investing in trucks, planes, rail cars, or ships and assume that we decide mode through thirdparty carrier. Calculations do not reflect costs of demand variability since they assume demand occurs at a constant, fixed rate. In distribution systems, variable demand can lead to significant overage costs in periods when demand exhausts capacity. These costs become more significant after transport mode decision, when deciding fleet capacity for a particular mode.

10 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs Eastern Electric (EE) purchases all the motors for its appliance from Westview. EE currently purchases 120,000 motors each year from Westview at a price of $120 per motor. Demand has been relatively constant for several years and is expected to stay this way Each motor averages about 10 kg and EE has traditionally purchased in lots of 3,000 motors Westview ships each EE order within a day of receiving it At its assembly plant, EE carries a safety inventory equal to 50 percent of the average demand for motors during the delivery lead time The plant manager at EE has received several proposals for transportation and must decide on the one to accept

11 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs The details of various proposals for Eastern Electric Carrier Range of Quantity Shipped (100 kg) Shipping Cost ($/100 kg) AM Railroad Northeastern Trucking Golden Freightways Golden Freightways Golden Freightways New Proposal: Golden Freightways

12 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs Annual cost of holding inventory=25% Annual holding cost is, therefore, H= $120(price)*0.25=$30/motor Shipments by rail requires 5 days Shipments by truck requires 3 days

13 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs The transportation decision affects the cycle inventory, safety inventory, and in-transit inventory. The AM Railroad proposal requires a minimum shipment of 20,000 kg(20 tons) which corresponds to 2000 motors The replenishment lead time = 5+1=6days For Q=2000 motors, the plant manager obtains the following: Cycle inventory = Q/2 = 2,000/2 = 1,000 motors Safety inventory = L/2 days of demand (6/2)(120,000/365) = 986 motors In-transit Inventory = 120,000(5/365) = 1,644 motors Total Average Inventory = 1, ,644 = 3,630 motors Annual holding cost using AM Rail = 3, = $108,900

14 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs The transportation cost: AM Rail charges $6.50 per 100 kg, resulting in a transportation cost of $0.65 per motor because each motor is about 10 kg. Annual transportation cost using AM Rail =120,000*0.65 = $78,000 The total annual cost using AM Rail = inventory holding cost + transportation cost = $108,900 + $78,000 = $186,900

15 Example 2 for the Choice of Transportation Mode Based on Cost Trade-Offs Alternative Lot Size (Motors) Transporta tion Cost Cycle Inventory Safety Inventory In-transit Inventory Inventory cost Total Cost AM Rail 2,000 $78,000 1, ,644 $108,900 $186,900 Northeaster n Trucking 1,000 $90, $64,320 $154,320 Golden 500 $96, $56,820 $152,820 Golden 1,500 $96, $71,820 $167,820 Golden 2,500 $86,400 1, $86,820 $173,220 Golden 3,000 $78,000 1, $94,320 $172,320 Golden (old proposal) Golden (new proposal) 4,000 4,000 $72,000 2, $109,320 $181,320 $67,500 2, $109,320 $176,820

16 Key Points When selecting a mode of transportation, managers must account for inventory costs. Modes with high transportation cost can be justified if they result in significantly lower inventories

17 Trade-off between Transportation Cost and Customer Responsiveness Temporal aggregation is the process of combining orders across time Temporal aggregation reduces transportation cost because it results in larger shipments and reduces variation in shipment sizes However, temporal aggregation reduces customer responsiveness

18 Temporal aggregation Example Alloy Steel is a steel service center. All orders are shipped to customers using Carrier charges $100 for an order an LTL shippments are charged 0.01x as variable cost a TL (40,000 pounds) shipment charged $350 (additional) x: number of pounds of steel shipped on the truck. Currently Alloy Steel ships orders on the day they are received. Allowing for two days in transit, this policy allows Alloy to achieve a response time of two days. What is the cost advantage of increasing the response time to three or four days? Demands over two-week period: Monday Tuesday Wednesday Thursday Friday Saturday Sunday Week 1 19,790 17,470 11,316 26,192 20,263 8,381 25,377 Week 2 39,171 2,158 20,633 23,370 24,100 19,603 18,442

19 Temporal aggregation Example Two-day response Three-day Four-day response response Day Demand Quantity Quantity Cost ($) Cost ($) Shipped Shipped Quantity Cost ($) Shipped , , , , , , , , , , , , , , , , , , , , , , , , , ,00 Total Cost 4120, , ,21

20 Vehicle Routing Seperate and Single Origin and Destination Points Multiple Origin and Destination points Routing with a Coincident Origin/Destination Point

21 Seperate and Single O/D Points Determine the best path between origin and destination points over a network of routes Origin Amarillo A B E 90 minutes Oklahoma City I C F H D Note : All link times are in minutes 48 G 150 J Destination Fort Worth

22 Seperate and Single O/D Points Shortest route method is efficient for finding the minimal cost route Consider a time network between Amarillo and Fort Worth. Find the minimum travel time. The procedure can be paraphrased as: Find the closest unsolved node to a solved node Calculate the cost to the unsolved node by adding the accumulated cost to the solved node to the cost from the solved node to the unsolved node. Select the unsolved node with the minimum time as the new solved node. Identify the link. When the destination node is solved, the computations stop. The solution is found by backtracking through the connections made.

23 Seperate and Single O/D Points

24 Multiple origin and destination points 4 a Supplier A Supply Plant 1 Requirements = Supplier B Supply Plant 2 Requirements = Supplier C Supply Plant 3 Requirements = 300 a The transportation rate in $ per ton for an optimal routing between. supplier A and plant 1 Use Excel Solver to solve the problem!

25 Multiple origin and destination points Solving via Excel Solver

26 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. D Depot (a) Poor routing-- paths cross D Depot (b) Good routing-- no paths cross

27 Single Route solved as a Travelling Sales Person Problem Y coordinates 8 Y coordinates D D X coordinates (a) Location of beverage accounts and distribution center (D) with grid overlay X coordinates (b) Suggested routing pattern

28 Multi-Vehicle Routing and Scheduling A problem similar to the single-vehicle routing problem except that a number of restrictions are placed on the problem. Chief among these are: A mixture of vehicles with different capacities Time windows on the stops Pickups combined with deliveries Total travel time for a vehicle

29 Practical Guidelines for Good Routing and Scheduling 1. Load trucks with stop volumes that are in closest proximity to each other Stops D Depot (a) Weak clustering D Depot (b) Better clustering

30 Guidelines (Cont d) 2. Stops on different days should be arranged to produce tight clusters F T F T F F F Stop T T T F F T T T F F F F F T T T T T T F F D Depot (a) Weak clustering-- routes cross May need to coordinate with sales to achieve clusters D Depot (b) Better clustering

31 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)

32 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?

33 Sweep Method Solution Sweep direction is arbitrary Route #1 10,000 units 2,000 1,000 3,000 4,000 Route #3 8,000 units 2,000 3,000 3,000 2,000 1,000 2,000 2,000 Route #2 9,000 units Depot 2,000

34 The Savings Method for VRP Stop d A,0 A d 0,A A d 0,B d 0,A 0 0 d A,B Depot d B,0 B Depot d B,0 B Stop (a) Initial routing Route distance = d 0,A +d A,0 +d 0,B + d B,0 (b) Combining two stops on a route Route distance = d 0,A +d A,B +d B,0 Savings is better than Sweep method has lower average error

35 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

36 Next Class Midterm Exam The Exam will be held in the classroom 20 questions multiple choice 2 or 3 questions problem solving (notes/books are open)