INTRODUCTION TO
MANAGEMENT
SCIENCE, 13e
Anderson
Sweeney
Williams
Martin
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slides by
JOHN
LOUCKS
St. Edward’s
University
Slide 1
Chapter 6, Part A
Distribution and Network Models





Transportation Problem
Assignment Problem
Transshipment Problem
Shortest Route Problem
Maximum Flow Problem
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2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 2
Transportation Problems (p.262)





Origin or Source node
Destination node
arcs
Minimize the total transportation cost meeting the
supply availability and demand requirements.
Example (p.261)
• Production capacity of 3 plants
• Demand forecasts for 4 distribution centers
• Network representation (p.262)
• Transportation costs for each route.
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 3
Transportation Problem

Linear Programming Formulation
Using the notation:
xij = number of units shipped from
origin i to destination j
cij = cost per unit of shipping from
origin i to destination j
si = supply or capacity in units at origin i
dj = demand in units at destination j
continued
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 4
Transportation Problem

Linear Programming Formulation (p.267)
Min
m
n
 c x
i 1 j 1
n
x
j 1
m
x
i 1
ij ij
ij
 si
i 1, 2,
,m
Supply
ij
 dj
j 1, 2,
,n
Demand
xij > 0 for all i and j
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 5
Transportation Problems





LP formulation (p.264)
Output (p.264)
What is the solution?
How much is the total transportation cost
Is it dual solution or degenerate solution?
Variable
X11
X12
X13
X14
X21
X22
X23
X24
X31
X32
X33
X34
Value
3500.000
1500.000
0.000000
0.000000
0.000000
2500.000
2000.000
1500.000
2500.000
0.000000
0.000000
0.000000
Reduced Cost
0.000000
0.000000
8.000000
6.000000
1.000000
0.000000
0.000000
0.000000
0.000000
4.000000
6.000000
6.000000
Row
1
2
3
4
5
6
7
8
Slack or Surplus Dual Price
39500.00
-1.000000
0.000000
3.000000
0.000000
0.000000
0.000000
4.000000
0.000000
-6.000000
0.000000
-5.000000
0.000000
-2.000000
0.000000
-3.000000
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 6
Transportation Problems


If total supply is greater than the total demand?
• No change
If total supply is less than the total demand?
• Use dummy origin with 0 cost
example : Lexington demand is 2500
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 7
Transportation Problem

LP Formulation Special Cases
• The objective is maximizing profit or revenue:
Solve as a maximization problem.
• Minimum shipping guarantee from i to j:
(p.266)
xij > Lij
• Maximum route capacity from i to j:
xij < Lij
• Unacceptable route:
Remove the corresponding decision variable.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 8
Assignment Problem (p.268)





Assign m workers to m jobs.
Minimize the total assignment cost
Cost info (p.268)
Assignment problem is a special case of
transportation problem with all RHS are 1.
Network representation (p.269)
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 9
Assignment Problem

Linear Programming Formulation
Using the notation:
xij =
1 if agent i is assigned to task j
0 otherwise
cij = cost of assigning agent i to task j
continued
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©
2008 Thomson South-Western. All Rights Reserved
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Slide 10
Assignment Problem

Linear Programming Formulation (continued)
Min
m
n
 c x
i 1 j 1
n
x
j 1
m
x
i 1
ij ij
ij
 1 i 1, 2,
,m
Agents
ij
 1 j 1, 2,
,n
Tasks
xij > 0 for all i and j
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 11
Assignment Problem





Variable
X11
X12
X13
X21
X22
X23
X31
X32
X33
LP formulation (p.270)
Output (p.270)
What is the solution?
What is the total completion time?
Is the solution dual solutions or degenerate?
Value
0.000000
1.000000
0.000000
0.000000
0.000000
1.000000
1.000000
0.000000
0.000000
Reduced Cost
2.000000
0.000000
4.000000
1.000000
3.000000
0.000000
0.000000
1.000000
0.000000
Row Slack or Surplus Dual Price
1
26.00000
-1.000000
2
0.000000
0.000000
3
0.000000
0.000000
4
0.000000
2.000000
5
0.000000
-8.000000
6
0.000000
-15.00000
7
0.000000
-5.000000
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 12
Assignment Problem

LP Formulation Special Cases
•Number of agents exceeds the number of tasks:
Extra agents simply remain unassigned.
•Number of tasks exceeds the number of agents:
Add enough dummy agents to equalize the
number of agents and the number of tasks.
The objective function coefficients for these
new variable would be zero.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 13
Assignment Problem

LP Formulation Special Cases (continued)
•The assignment alternatives are evaluated in terms
of revenue or profit:
Solve as a maximization problem.
•An assignment is unacceptable:
Remove the corresponding decision variable.
•An agent is permitted to work t
n
x
j 1
ij
t
i 1, 2,
,m
tasks:
Agents
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 14
Assignment Problem

If there are two more tasks with completion times
(10, 18, 3) and (9, 14, 9)
Min=10*x11+15*x12+9*x13+9*x21+18*x22+5*x23+6*x31+14*x32+3*x33
+10*x14+9*x15+18*x24+14*x25+3*x34+9*x35;
x11+x12+x13
+x14 +x15 <= 1;
x21+x22+x23
+x24 +x25 <= 1;
x31+x32+x33
+x34 +x35 <= 1;
x11
+x21
+x31
+xd1 = 1;
x12
+x22
+x32
+xd2 = 1;
x13
+x23
+x33
+xd3 = 1;
x14
+x24
+x34
+xd4 = 1;
x15
+x25
+x35 +xd5 = 1;
xd1 +xd2 +xd3 +xd4 +xd5
<= 2;

If task 1, 2 should be assigned
remove xd1, xd2 in constraints 4 and 5.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 15
Assignment Problem

If there are two more tasks with completion times
Variable Value
Reduced Cost
(10, 18, 3) and (9, 14, 9)
Objective value = 17
X11
X12
X13
X21
X22
X23
X31
X32
X33
X14
X15
X24
X25
X34
X35
XD1
XD2
XD3
XD4
XD5
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
1.000000
0.000000
1.000000
1.000000
0.000000
0.000000
0.000000
1.000000
6.000000
4.000000
0.000000
9.000000
0.000000
0.000000
8.000000
1.000000
4.000000
0.000000
12.00000
5.000000
0.000000
3.000000
0.000000
0.000000
4.000000
3.000000
0.000000
Slide 16
Transshipment Problem




A special case of transportation problem with
transhipement nodes.
For transhipment nodes
Flow out = Flow in
 Flow out – Flow in = 0
Network representation (p.274)
Unit transportation cost (p.274, Table 6.5)
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 17
Transshipment Problem

Linear Programming Formulation
Using the notation:
xij = number of units shipped from node i to node j
cij = cost per unit of shipping from node i to node j
si = supply at origin node i
dj = demand at destination node j
continued
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 18
Transshipment Problem

Linear Programming Formulation (p.279)
Min
cx
ij ij
all arcs
s.t.

xij 

xij 
arcs out
ij
 si
Origin nodes i
x
ij
0
Transhipment nodes
arcs in
arcs out
x
arcs in
x
arcs in
ij


xij  d j Destination nodes j
arcs out
xij > 0 for all i and j
continued
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 19
Transshipment Problem







LP formulation (p.276)
Output (p.276)
What is the solution?
How much is the total transportation cost?
Modified Example (p.277, Figure 6.9)
Formulation for modified example (p.278)
Output (p.278)
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 20
Transshipment Problem

LP Formulation Special Cases
• Total supply not equal to total demand
• Maximization objective function
• Route capacities or route minimums
• Unacceptable routes
The LP model modifications required here are
identical to those required for the special cases in
the transportation problem.
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 21
Shortest-Route Problem (p.281)

Finding the shortest path in a network from one
node (or set of nodes) to another node (or set of
nodes).

All arcs are two-way (Figure 6.12)
Some arcs are two-way (Figure 6.13)

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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 22
Shortest-Route Problem

Linear Programming Formulation
Using the notation:
xij =
1 if the arc from node i to node j
is on the shortest route
0 otherwise
cij = distance, time, or cost associated
with the arc from node i to node j
continued
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 23
Shortest-Route Problem

Linear Programming Formulation (p.283)
Min
cx
ij ij
all arcs
s.t.

xij

xij 
 1 Origin node i
arcs out
arcs out
x
ij
0
x
ij
 1 Destination node j
Transhipment nodes
arcs in
arcs in
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 24
Shortest-Route Problem (p.281)





LP Formulation for Figure 6.13 (p.282)
Output (p.283)
What is the solution?
1–3–2–4–6
What is the total distance?
For a single source shortest-route problem,
there is Dijkstra’s algorithm.
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©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 25
Shortest-Route Problem (p.281)
Dijkstra’s algorithm
Start from the source
Fro each node
[min dist, previous node]
1. {1} 2 [25, 1]
3 [20, 1]***
2. {1, 3} 2 [23, 3]***
5 [26, 3]
3. {1, 3, 2} 4 [28, 2], 5 [26, 3]***, 6 [37, 2]
4. {1, 3, 2, 5} 4 [28, 2]***, 6 [32, 4]
5. {1, 3, 2, 5, 4} 6 [32, 4]***
6–4–2–3–1

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 26
Minimum spanning tree (p.281)
Connect all nodes with minimum total distance
 Prim’s algorithm
Start with any node
Connect the closest node to connected set
1. {2} 1 [25, 2], 3 [3, 2]**, 4 [5, 2], 6 [14, 2]
2. {2, 3} 1 [20, 3], 4 [5, 2]**,
5 [6, 3], 6 [14, 2]
3. {2, 3, 4} 1 [20, 3], 5 [4, 4]**,
6 [4, 4]**
4. {2, 3, 4, 5, 6} 1 [20, 3]**
5. {1, 2, 3, 4, 5, 6}
1 – 3, 5 – 4 – 6, 4 – 2, 3 – 2

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 27

Minimum spanning tree (p.281)
LP formulation
Min = 25*x12 +20*x13 +3*x23 +5*x24 +14*x26 +6*x35 +4*x45 +4*x46 +7*x56;
x12 +x13 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 5;
! maximum arcs is 5;
x13 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 4;
! connect 1-2;
x12
+x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 4;
! connect 1-3;
x12 +x13
+x24 +x26 +x35 +x45 +x46 +x56 >= 4;
! connect 2-3;
x12 +x13 +x23
+x26 +x35 +x45 +x46 +x56 >= 4;
! connect 2-4;
x12 +x13 +x23 +x24
+x35 +x45 +x46 +x56 >= 4;
! connect 2-6;
x12 +x13 +x23 +x24 +x26
+x45 +x46 +x56 >= 4;
! connect 3-5;
x12 +x13 +x23 +x24 +x26 +x35
+x46 +x56 >= 4;
! connect 4-5;
x12 +x13 +x23 +x24 +x26 +x35 +x45
+x56 >= 4;
! connect 4-6;
x12 +x13 +x23 +x24 +x26 +x35 +x45 +x46
>= 4;
! connect 5-6;
x24 +x26 +x35 +x45 +x46 +x56 >= 3;
! connect (123, 4, 5,
x13 +x23
+x26 +x35 +x45 +x46 +x56 >= 3;
! connect (1-2, 2-4),
x13 +x23 +x24
+x35 +x45 +x46 +x56 >= 3;
! connect (1-2, 2-6),
x12
+x23 +x24 +x26
+x45 +x46 +x56 >= 3;
! connect (1-3, 3-5),
x12 +x13
+x26 +x35 +x45 +x46 +x56 >= 3;
! connect (2-3, 2-4),
x12 +x13
+x24 +x26
+x45 +x46 +x56 >= 3;
! connect (2-3, 3-5),
x12 +x13
+x24
+x35 +x45 +x46 +x56 >= 3;
! connect (2-3, 2-6),
x12 +x13 +x23
+x26 +x35
+x46 +x56 >= 3;
! connect (2-4, 4-5),
x12 +x13 +x23
+x35 +x45
+x56 >= 3;
! connect (246, 1, 3,
x12 +x13 +x23 +x24
+x35 +x45 +x46
>= 3;
! connect (2-6, 5-6),
x12 +x13 +x23 +x24 +x26
+x46 +x56 >= 3;
! connect (3-5, 4-5),
x12 +x13 +x23 +x24 +x26
+x45 +x46
>= 3;
! connect (3-5, 5-6),
x12 +x13 +x23 +x24 +x26 +x35
>= 3;
! connect (456, 1, 2,
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
6);
(124,
(126,
(135,
(234,
(235,
(236,
(245,
5);
(256,
(345,
(356,
3);
3,
3,
2,
1,
1,
1,
1,
5,
4,
4,
5,
4,
4,
3,
6);
5);
6;
6);
6);
5);
6);
1, 3, 4);
1, 2, 6);
1, 2, 4);
Slide 28
Minimum spanning tree (p.281)

LP formulation (continued)
x26 +x35 +x45
x24 +x26
+x45
x24
+x35 +x45
+x26 +x35
+x35 +x45
+x24
+x35 +x45
+x24 +x26
+x24 +x26
+x45
+x26
+x35 +x45
+x24
+x45
+x35
+x24 +x26
x26
x35 +x45
x24
+x45
+x35
+x24 +x26
+x46
+x46
+x46
+x46
+x56
+x56
+x56
+x56
+x56
>= 2;
! connect (1234, 5, 6);
>= 2;
! connect (1235, 4, 6);
>= 2;
! connect (1236, 4, 5);
x13 +x23
>= 2;
! connect (1245, 3, 6);
x13 +x23
>= 2;
! connect (1246, 3, 5);
x13 +x23
+x46
>= 2;
! connect (1256, 3, 4);
x12
+x23
+x46 +x56 >= 2;
! connect (1345, 2, 6);
x12
+x23
+x46
>= 2;
! connect (1356, 2, 4);
x12 +x13
+x46 +x56 >= 2;
! connect (2345, 1, 6);
x12 +x13
+x56 >= 2;
! connect (2346, 1, 5);
x12 +x13
+x46
>= 2;
! connect (2356, 1, 4);
x12 +x13 +x23
>= 2;
! connect (2456, 1, 3);
x12 +x13 +x23
>= 2;
! connect (3456, 1, 2);
+x46 +x56 >= 1;
! connect (12345, 6);
+x56 >= 1;
! connect (12346, 5);
+x46
>= 1;
! connect (12356, 4);
x13 +x23
>= 1;
! connect (12456, 3);
x12
+x23
>= 1;
! connect (13456, 2);
x12 +x13
>= 1;
! connect (23456, 1);
! choose arcs from unselected nodes to any of selected nodes
! for nodes that can not be connected to any of selected nodes, choose all the
outgoing arcs.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 29
Minimum spanning tree (p.281)

LP output
Objective value:
36.00000
Variable
X12
X13
X23
X24
X26
X35
X45
X46
X56
Row
1
2
3
4
5
6
7
8
Value
0.000000
1.000000
1.000000
1.000000
0.000000
0.000000
1.000000
1.000000
0.000000
Slack or Surplus
36.00000
0.000000
1.000000
0.000000
0.000000
0.000000
1.000000
1.000000
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Reduced Cost
5.000000
0.000000
0.000000
0.000000
9.000000
1.000000
0.000000
0.000000
3.000000
Dual Price
-1.000000
-2.000000
0.000000
0.000000
-2.000000
0.000000
0.000000
0.000000
Slide 30
Maximal Flow Problem



Maximal Flow Problem (p.284)
The maximal flow problem is concerned with
determining the maximal volume of flow from one
node (called the source) to another node (called the
sink).
In the maximal flow problem, each arc has a
maximum arc flow capacity which limits the flow
through the arc.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 31
Example: Maximal Flow




A capacitated transshipment model can be developed
for the maximal flow problem.
We will add an arc from the sink node back to the
source node to represent the total flow through the
network.
There is no capacity on the newly added sink-tosource arc.
We want to maximize the flow over the sink-to-source
arc.
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 32
Maximal Flow Problem

LP Formulation
(as Capacitated Transshipment Problem)
• There is a variable for every arc.
• There is a constraint for every node; the flow out
must equal the flow in.
• There is a constraint for every arc (except the
added sink-to-source arc); arc capacity cannot be
exceeded.
• The objective is to maximize the flow over the
added, sink-to-source arc.
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2008 Thomson South-Western. All Rights Reserved
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Slide 33
Maximal Flow Problem

LP Formulation
(as Capacitated Transshipment Problem)
Max xk1
s.t.
(k is sink node, 1 is source node)
xij - xji = 0
(conservation of flow)
xij < cij
(cij is capacity of ij arc)
i
j
xij > 0, for all i and j
(non-negativity)
(xij represents the flow from node i to node j)
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 34
Maximal Flow Problem




LP Formulation (pp.284–285)
Computer output (p.286)
What is the solution?
What is the maximum flow?
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 35
A production and Inventory application





Problem (p.288)
to determine how many yards of carpet to produce each
quarter to minimize the total production and inventory
cost
Minimizing or maximizing?
What are the constraints?, How many?
Network representation (p.289)
What problem is this network representation?
Decision variable?
𝑥𝑖𝑗 : yards of carpets from node i to j.
𝑥15 , 𝑥26 , 𝑥37 , 𝑥48 : production in quarters 1,2,3,4
𝑥56 , 𝑥67 , 𝑥78 : inventory of quarter i
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 36
A production and Inventory application




LP formulation (p.289)
Computer output (p.290)
What is the solution?
What is the total production and inventory cost?
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 37
End of Chapter 6, Part A
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
©
2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 38