LESSON 6: SUPPLY CHAIN MANAGEMENT
Outline
• Supply Chain and Supply Chain Management
• Supply Chain Database and Planning
• Supply Chain Management and Competitive
Advantage
• Investment in Supply Chain
• Transportation Problem
1
Supply Chain
• A supply chain represents all the stages at which value is
added in producing and delivering a product or service from
suppliers (and their suppliers) to customers (and their
customers).
• A conceptual view of the supply chain network is shown on
the next slide. In practice, the supply chain network can be
quite large and complex.
2
Supply Chain
Customer
Customer
Customer
Distribution
Center
Customer
Distribution
Center
Manufacturer
Supplier
Supplier
Supplier
Supplier
Tier 1
Supplier
Tier
2
3
Supply Chain Management
• Supply chain management is the coordination of the following
functions and activities along the supply chain:
– Planning and managing of supply and demand
– Acquiring material
– Warehousing
– Inventory control and distribution
– Producing and scheduling the product or service
– Delivery and customer service
4
Supply Chain Database and Planning
A supply chain database contains information on flow of items
5
through the supply chain and cater to other planning decisions.
Supply Chain Management and Competitive
Advantage
• Competitive advantage may be obtained with a total
systems approach to managing flow of information,
materials and services along the supply chain
• Text p. 307 discusses how Wal-Mart obtained a
competitive advantage and won market-share over KMart with an investment in information technology
towards an efficient supply chain management.
6
Investment in Supply Chain
• For high profit margin (high tech products, innovative
products) products, it’s important to create a responsive
supply chain. The goals of investment are:
– Reduce stockout
– Reduce lead time along the supply chain
7
Investment in Supply Chain
• For low profit margin (staple products, functional products)
products, it’s important to create an efficient supply chain.
The goals of investment are:
– Reduce cost
– Increase resource utilization, minimize inventory,
– select vendors on the basis of cost and quality, design
products that can be produced depending on such
vendors
8
Investment in Supply Chain
Functional
Innovative
Products
Products
Efficient
SupplyChain
Match
Mismatch
Mismatch
Match
Responsive
Supply-Chain
9
Transportation Problem
• The transportation problem minimizes the total
shipping costs of transporting goods from m origins
(each with a supply si) to n destinations (each with
a demand dj), when the unit shipping cost from an
origin, i, to a destination, j, is cij.
• The network representation for a transportation
problem with two sources and three destinations is
given on the next slide.
10
Transportation Problem
• Network Representation
s1
1
c11
1
d1
2
d2
3
d3
c12
c13
c21 c22
s2
2
c23
SOURCES
DESTINATIONS
11
Transportation Problem
Example
The Riverside Disk Drive Corporation produces
Ambassador drives for personal computers in two
plants located in Waterloo and Toronto. Periodically,
shipments are made from these production facilities
to three warehouses located in Windsor, London, and
Ottawa.
12
Transportation Problem
Example
Over the next month, the production quantities of the
80 GB drives are expected to be (in thousands of
units)
Plant
Anticipated Production (000 units)
Waterloo
70
Toronto
110
So, the total production = 70+110 = 180 (000 units)
13
Transportation Problem
Example
The amounts shipped to the three warehouses will be
Warehouse
Shipment Quantity (000 units)
Windsor
35
London
55
Ottawa
90
So, the total shipment = 35+55+90 = 180 (000 units)
= total production.
As it is here, transportation problems generally require total
production (supply) = total shipment (demand).In case of
inequality a dummy plant or warehouse can be used.
14
Transportation Problem
Example
The unit costs for shipping 1,000 units from each plant
to each warehouse is given below:
Plant
Windsor
London
Ottawa
Waterloo
420
280
580
Toronto
990
640
820
Determine a pattern of shipping that minimizes total
transportation cost.
15
Transportation Problem
Example: Network Representation
35
Windsor
55
London
90
Ottawa
420
Waterloo
70
280
580
990 640
Toronto
110
820
SOURCES
DESTINATIONS
16
Transportation Problem
A Greedy Heuristic
• The transportation problem will be solved using a
greedy heuristic and Excel Solver.
• The heuristic is easy to use, but does not
guarantee an optimal solution. The heuristic
provides an intuition on the structure of the
problem.
• An optimal solution can be obtained by solving the
problem as a linear programming problem. Excel
Solver will be used to solve the linear programming
problem.
17
Transportation Problem
A Greedy Heuristic
• The greedy heuristic discussed here is an iterative
approach.
• In each iteration, a minimum unit cost
transportation alternative is chosen and the
maximum possible units is transported using that
alternative.
• Note that the maximum units to transport is actually
minimum of production available and shipment
required.
18
Transportation Problem
Example: Greedy Heuristic
Windsor
420
London
Ottawa
280
580
Waterloo
70
990
640
820
Toronto
110
90
55
35
The initial tableau shows all production volumes (70, 110),
shipments required (90, 55, 35) and all unit costs (in boxes)
19
Transportation Problem
A Greedy Heuristic
• Final solution and cost:
• From Waterloo transport
• 15 units to Windsor, cost = 15(420)
= $6,300
• 55 units to London, cost = 55(280)
= $15,400
• From Toronto transport
• 75 units to Windsor, cost = 75(990)
= $74,250
• 35 units to Ottawa, cost = 35(820)
= $28,700
• Total cost
=$124,650
• As you will see in the next few slides, linear programming
technique provides a better solution with a cost of
20
$113,100 when the problem is solved using Excel Solver.
Excel Solution
• Set up the spreadsheet as shown on the next slide.
• The cells with question marks are the decision variables.
Excel will find numbers to these cells.
• The cells with numbers are input cells.
• Blank cells B10, C10, D10, E8 and E9 are to be filled up
by formula. Cell B10 will contain total amount shipped to
Windsor as suggested by the Excel solution. So, the
formula for this cell is =B8+B9. Similarly, C10 and D10 will
contain total amount shipped to London and Ottawa
respectively. Cells E8 and E9 will contain total amount
shipped from Waterloo and Toronto respectively. All the
formula of these cells are listed on Slide 27.
21
Excel Solution
A
B
C
D
E
F
1 Costs
2
Windsor London Ottawa
3 Waterloo
420
280
580
4 Toronto
990
640
820
5
6 Shipment
7
Windsor London Ottawa Shipped Production
8 Waterloo
?
?
?
70
9 Toronto
?
?
?
110
10 Received
11 Demand
90
55
35
12
13 Total Cost
22
Excel Solution
Cell Formula
E8
=SUM(B8:D8)
E9
=SUM(B9:D9)
B10
=SUM(B8:B9)
C10
=SUM(C8:C9)
D10
=SUM(D8:D9)
B13
=SUMPRODUCT(B3:D4,B8:D9)
23
Excel Solution
• After the Excel spreadsheet is set up with input values,
formula and blank cells (or, ? cells) for the decision
variables,
– click Tools, Solver.
– complete the
• Solver Parameter dialog box,
• Add Constraint dialog box (to access this dialog
box, click Add button) and
• Solver Options dialog box (to access this dialog
box, click Options button) as shown in the next
three slides and
– click Solve button.
24
Excel Solution
25
Excel Solution
26
Excel Solution
27
Transportation Problem
A Greedy Heuristic
• As shown on the next slide, the Excel solution and costs
are as follows:
• From Waterloo transport
• 70 units to Windsor, cost = 70(420)
= $29,400
• From Toronto transport
• 20 units to Windsor, cost = 20(990)
= $19,800
• 55 units to London, cost = 55(280)
= $35,200
• 35 units to Ottawa, cost = 35(820)
= $28,700
• Total cost
=$113,100
• This solution is better than the $124,650 solution given by
28
the heuristic.
Excel Solution
A
B
C
D
E
F
1 Costs
2
Windsor London Ottawa
3 Waterloo
420
280
580
4 Toronto
990
640
820
5
6 Shipment
7
Windsor London Ottawa Shipped Production
8 Waterloo
70
0
0
70
70
9 Toronto
20
55
35
110
110
10 Received
90
55
35
11 Demand
90
55
35
12
13 Total Cost 113100
29
READING AND EXERCISES
Lesson 6
Reading:
– Section 6.1 pp. 305-312 (4th Ed.), pp. 294-303 (5th
Ed.)
Exercises:
– 6.1 pp. 319-320 (4th Ed.), p. 310 (5th Ed.)
30
LESSON 7: VEHICLE SCHEDULING
Outline
•
•
•
•
•
•
•
•
•
Vehicle Scheduling Problem
Savings Matrix Method
Identify Distance Matrix
Identify Savings Matrix
Rank Savings
Assign Customers to Vehicles
Sequence Customers within Routes
Construction and Improvement Procedures
Example
31
Vehicle Scheduling Problem
• Goods and items are delivered from one city to
another, one province to another and one country to
another. In this lesson, we shall discuss a procedure
used to deliver items by a fleet of vehicles.
• Suppose that a company owns several trucks. The
company needs to deliver items to various customers
within the Windsor City. The manager takes a map of
the city, plots all the customers on the map and finds
that the customers are scattered over the entire
Windsor area.
32
Vehicle Scheduling Problem
• The manager finds that two vehicles will provide a
sufficient capacity for all orders. One vehicle will
serve the East side and the other the West side. To
find out if each vehicle will be able to finish all the
deliveries before their due dates, the manager first
needs a possible route of each vehicle. Different
route provides a different speed limit and risk levels.
The manager draws a fast and safe route for each
vehicle and estimates delivery times. It turns out that
a large number of customers would receive their
orders late.
33
Vehicle Scheduling Problem
• The manager then repeats the process with three
vehicles, one to the South side, one to the East side,
and one to the West side. This plan appeared
suitable to meet all the due dates.
• Such problems may appear in case of e.g., an on-line
grocery stores such as http://www.peapod.com. See
the next slide for a picture of their website. The
orders are taken 24 hours on-line. When the
manager starts working in the morning, the manager
has to decide number of vehicles and their routes.
34
35
Vehicle Scheduling Problem
• Assume that
– There are orders from 5 different customers
– There are 2 trucks each capable of carrying 200
units
36
Vehicle Scheduling Problem
• The manager must solve two sub-problems
1. Split the city into several smaller regions, each of
which will be served by one vehicle. This can be
done by considering a customer first, assigning
the customer to a vehicle, and then assigning
other nearby customers to the same vehicle. So,
this sub-problem will be called assigning
customers to vehicles.
2. Sequence customers served by the same vehicle.
37
Vehicle Scheduling Problem
• The objectives of the vehicle scheduling problem can
be many.
• Following are some examples:
– Minimize number of vehicles required
– Minimize total distance traveled
– Minimize total travel time
– Minimize cost
• We shall consider the problem with the objective of
minimizing total distance.
38
Vehicle Scheduling Problem
X Coordinate
W
1
2
3
4
5
0
0
6
7
9
15
Y Coordinate
0
12
5
15
12
3
Order Size
48
60
43
92
80
Assume that the customer locations and order sizes
are as shown above. The locations are plotted on the
39
next slide.
Vehicle Scheduling Problem
3
4
1
5
2
W
Location of Warehouse and Customers
The customer locations shown on the previous slide
are plotted above.
40
Savings Matrix Method
•
Following are the steps of the Savings Matrix
Method:
1. Identify distance matrix
2. Identify savings Matrix
3. Rank savings
4. Assign customers to vehicles
5. Sequence customers within routes
41
Identify Distance Matrix
•
First, the Euclidean distances are computed. The
formula and a sample computation is shown below.
The other distances are computed similarly and
shown on the next slide.
Dist( A, B ) 
x A  xB 2   y A  y B 2
Example :
Dist(1,2) 
42
Identify Distance Matrix
Warehouse
Customer 1
Customer 2
Customer 3
Customer 4
Customer 5
Distance Matrix
W Cust 1 Cust 2 Cust 3 Cust 4 Cust 5
0 12.0
7.8 16.6 15.0
15.3
0
9.2
7.6
9.0
17.5
0 10.0
7.6
9.2
0
3.6
14.4
0
10.8
0
43
Identify Savings Matrix
•
•
•
Instead of serving two different customers by two
different vehicles, if a single vehicle is used to
serve both the customers, then some traveling
distance is saved.
Savings S(A,B) represents the savings in traveling
distance obtained by assigning Customers A and B
to the same vehicle instead of assigning them to
two different vehicles.
The capacity constraint is not considered at the
time of computation of savings. In other words,
when computing savings, assume an infinite
capacity of the vehicles.
44
Identify Savings Matrix
•
The savings are computed for all pairs of
customers using the data from the distance matrix.
The formula and a sample computation is shown
below. The other savings are computed similarly
and shown on the next slide.
S ( A, B)  Dist( A,W )  Dist( B, W )  Dist( A, B)
Example :
S (1,2) 
45
Identify Savings Matrix
Savings Matrix
Cust 1 Cust 2 Cust 3 Cust 4 Cust 5
Customer 1
0 10.6 20.9 18.0
9.8
Customer 2
0 14.3 15.2 13.9
Customer 3
0 27.9 17.4
Customer 4
0 19.5
Customer 5
0
46
Rank Savings
•
•
•
•
•
The next step is to rank the savings.The idea is to
merge those two customers to the same vehicle,
whose merging gives the highest savings.
The savings are ranked from high to low.
From the savings matrix shown on the previous
slide, the highest savings of 27.9 is obtained by
merging Customers 3 and 4 to the same vehicle.
Next highest savings of 20.9 is obtained by
merging Customers 1 and 3 to the same vehicle.
Similarly the other savings are ranked and shown
on the next slide.
47
Rank Savings
Savings Matrix
Cust 1 Cust 2 Cust 3 Cust 4 Cust 5
Customer 1
0 10.6 20.9 18.0
9.8
Customer 2
0 14.3 15.2 13.9
Customer 3
0 27.9 17.4
Customer 4
0 19.5
Customer 5
0
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
48
Assign Customers to Vehicles
Next, merge
the customers.
The pair giving
the highest
savings is
merged first if
the capacity is
available.
3
Customer
4
1
5
2
1
2
3
4
5
Order
Size
48
60
43
92
80
W
Location of Warehouse and Customers
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
49
Assign Customers to Vehicles
To merge the
lowest ranked
pair (3,4), the
capacity
1
required =
43+92= 135 <
200 = capacity
available. So,
merge 3 and 4. W
3
Customer
4
5
2
1
2
3
4
5
Order
Size
48
60
43
92
80
Location of Warehouse and Customers
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
50
Assign Customers to Vehicles
To merge the
next pair (1,3),
capacity
required =
43+92+40=
175 < 200 =
capacity
available. So,
merge 1 and 3
(and 4).
3
Customer
4
1
5
2
1
2
3
4
5
Order
Size
48
60
43
92
80
W
Location of Warehouse and Customers
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
51
Assign Customers to Vehicles
Merging (4,5),
(3,5), (2,4) and
(2,3) requires
more capacity
than available.
The pair (1,4)
is already
merged. So,
the pairs are
crossed out.
3
Customer
4
1
5
2
1
2
3
4
5
Order
Size
48
60
43
92
80
W
Location of Warehouse and Customers
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
52
Assign Customers to Vehicles
The next pair
(2,5) are
Order
3
merged and
Customer
Size
assigned to a
1
48
4
1
new vehicle as
2
60
the capacity
3
43
available =
4
92
5
200 > 60 + 80
2
5
80
= 140 =
W
capacity
required. Location of Warehouse and Customers
Rank
(3,4)
(2,4)
(1,3)
(2,3)
(4,5)
(2,5)
(1,4)
(1,2)
(3,5)
(1,5)
53
Sequence Customers
• The next step is sequencing customers assigned to
the same vehicle. A question is in what sequence will
the first vehicle visit customers 1, 3 and 4 and return
to the warehouse? Similarly, another question is in
what sequence will the other vehicle visit customers 2
and 5.
• This problem is popularly called the traveling
salesman problem.
• We shall use the nearest neighbor rule which states
that always visit the customer that is nearest.
54
Sequence Customers
First, consider the problem of
sequencing customers 1, 3 and
1
4 who are assigned to the
same vehicle. The relevant
distances are copied from the
distance matrix and shown
W
below.
3
4
5
2
Location
Distance Matrix
W Cust 1 Cust 3 Cust 4
Warehouse
0 12.0 16.6 15.0
Customer 1
0
7.6
9.0
Customer 3
0
3.6
Customer 4
0
55
Sequence Customers
Among 1, 3 and 4 is 1 is the
nearest to Warehouse. So, the
1
vehicle will first travel from
Warehouse to customer 1. The
row (=from) corresponding to
Warehouse and the column
W
(=to) corresponding to
customer 1 are crossed out.
3
4
5
2
Location
Distance Matrix
W Cust 1 Cust 3 Cust 4
Warehouse
0 12.0 16.6 15.0
Customer 1
0
7.6
9.0
Customer 3
0
3.6
Customer 4
0
56
Sequence Customers
Between 3 and 4 is 3 is the
nearest to 1. So, the vehicle will
1
travel from 1 to 3. The row
(=from) corresponding to
customer 1 and the column
(=to) corresponding to
W
customer 3 are crossed out.
3
4
5
2
Location
Distance Matrix
W Cust 1 Cust 3 Cust 4
Warehouse
0 12.0 16.6 15.0
Customer 1
0
7.6
9.0
Customer 3
0
3.6
Customer 4
0
57
Sequence Customers
The only possible tour is then
W-1-3-4-W. Next, consider the
1
problem of sequencing
customers 2 and 5 who are
assigned to the same vehicle.
The relevant distances are
copied from the distance matrix W
and shown below.
3
4
5
2
Location
Distance Matrix
W Cust 2 Cust 5
Warehouse
0
7.8 15.3
Customer 2
0
9.2
Customer 5
0
58
Sequence Customers
Between 2 and 5 is 2 is the
nearest to the Warehouse. So,
1
the vehicle will travel from the
Warehouse to customer 2. The
only tour is then W-2-5-W.
Note: due to symmetry both W2-5-W and W-5-2-W have the W
same distance traveled.
3
4
5
2
Location
Distance Matrix
W Cust 2 Cust 5
Warehouse
0
7.8 15.3
Customer 2
0
9.2
Customer 5
0
59
Construction and Improvement Procedures
• The nearest neighbor rule just discussed is a tour
construction procedure which can construct a tour
when there is no tour.
• The nearest neighbor rule is only a heuristic and
does not guarantee optimality. The tour obtained by
the heuristic may provide improvement opportunities.
• If a tour intersects its own path, the tour can be
improved. An improvement procedure will be
discussed now.
60
Construction and Improvement Procedures
• For example, consider the
locations and the tour
shown on the right.
• From the Warehouse, 1 is
the nearest. From 1, 2 is
the nearest, etc. So, the
nearest neighbor rule
produces the tour W-1-2-34-W.
• However, the tour
intersects itself. The arc
(1,2) intersects arc (4,W).
4
3
1
2
W
61
Construction and Improvement Procedures
•
•
The improvement
procedure has three
steps.
Step 1: Remove the
intersecting arcs. The
result is two disjointed
paths
4
3
1
2
W
62
Construction and Improvement Procedures
•
Step 2: Arbitrarily choose
one of the two disjointed
paths and reverse the
path. In this picture, 2-3-4
is reversed to get 4-3-2.
One could as well
reversed 1-W to W-1.
4
3
1
•
Step 3: There is only one
way to get a tour from the
two resulting paths.
Construct the tour.
(continued..)
2
W
63
Construction and Improvement Procedures
•
•
Step 3 continues: For
example, here the tour is
constructed by adding
arcs (1,4) and (2,W)
Note: The resulting tour
may include a new
intersection. In such a
case, apply the procedure
again!.
4
3
1
2
W
64
READING AND EXERCISES
Lesson 7
Reading:
– Section 6.6 pp. 325-330 (4th Ed.), pp. 315-320 (5th
Ed.)
Exercises:
– 6.16 p. 331 (4th Ed.), p. 321 (5th Ed.)
65