1. International Operations Management
MGMT 6367
Lecture 08
Instructor: Yan Qin

2. Outline Transportation Decisions Transport Service Selection
Basic cost trade-offs
Competitive considerations
Consolidation of Deliveries
Vehicle Routing
Separate single-origin and single-destination networks
Multiple-origin and multiple-destination networks
Coincident origin-and-destination-point networks
Principles for Vehicle Routing and Scheduling

3. Transport Service Selection
The selection of a mode of transportation or service depends on service characteristics such as:
Freight rates;
Transit time;
Transit time variability;
Loss, damage, claims processing, and tracing;
Shipper market considerations;
Carrier considerations.
Source: Michael McGinnis, “The relative importance of cost and service in freight transportation choices: before and after deregulation.”

4. Basic Cost Selection
When transportation service is not used to provide a competitive advantage, the best service choice is found by trading off the cost of using a particular transport service with the indirect cost of inventory associated with the performance of the selected mode.
Transit time and Transit time variability affect both the shipper’s and the carrier’s inventory levels as well as the amount of inventory that is in transit between the origin(s) and the destination(s).

5. Example: Cost Tradeoff
Problem Description: Suppose a company stocks finished-goods inventory at the East Coast plant and ships to the company-owned West Coast warehouse by way of common carriers. Rail is currently used to ship between the East Coast plant and the West Coast warehouse. The current transit time is 21 days. And there is an average inventory level of 100,000 units at each of the two stocking points, the plant and the warehouse. The inventory carrying cost per unit per year is I = 30% of the corresponding unit inventory value.

6. Example: (Cont.)
Suppose the unit production cost is C = $30 in this case. Then the unit inventory value of the stock at the plant is C = $30 and the unit inventory value of the stock at the warehouse is (C + the unit freight rate). We assume that both the number of shipments per year and the transit time have an impact on the average inventory level (Q) at the two stocking points. The average inventory level decreases linearly as the number of shipments increases. Specifically, the average inventory level halves as the number of shipments per year doubles. The average inventory level can also be reduced by 1% for every day that transit time can be reduced from the current 21 days.

7. Example: (Cont.)
Suppose the annual demand is 700,000 units. Procurement costs and transit-time variability are assumed negligible. The transportation services available to the company and the respective transportation performances are summarized as follows:
Transportation Service
Rate, $/unit
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

8. Example: (Cont.)
The entire annual demand (D) spends some time in transit; this fraction of the year is represented by T/365 days, where T is the transit time. The annual carrying cost of the in-transit inventory is ICDT/365. The company wishes to select the mode of transportation that will minimize the total cost, which can be expressed as: The total cost = Freight transportation cost + Inventory holding cost at plant + Inventory holding cost at warehouse + In-transit inventory holding cost But how to calculate each of the four cost terms involved in this example?

9. Example: Formulas 1. Transportation cost = Freight Rate * D
2. Inventory holding cost at plant
= Unit inventory holding cost at plant * Average inventory level at plant
= (I*C)*Q
3. Inventory holding cost at warehouse
= Unit inventory holding cost at warehouse * Average inventory level at warehouse
= I*(C + Freight Rate) * Q

10. Example: Formulas 4. In-transit inventory holding cost
= Unit inventory holding cost at plant * Average inventory level in transit * transit time
= ICDT/365

11. Example: Calculation For Rail, which is the benchmark,
Transit time 𝑇 𝑅 = 21 days (given)
The number of shipments per year = 10 (given)
Average inventory level 𝑄 𝑅 = 100,000 units (given)
Now how to calculate the average inventory level under another transport mode in this case?

12. Example: Calculations (Cont.)
Here is the formula:
The average inventory level under another transport mode
= 𝑄 𝑅 * (# of shipments for rail / # of shipments under another mode) * (100 − (𝑇 𝑅 - Transit time under another mode))%
This is because it has been assumed in the problem that
The average inventory level decreases linearly as the number of shipments increases; and
The average inventory level can be reduced by 1 % for each day saved from the current 21 days.
In the EOQ situation, Q*N = D/2
Simple Vs. Compound

13. Example: Calculations (Cont.)
Then for Piggyback,
Transit time = 14 days (given)
The number of shipments per year = 20 (given)
Average inventory level
𝑄 𝑃 = 𝑄 𝑅 * (10/20) * (100- (21 – 14))%
= 𝑄 𝑅 * 0.5 * 93%
= 46,500 units

14. Example: Calculations (Cont.)
For Truck,
Transit time = 5 days (given)
The number of shipments per year = 20 (given)
Average inventory level 𝑄 𝑇 = 42,000 units
For Air,
Transit time = 2 days (given)
The number of shipments per year = 40 (given)
Average inventory level 𝑄 𝐴 = 20,250 units

15. Example: Calculations (Cont.)
Cost Type:
Transportation
In-Transit
Plant inventory
Warehousing
Formula
Rate * D
I*C*(D/365)*T
I*C*Q
I*(C+Rate)*Q
Rail
$0.10*700,000
0.30*$30*(700,000/365)*21
0.30*$30* 100,000
0.30*($30+$0.10)* 100,000
PiggyBack
$0.15*700,000
0.30*$30*(700,000/365)*14
0.30* $30 * 46,500
0.30*($30 + $0.15)*46,500
Truck
$0.20*700,000
0.30*$30*(700,000/365)*5
0.30* $30 * 42,000
0.30 * ($ ) * 42,000
Air
$1.40*700,000
0.30* $30 * (700,000/365)*2
0.30 * $30 * 20,250
0.30 * ($30 + $1.40) * 20,250

16. Example: Calculations (Cont.)
Now let’s compare the total costs for the four modes:
Therefore, the company should use trucks to transport the products since this mode leads to the lowest total cost.
Rail
PiggyBack
Truck
Air
$2,235,466
$1,185,737
$984,821
$1,387,526

17. Competitive Considerations
A rational buyer may be willing to offer more business to a supplier with the preferred transport service, as the transport service usually has an impact on the buyer’s inventory level and/or operating schedule.
Transportation service thus becomes part of a company’s competitive consideration.
When a buyer in a supply chain purchases goods from more than one supplier, the transport service offered by the supplier, as well as price, influences the buyer’s supplier selection.

18. Example: Patronage
Suppose an appliance manufacturer purchases 3,000 cases of plastic parts valued at $100 per case from two suppliers. Purchases are currently divided equally between the suppliers. Each supplier uses rail transport and achieves the same average delivery time.
However, for each day that a supplier can reduce in the average transit time, the manufacturer is willing to shift 5% of its total purchase, or 150 cases, to the supplier offering the premium service. A supplier earns a margin of 20% on each case before transportation cost is considered.
Supplier A is considering whether it is worthwhile to upgrade the transport service.

19. Example: (Cont.)
Mode
Transport Rate
Delivery Time
Rail
$2.50 / case
7 days
Truck
$6.00 / case
4 days
Air
$10.35 / case
2 days
Question: Which mode should Supplier A choose to maximize its profit?

20. Example: Calculation
Mode
Sales (Supplier A)
Gross Profit -
Transport cost =
Net Profit
Rail
1,500
$20 * 1,500
$2.50 * 1,500
$26,250.00
Truck
1,950
$20 * 1,950
$6.00 * 1,950
$27,300.00
Air
2,250
$20 * 2,250
$10.35 * 2,250
$21,712.50
Based on the calculation results, Supplier A should switch to Truck to gain 450 more units in sales from the manufacturer in order to maximize its profit.

21. Consolidation of deliveries
Consolidation of deliveries: coordinating small consignments into larger flows to benefit from the economies of scale effect in transportation.
It can be carried out in the following ways:
Larger deliveries to storage points
Fixed delivery days: Not suitable when there is high expectation of customer service
Balanced flows
Milk Runs
Consolidated distribution

22. Consolidation of deliveries (Cont.)
Breakpoint distribution: Large loads are broken down at a breakpoint.
Hub-and-Spoke System: A large number of shipments are broken down and consolidated based on destination at a hub.

23. Vehicle Routing
To reduce transportation costs and to improve customer service, finding the best paths that a vehicle should follow through a network of roads, rail lines, shipping lanes, or air navigational routes that will minimize time or distance is a frequent decision problem, which is called Shortest Path problem.

24. Simple Route Planning methods
“Sweep” method
Used when there is a single origin point and multiple destinations in the transportation network.
Route 1
Origin
Destinations
Route 2
Start “sweeping” clockwise in relation to the origin point and continue until the vehicle is full or maximum driving time is reached.

25. Simple Route Planning methods
Savings Matrix method (Will not be tested)
Step 1: Identify Full-Truck-Load deliveries. Demand fulfilled with FTL is excluded from the following iterative route planning procedure.
Step 2: Develop a distance matrix. The distances can be the geographical distances or the costs to travel from the origin to each destination and from one destination to another.
Step 3: Develop a savings matrix. For each pair of destinations, say Destination A and Destination B, the saving from consolidating shipments to A and B = Distance from origin to A + Distance from origin to B – Distance between A and B.

26. Simple Route Planning methods
Savings Matrix method (Cont.)
Step 4: Rank the pairs of destinations based on the savings from the highest to the lowest. Start consolidating by combining shipments that result in highest savings and stop when there is no positive savings resulted.

27. Example: Savings matrix method
The Farmer Association distributes fodder and other supplies from a depot to six farms (customers). They want to use the savings matrix method to determine a route schedule.
Farm
Quantity (Tons) 1
1.2 2
2.0 3
1.8 4
1.5 5
2.5 6

28. Example (Cont.)
Consider the vehicle capacity restriction of 12 tons. No maximum driving time limitation, however.
Distances in kilometers between the depot and farms are:
Depot
Farm 1
Farm 2
Farm 3
Farm 4
Farm 5
Farm 6 - 27 15 21 24 51 34 39 18 30 28 13 41 14 29 12 53 10

29. Vehicle Routing problems
There are three types of vehicle routing (shortest path) problems:
Separate single-origin and single-destination problems
Multiple-origin and multiple-destination problems
Coincident origin- and destination-point problems

30. Separate single-origin and single-destination
We will introduce Dijkstra's Algorithm that can be used to solve for an optimal route for a given transportation and/or distribution network.
PC*Miller and IntelliRoute are examples of commercial software for finding the most desired routes.

31. Dijkstra's Algorithm
Consider a network consisting of nodes and arcs, where the nodes may represent the locations of various physical facilities and the arcs represent the costs between nodes.
Then Dijkstra’s algorithm can be described as follows:
Call the node representing the origin point the initial node. Let the distance of node Y be the distance from the initial node to Y. Dijkstra's algorithm first assigns some initial distance values to each node and then tries to improve them step by step.
* Modified based on contents from

32. Steps of Dijkstra's Algorithm
Step 1: Assign to every node a tentative distance value: set it to zero for our initial node and to infinity for all other nodes. Step 2: Mark all the nodes as unvisited. Set the initial node as current. Create a set consisting of all the unvisited nodes. Call the set the Unvisited set. Step 3: For the current node, consider all the unvisited nodes connected to it via exactly one arc. Calculate and update the neighbors’ tentative distances.

33. Steps of Dijkstra's Algorithm
Step 4: When we are done considering all the neighbors of the current node, mark the current node as visited and remove it from the Unvisited set. A visited node will never be checked again and its distance recorded now is final and minimal. Step 5: Pick a new current node and repeat Step 3 and Step 4. The new current node should be the node with the lowest distance value in the Unvisited set. Step 6: Stop when the Unvisited set is empty.

34. Example: Dijkstra's
Check out this interactive example: server.sce.carleton.ca/POAnimations2007/DijkstrasAlgo.html

35. Multi-origin and Multi-destination networks
When there are multiple source points that serve multiple destination points, there is a problem of assigning destinations to sources as well as finding the best route between them.
A simple problem can be solved manually using Linear Programming. Logware is one of the common software for solving this type of problem.

36. Example A Multi-origin and Multi-destination Network: Plant 1
Warehouse 1
400 units
300 units
Plant 2
100 units
200 units
Warehouse 2
300 units
Plant 3
400 units

37. Coincident Origin-and-Destination-point
The objective is to find the sequence in which the points should be visited to minimize the total travel time or distance.
Examples include:
Beverage delivery to bars and restaurants
Home appliance repair, service, and delivery
School bus routing
Newspaper delivery
Wholesale distribution from warehouses to retailers.

38. Coincident Origin-and-Destination-point
This type of routing problem is called “travelling salesman” problem. The classic description is the following:
Salesperson has to visit n cities and return to starting city;
Find a tour of the cities so that
Each city is visited only once;
The total distance traveled is minimized.
It is usually impractical to solve a “travelling salesman” problem manually when there are many points included.

39. Vehicle routing and scheduling
Vehicle routing and scheduling is an extension of the basic vehicle routing problem.
Realistic restrictions are now included such as:
Each stop may have volume to be picked up as well as delivered;
Multiple vehicles may be used having different capacity limitations;
A maximum total driving time is allowed on a route;
Some stops may allow pickup and deliveries only at certain times;
Drivers may be allowed to take short breaks.

40. Principles for Good Routing and Scheduling
Load trucks at stop points that are in the closest proximity to each other. Form clusters but avoid overlaps.
Stop point to be served on different days should be arranged to produce tight clusters.
Build routes beginning with the farthest stop from the origin.

41. Principles for Good Routing and Scheduling
The sequence of stops on a truck route should form a teardrop pattern. Stops should be sequenced so that no route paths cross.
Start point
Start point
Poor routing – Path cross
Good routing

42. Principles for Good Routing and Scheduling
The most efficient routes are built using the largest vehicles available, as using a vehicle large enough to handle all stops in one rout will minimize total distance, or time, traveled to serve the stops.
Pickups should be mixed into delivery routes.
A stop point that is remote from a route cluster is a good candidate for an alternate means of delivery, such as using small trucks or outsource.
Narrow stop point time windows should be avoided.

43. Next Week Distribution network Importance of distribution channels
Pros and Cons of using channel members
Two key decisions in designing distribution channels