International Operations
Management
MGMT 6367
Lecture 08
Instructor: Yan Qin
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
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.”
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).
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.
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.
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 Rate,
Service
$/unit
Transit time, No. of shipments
Days
per year
Rail
0.10
21
10
Piggyback
0.15
14
20
Truck
0.20
5
20
Air
1.40
2
40
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?
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
Example: Formulas
4. In-transit inventory holding cost
= Unit inventory holding cost at plant * Average
inventory level in transit * transit time
= ICDT/365
Example: Calculation

For Rail, which is the benchmark,
◦ Transit time 𝑇𝑅 = 21 days (given)
◦ The number of shipments per year = 10 (given)
◦ Average inventory level 𝑄𝑅 = 100,000 units (given)

Now how to calculate the average inventory level
under another transport mode in this case?
Example: Calculations (Cont.)

Here is the formula:
The average inventory level under another transport mode
= 𝑄𝑅 * (# of shipments for rail / # of shipments under
another mode) * (100 − (𝑇𝑅 - Transit time under another
mode))%

This is because it has been assumed in the problem that
1. The average inventory level decreases linearly as the
number of shipments increases; and
2. The average inventory level can be reduced by 1 % for
each day saved from the current 21 days.
Example: Calculations (Cont.)

Then for Piggyback,
◦ Transit time = 14 days (given)
◦ The number of shipments per year = 20 (given)
◦ Average inventory level
𝑄𝑃 = 𝑄𝑅 * (10/20) * (100- (21 – 14))%
= 𝑄𝑅 * 0.5 * 93%
= 46,500 units
Example: Calculations (Cont.)

For Truck,
◦ Transit time = 5 days (given)
◦ The number of shipments per year = 20 (given)
◦ Average inventory level 𝑄𝑇 = 42,000 units

For Air,
◦ Transit time = 2 days (given)
◦ The number of shipments per year = 40 (given)
◦ Average inventory level 𝑄𝐴 = 20,250 units
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*(700,00 0.30*$30*
0/365)*21
100,000
0.30*($30+$0.10)
* 100,000
PiggyBack
$0.15*700,000
0.30*$30*(700,00 0.30* $30 *
0/365)*14
46,500
0.30*($30 +
$0.15)*46,500
Truck
$0.20*700,000
0.30*$30*(700,00 0.30* $30 *
0/365)*5
42,000
0.30 * ($30 +
0.20) * 42,000
Air
$1.40*700,000
0.30* $30 *
(700,000/365)*2
0.30 * ($30 +
$1.40) * 20,250
0.30 * $30 *
20,250
Example: Calculations (Cont.)


Now let’s compare the total costs for the four modes:
Rail
PiggyBack
Truck
Air
$2,235,466
$1,185,737
$984,821
$1,387,526
Therefore, the company should use trucks to transport
the products since this mode leads to the lowest total
cost.
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.
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.
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?
Example: Calculation
Mode
Sales
Gross
(Supplier A) 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.
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
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.
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.
Simple Route Planning methods

“Sweep” method
◦ Used when there is a single origin point and multiple
destinations in the transportation network.
Route 1
Destinations
Destinations
Origin
Route 2
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.
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.
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
2.0
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
Depot
-
Farm 1
27
-
Farm 2
15
21
-
Farm 3
24
51
34
-
Farm 4
27
39
18
30
-
Farm 5
28
27
13
41
14
-
Farm 6
29
12
14
53
29
10
Farm 6
-
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
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.
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 www.wikipedia.org
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.
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.
Example: Dijkstra's

Check out this interactive example: http://optlabserver.sce.carleton.ca/POAnimations2007/DijkstrasAlgo.html
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.
Example

A Multi-origin and Multi-destination Network:
Plant 1
Warehouse 1
400 units
300 units
Plant 2
100 units
200 units
Plant 3
300 units
400 units
Warehouse 2
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.
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
◦ http://en.wikipedia.org/wiki/Travelling_salesman_problem
 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.
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.
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.
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
Poor routing – Path cross
Start point
Good routing
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.
Next Week

Distribution network
◦ Importance of distribution channels
◦ Pros and Cons of using channel members
◦ Two key decisions in designing distribution channels