Finite Mathematics
Chapter 10
Name ________________________________ Date ______________ Class ____________
Section 10-1 Strictly Determined Games
Goal: To solve problems involving game theory
Locating Saddle Values:
1. Circle the minimum value in each row (it may occur in more than one place)
2. Place squares around the maximum value in each column (it may occur in more
than one place)
3. Any entry with both a circle and a square around it is a saddle value.
Definition:
Strictly determined matrix games
A matrix game is said to be strictly determined if it has a saddle value. The optimal
strategies would be for player R to choose the row containing the saddle value and
player C to choose the column containing the saddle value. The game is fair if the
saddle value is zero.
For Problems 1–6
a) Determine if the game is strictly determined or nonstrictly determined.
If the game is strictly determined then answer the following:
b) Locate the saddle value.
c) Is the game fair?
d) Find the optimal strategies for R and C.
e) Find the value of the game.
é- 2 4 ù
1. ê
ú
ë 6 - 4û
é2 3 ù
2. ê
ú
ë1 - 4 û
Circles would be placed around –2 and –4.
Squares would be placed around 6 and 4.
Since there is no number with both a
circle and a square, it is not strictly
determined.
Circles would be placed around 2 and –4.
Squares would be placed around 2 and 3.
Since the 4 would have a circle and a square,
it is strictly determined. The saddle value is 2.
The game is not fair (SV=2). Optimal strategy is
row 1 column 1 and the value of the game is 2.
10-1
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Finite Mathematics
Chapter 10
é1 0 2ù
4. ê 3 2 - 1ú
ê
ú
ëê- 1 2 0 ú
û
é0 - 5ù
3. ê
ú
ë1 2 û
Circles would be placed around –5 and 1.
Squares would be placed around 1 and 2.
Since the 1 would have a circle and a
square, it is strictly determined. The saddle
value is 1. The game is not fair. Optimal
strategy is row 2 column 1 and the value
of the game is 1.
é - 4 1 1ù
5. ê - 3 4 7 ú
ê
ú
êë- 10 3 5ú
û
Circles would be placed around 0, –1 and –1.
Squares would be placed around 3, 2, and 2.
Since there is no number with both a circle
and a square, it is not strictly determined.
é 3 0 - 4 - 3ù
6. ê 2 3 1 2 ú
ê
ú
êë- 4 2 - 1 3 ú
û
Circles would be placed around –4, –3, –10.
Squares would be placed around –3, 4, 7.
Since the –3 would have a circle and a
square, it is strictly determined. The saddle
value is –3. The game is not fair. Optimal
strategy is row 2 column 1 and the value
of the game is –3.
Circles would be placed around –4, 1, –4.
Squares would be placed around 3, 3, 1, 3.
Since the 1 would have a circle and a square,
it is strictly determined. The saddle value is
1. The game is not fair. Optimal strategy is
row 2 column 3 and the value of the game
is 1.
7. Two computer companies, Spitfire and Computer World, are each planning to open a
new store in a large metropolitan area. Each must decide whether to open the new store in a
downtown location or in a suburban mall. The following matrix illustrates the gain or loss in
revenue based on the decisions made by the two companies. The positive numbers indicate a
gain in revenue for Spitfire over Computer World and negative numbers in the matrix
indicate a loss in revenue for Spitfire over Computer World.
Computer World
Downtown Mall
Downtown é2 - 5ù
Spitfire
Mall êë7 - 1ú
û
Is there a best site for the two companies to locate their new stores?
Circles would be placed around –5 and –1. Squares would be placed around 7 and –1. Since
the –1 would have both a circle and a square, the best location is a suburban mall.
10-2
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Name ________________________________ Date ______________ Class ____________
Section 10-2 Mixed Strategy Games
Goal: To solve problems using mixed strategy game theory
If each element of a row (dominant row) is > each corresponding element of another row
(recessive row), then the recessive row can be deleted. (Delete the row that has the smaller
numbers.)
If each element of a column (dominant column) is < each corresponding element of another
column (recessive column), then the recessive column can be deleted. (Delete the column
that has the larger numbers.)
For the nonstrictly determined game:
éa b ù
M =ê
ú
ëc d û
the optimal strategies P* and Q* and the value of the game are given by
éd - c
P* = é p1* p2* ù = ê
ë
û ë D
v=
ad - bc
D
a - bù
D ú
û
éd - b ù
éq1* ù ê D ú
Q* = ê ú = ê
ú
a- cú
êëq2* ú
û ê
ú
ëê D û
where D = (a + d ) - (b + c )
In Problems 1 and 2 which rows and columns of the game matrix are recessive?
é 3 - 4 - 2ù
1. ê 1 - 5 - 5ú
ê
ú
3ú
ëê- 2 1
û
é 0 2 4ù
2. ê 1 - 1 3ú
ê
ú
ëê- 3 - 4 0 úû
Based on the definitions above R2 and
C3 are recessive.
Based on the definitions above R3 and
C3 are recessive.
10-3
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
In Problems 3–8 solve the matrix games, indicating optimal strategies P* and Q* for R and C,
respectively, and the value v of the game.
é10 - 18ù
3. ê
ú
ë- 8 16 û
Based on the matrix, a = 10, b = - 18, c = - 8, d = 16 and
D = (a + d ) - (b + c) = (10 + 16) - (- 18 + - 8) = 52
éd - c
P* = ê
ë D
a - b ù é16 - (- 8)
=
D ú
û êë 52
10 - (- 18) ù é 24
=
ú
52
û êë 52
28 ù é 6 7 ù
=
52 ú
û êë13 13 ú
û
é d - b ù é16 - (- 18) ù é34 ù é17 ù
ú ê 52 ú ê 26 ú
ê D ú ê
52
Q* = ê
ú= ê ú= ê ú
ú= ê
a
c
10
(
8)
ú ê18 ú ê 9 ú
ê
ú ê
êë D ú
ê
ú
û ë
52
û êë 52 ú
û êë 26 ú
û
v=
ad - bc (10)(16) - (- 18)(- 8) 160 - 144 16 4
=
=
=
=
D
52
52
52 13
é- 3 0 ù
4. ê
ú
ë 8 2û
Based on the matrix, a = - 3, b = 0, c = 8, d = 2 and
D = (a + d ) - (b + c) = (- 3 + 2) - (0 + 8) = - 9
éd - c
P* = ê
ë D
a - b ù é2 - 8
=
D ú
û êë - 9
- 3 - 0 ù é- 6
=
-9 ú
û êë - 9
- 3 ù é2
=
- 9ú
û êë 3
1ù
3ú
û
éd - b ù é 2 - 0 ù é 2 ù é 2 ù
ê D ú ê - 9 ú ê - 9 ú ê- 9 ú
*
Q =ê
ú= ê
ú= ê ú
ú= ê
ê a - c ú ê - 3 - 8 ú ê - 11ú ê 11 ú
ú ëê - 9 û
ú ëê - 9 û
ú ëê 9 û
ú
ëê D û
v=
ad - bc (- 3)(2) - (0)(8) - 6 - 0 - 6 2
=
=
=
=
D
-9
-9
-9 3
10-4
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Chapter 10
é 4 2ù
5. ê
ú
ë- 8 3 û
Based on the matrix, a = 4, b = 2, c = - 8, d = 3 and
D = (a + d ) - (b + c) = (4 + 3) - (2 + (- 8)) = 13
éd - c
P* = ê
ë D
a - b ù é 3 - (- 8)
=
D ú
û êë 13
4 - 2 ù é11 2 ù
=
13 ú
û êë13 13 ú
û
éd - b ù é 3 - 2 ù é 1 ù
ê D ú ê 13 ú ê13 ú
Q* = ê
ú= ê ú
ú= ê
ê a - c ú ê 4 - (- 8) ú ê12 ú
ú ëê 13 û
ú ëê13 û
ú
ëê D û
v=
ad - bc (4)(3) - (2)(- 8) 12 - (- 16) 28
=
=
=
D
13
13
13
é 2 - 3ù
6. ê
ú
ë- 3 4 û
Based on the matrix, a = 2, b = - 3, c = - 3, d = 4 and
D = (a + d ) - (b + c) = (2 + 4) - (- 3 + (- 3)) = 12
éd - c
P* = ê
ë D
a - b ù é 4 - (- 3)
=
D ú
û êë 12
2 - (- 3) ù é 7
5ù
=
12 ú
û êë12 12 ú
û
é d - b ù é 4 - (- 3) ù é 7 ù
ê D ú ê 12 ú ê12 ú
Q* = ê
ú= ê
ú= ê ú
a
c
2
(
3)
ê
ú ê
ú ê5ú
êë D ú
ê
û ë 12 ú
û êë12 ú
û
v=
ad - bc (2)(4) - (- 3)(- 3) 8 - (9) - 1
=
=
=
D
12
12
12
10-5
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Chapter 10
7. Matrix game in Problem 1
Since there is a recessive row and column in the original matrix, they can both be removed
and based on the new matrix, a = 3, b = - 4, c = - 2, d = 1 and
D = (a + d ) - (b + c) = (3 + 1) - (- 4 + (- 2)) = 10
éd - c
P* = ê
ë D
a- b
D
ù é1 - (- 2)
0ú = ê
û ë 10
3 - (- 4)
10
ù é3 7
0ú = ê
û ë10 10
ù
0ú
û
é d - b ù é1 - (- 4) ù é 5 ù é 1 ù
ê D ú ê 10 ú ê10 ú ê 2 ú
ú ê ú ê ú
ê
ú ê
Q* = ê 0 ú = ê 0 ú = ê 0 ú = ê 0 ú
ê a - c ú ê 3 - (- 2) ú ê 5 ú ê 1 ú
ú ê ú ê ú
ê
ú ê
ë D û ë 10 û ë10 û ë 2 û
v=
ad - bc (3)(1) - (- 4)(- 2) 3 - 8 - 5
1
=
=
=
=D
10
10
10
2
8. Matrix game in Problem 2
Since there is a recessive row and column in the original matrix, they can both be removed
and based on the new matrix, a = 0, b = 2, c = 1, d = - 1 and
D = (a + d ) - (b + c) = (0 + - 1) - (2 + 1) = - 4
éd - c
P* = ê
ë D
a- b
D
ù é- 1- 1
0ú = ê
û ë -4
0- 2
-4
ù é- 2
0ú = ê
û ë- 4
-2
-4
ù é1
0ú = ê
û ë2
éd - b ù é- 1- 2 ù é- 3 ù é3 ù
ê D ú ê - 4 ú ê- 4 ú ê4 ú
ú ê
ú ê ú ê ú
* ê
Q = ê a - c ú = ê 0 - 1 ú = ê - 1 ú = ê1 ú
ê D ú ê - 4 ú ê- 4 ú ê4 ú
ê 0 ú ê 0 ú ê 0 ú ê0 ú
ë
û ë
û ë û ë û
v=
ad - bc (0)(- 1) - (2)(1) 0 - 2 - 2 1
=
=
=
=
D
-4
-4
-4 2
10-6
Copyright © 2015 Pearson Education, Inc.
1
2
ù
0ú
û
Finite Mathematics
Chapter 10
9. In an election there are two candidates for sheriff, Gerry Gunner and Remington Rifler.
The candidates’ steering committees have identified two issues important to the electorate:
speeding cars and teenagers out late at night. The units assigned to each candidate’s strategy
are given in the table. A positive entry indicates a strength for Colt Gunner and a negative
entry indicates a weakness. Assume that a strength of one candidate equals a weakness of
the other. What is the best strategy for each candidate? What is the value of the game?
Remington Rifler
Speed Teenagers
Colt Gunner
é4 - 1ù
ú
Teenagers ê
ë0 3 û
Speed
Based on the matrix, a = 4, b = - 1, c = 0, d = 3 and D = (a + d ) - (b + c) = (4 + 3) - (- 1 + 0) = 8
éd - c
P* = ê
ë D
a - b ù é3 - 0
=
D ú
û êë 8
4 - (- 1) ù é 3
=
8 ú
û êë8
5ù
= [0.375 0.625]
8ú
û
é d - b ù é 3 - (- 1) ù é 4 ù
ê D ú ê 8 ú ê 8 ú é0.5ù
Q* = ê
ú= ê ú= ê ú
ú= ê
ê a - c ú ê 4 - 0 ú ê 4 ú ë0.5û
ú ëê 8 û
ú ëê 8 û
ú
ëê D û
v=
ad - bc (4)(3) - (- 1)(0) 12 - 0 12
=
=
=
= 1.5
D
8
8
8
Therefore the best strategy is for Colt Gunner to spend 37.5% of her time on speeding issues
and 62.5% of her time on teenagers staying out late at night. Reminton Rifler should spend
his time equally on the issues. However, no matter what Reminton Rifler does, Colt Gunner
gains at least 1.5 units by employing her best strategy.
10-7
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Finite Mathematics
Chapter 10
10-8
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Name ________________________________ Date ______________ Class ____________
Section 10-3 Linear Programming and 2×2
Games: A Geometric Approach
Goal: To solve problems using a geometric approach to linear programming
Given the nonstrictly determined matrix game:
éa b ù
*
M =ê
ú to find P = [ p1
c
d
ë
û
proceed as follows:
é q1 ù
p2 ], Q* = ê ú,
ëq2 û
and v,
1. Find matrix M1 . (If M does not have all positive entries, then add a
number k such that all entries of M are positive).
e = a+k f =b+k
ée f ù
M1 = ê
ú
g =c+k h = d +k
ëg h û
This new matrix game, M1 , has the same optimal strategies
P* and Q* as M . However, if v1 is the value of the game M1 , then
is the value of the original game M.
v = v1 - k
2. Set up the two corresponding linear programming problems:
A) Minimize y = x1 + x2
ì ex1 + gx2 ³ 1
ï
subject to í fx1 + hx2 ³ 1
ï x ,x ³ 0
î 1 2
B) Maximize y = z1 + z2
ì ez1 + fz2 £ 1
ï
subject to í gz1 + hz2 £ 1
ï z ,z ³ 0
î 1 2
3. Solve each linear programming problem geometrically.
4. Use the solutions in step 3 to find the value v1 for the game M1 and
the optimal strategies and value v for the original game M:
v1 =
1
1
=
y x1 + x2
or
v1 =
1
1
=
y z1 + z2
10-9
Copyright © 2015 Pearson Education, Inc.
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Chapter 10
P* = [ p1
p2 ] = [v1x1 v1x2 ]
é q1 ù é v1z1 ù
Q* = ê ú = ê
ú
ëq2 û ëv1z2 û
v = v1 - k
5. A further check of the solution is provided by showing that
P*MQ* = v
In Problems 1–6, solve the matrix games using a geometric linear programming approach.
é2 4 ù
1. ê
ú
ë3 1 û
Step 1: Since all the values are positive, there is no need to change the matrix.
Step 2: Set up the two linear programming problems:
ì 2 x1 + 3 x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í 4 x1 + x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 2 z1 + 4 z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í 3 z1 + z2 £ 1
ï z ,z ³ 0
î 1 2
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
æ1 ö
çè , 0 ÷
2 ø
æ1 1 ö
çè , ÷
5 5ø
(0,1)
y = x1 + x2
0.5
0.4
1
æ1 1 ö
Minimum value occurs at ç , ÷.
è5 5 ø
10-10
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Finite Mathematics
Chapter 10
B)
Corner points
y = z1 + z2
0
1
3
2
5
1
4
(0, 0)
æ1 ö
çè , 0 ÷
3 ø
æ3 1 ö
çè , ÷
10 10 ø
æ 1ö
çè0, ÷
4ø
æ3 1 ö
Maximum value occurs at ç , ÷.
è10 10 ø
Step 4:
v1 =
1 1 5
=
=
y 25 2
P* = [v1x1 v1x2 ] = [(2.5)(0.2) (2.5)(0.2) ] = [0.5 0.5]
é v1z1 ù é(2.5)(0.3) ù é0.75ù
Q* = ê
ú= ê
ú= ê
ú
ëv1z2 û ë(2.5)(0.1) û ë0.25û
v = v1 - k =
2.
5
5
- 0=
2
2
é- 1 3ù
ê 5 0ú
ë
û
Step 1: Since there is a negative value in the matrix, we will add 2 to all values. Therefore
the new matrix will be written as follows:
é1 5 ù
M1 = ê
úand k = 2
ë7 2û
10-11
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Finite Mathematics
Chapter 10
Step 2: Set up the two linear programming problems:
ì x1 + 7 x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í 5 x1 + 2 x2 ³ 1
ï x ,x ³ 0
î 1 2
ì z1 + 5 z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í 7 z1 + 2 z2 £ 1
ï z ,z ³ 0
î 1 2
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
(1, 0)
y = x1 + x2
1
æ5 4 ö
çè , ÷
33 33 ø
æ 1ö
çè0, ÷
2ø
3
11
1
2
æ5 4 ö
Minimum value occurs at ç , ÷.
è 33 33 ø
B)
Corner points
(0, 0)
æ1 ö
çè , 0 ÷
7 ø
æ1 2 ö
çè , ÷
11 11 ø
æ 1ö
çè0, ÷
5ø
y = z1 + z2
0
1
7
3
11
1
5
æ1 2 ö
Maximum value occurs at ç , ÷.
è11 11 ø
10-12
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Finite Mathematics
Chapter 10
Step 4:
v1 =
1
1 11
=3 =
y
3
11
éæ11 öæ 5 ö æ11 öæ 4 öù é 5
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ëè 3 øè 33 ø è 3 øè 33 øû ë9
4ù
9ú
û
éæ11 öæ1 öù é 1 ù
êç ÷ç ÷ú ê ú
3
* é v1z1 ù êè 3 øè11 øú
Q =ê
=
=ê ú
ú
ëv1z2 û êæ11 öæ 2 öú ê 2 ú
êçè 3 ÷ç
ú
øè11 ÷
øú
ë
û ëê 3 û
v = v1 - k =
11
5
- 2=
3
3
é 1 - 1ù
3. ê
ú
ë- 2 3 û
Step 1: Since there is a negative value in the matrix, we will add 3 to all values. Therefore
the new matrix will be written as follows:
é4 2ù
M1 = ê
úand k = 3
ë1 6 û
Step 2: Set up the two linear programming problems:
ì 4 x1 + x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í 2 x1 + 6 x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 4 z1 + 2 z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í z1 + 6 z2 £ 1
ï z ,z ³ 0
î 1 2
10-13
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Chapter 10
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
(1, 0)
y = x1 + x2
1
æ5 1 ö
çè , ÷
22 11 ø
7
22
(0,1)
1
æ5 1 ö
Minimum value occurs at ç , ÷.
è 22 11 ø
B)
Corner points
(0, 0)
æ1 ö
çè , 0 ÷
4 ø
æ2 3 ö
çè , ÷
11 22 ø
æ 1ö
çè0, ÷
6ø
y = z1 + z2
0
1
4
7
22
1
6
æ2 3 ö
Maximum value occurs at ç , ÷.
è11 22 ø
Step 4:
v1 =
1
1
22
=7 =
y
7
22
éæ22 öæ 5 ö æ22 öæ1 öù é 5
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ëè 7 øè 22 ø è 7 øè11 øû ë 7
2ù
7ú
û
10-14
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
é æ22 öæ 2 ö ù é 4 ù
ç ÷ç ÷
é v1z1 ù ê è 7 øè11 ø ú ê 7 ú
ê
ú= ê ú
Q* = ê
=
ú
ëv1z2 û êæ22 öæ 3 öú ê 3 ú
êçè 7 ÷ç
øè 22 ÷
øú
û
ë
û êë 7 ú
v = v1 - k =
22
1
- 3=
7
7
é 3 - 2ù
4. ê
ú
ë- 1 2 û
Step 1: Since there is a negative value in the matrix, we will add 3 to all values. Therefore
the new matrix will be written as follows:
é6 1ù
M1 = ê
ú and k = 3
ë2 5û
Step 2: Set up the two linear programming problems:
ì 6 x1 + 2 x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í x1 + 5 x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 6 z1 + z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í 2 z1 + 5 z2 £ 1
ï z ,z ³ 0
î 1 2
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
(1, 0)
y = x1 + x2
1
æ3 5 ö
çè , ÷
28 28 ø
æ 1ö
çè0, ÷
2ø
2
7
1
2
æ3 5 ö
Minimum value occurs at ç , ÷.
è 28 28 ø
10-15
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Finite Mathematics
Chapter 10
B)
Corner points
y = z1 + z2
0
1
6
2
7
1
5
(0, 0)
æ1 ö
çè , 0 ÷
6 ø
æ1 1 ö
çè , ÷
7 7ø
æ 1ö
çè0, ÷
5ø
æ1 1 ö
Maximum value occurs at ç , ÷.
è7 7 ø
Step 4:
v1 =
1 1 7
=
=
y 27 2
éæ7 öæ 3 ö æ7 öæ 5 öù é 3
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ëè 2 øè 28 ø è 2 øè 28 øû ë8
5ù
8ú
û
éæ7 öæ1 öù é 1 ù
ç ÷ç ÷
é v1z1 ù êè 2 øè 7 øú ê 2 ú
ê
ú= ê ú
Q* = ê
=
ú
ëv1z2 û êæ7 öæ1 öú ê 1 ú
÷ú
êçè 2 ÷ç
û
ë øè 7 øû êë 2 ú
v = v1 - k =
7
1
- 3=
2
2
é 2 - 1ù
5. ê
ú
ë- 1 4 û
Step 1: Since there is a negative value in the matrix, we will add 2 to all values. Therefore
the new matrix will be written as follows:
é4 1 ù
M1 = ê
ú and k = 2
ë1 6û
10-16
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 2: Set up the two linear programming problems:
ì 4 x1 + x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í x1 + 6 x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 4 z1 + z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í z1 + 6 z2 £ 1
ï z ,z ³ 0
î 1 2
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
(1, 0)
y = x1 + x2
1
æ5 3 ö
çè , ÷
23 23 ø
8
23
(0,1)
1
æ5 3 ö
Minimum value occurs at ç , ÷.
è 23 23 ø
B)
Corner points
(0, 0)
æ1 ö
çè , 0 ÷
4 ø
æ5 3 ö
çè , ÷
23 23 ø
æ 1ö
çè0, ÷
6ø
y = z1 + z2
0
1
4
8
23
1
6
æ5 3 ö
Maximum value occurs at ç , ÷.
è 23 23 ø
10-17
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 4:
v1 =
1
1
23
=8 =
y
8
23
éæ23 öæ 5 ö æ23 öæ 3 öù é5
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ëè 8 øè 23 ø è 8 øè 23 øû ë8
3ù
8ú
û
éæ23 öæ 5 öù é 5 ù
êç ÷ç ÷ú ê ú
8
* é v1z1 ù êè 8 øè 23 øú
Q =ê
=
=ê ú
ú
ëv1z2 û êæ23 öæ 3 öú ê 3 ú
êçè 8 ÷ç
ú
øè 23 ÷
øú
ë
û ëê8 û
v = v1 - k =
23
7
- 2=
8
8
é 5 4ù
6. ê
ú
ë- 3 7 û
Step 1: Since there is a negative value in the matrix, we will add 4 to all values. Therefore
the new matrix will be written as follows:
é9 8 ù
M1 = ê
ú and k = 4
ë1 11û
Step 2: Set up the two linear programming problems:
ì 9 x1 + x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í 8 x1 + 11x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 9 z1 + 8 z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í z1 + 11z2 £ 1
ï z ,z ³ 0
î 1 2
10-18
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
æ1 ö
çè , 0 ÷
8 ø
æ10 1 ö
çè , ÷
91 91 ø
(0,1)
y = x1 + x2
1
8
11
91
1
æ10 1 ö
Minimum value occurs at ç , ÷.
è 91 91 ø
B)
Corner points
(0, 0)
æ1 ö
çè , 0 ÷
9 ø
æ3 8 ö
çè , ÷
91 91 ø
æ 1ö
çè0, ÷
11 ø
y = z1 + z2
0
1
9
11
91
1
11
æ3 8 ö
Maximum value occurs at ç , ÷.
è 91 91 ø
Step 4:
v1 =
1
1
91
=
=
y 1191 11
éæ91 öæ10 ö æ91 öæ 1 öù é10 1 ù
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ú
ëè 11 øè 91 ø è 11 øè 91 øû ë11 11û
10-19
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
éæ91 öæ 3 öù é 3 ù
ç ÷ç ÷
é v1z1 ù êè 11 øè 91 øú ê11ú
ê
ú= ê ú
Q* = ê
=
ú
ëv1z2 û êæ91 öæ 8 öú ê 8 ú
êçè 11 ÷ç
øè 91 ÷
øú
û
ë
û êë11ú
v = v1 - k =
91
47
- 4=
11
11
7. In an election there are two candidates for sheriff, Gerry Gunner and Remington
Rifler. The candidates’ steering committees have identified two issues important to the
electorate: speeding cars and teenagers out late at night. The units assigned to each
candidate’s strategy are given in the table. A positive entry indicates a strength for Colt
Gunner and a negative entry indicates a weakness. Assume that a strength of one candidate
equals a weakness of the other. What is the best strategy for each candidate? What is the
value of the game?
Remington Rifler
Speed Teenagers
Colt Gunner
é4 - 1ù
ú
Teenagers ê
ë0 3 û
Speed
Step 1: Since there is a negative value in the matrix, we will add 2 to all values. Therefore
the new matrix will be written as follows:
é6 1ù
M1 = ê
ú and k = 2
ë2 5û
Step 2: Set up the two linear programming problems:
ì 6 x1 + 2 x2 ³ 1
ï
A) Minimize y = x1 + x2 subject to í x1 + 5 x2 ³ 1
ï x ,x ³ 0
î 1 2
ì 6 z1 + z2 £ 1
ï
B) Maximize y = z1 + z2 subject to í 2 z1 + 5 z2 £ 1
ï z ,z ³ 0
î 1 2
10-20
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the problems above geometrically. Use a graphing utility to find the corner
points for both parts A and B above.
A)
Corner points
(1, 0)
y = x1 + x2
1
æ3 5 ö
çè , ÷
28 28 ø
æ 1ö
çè0, ÷
2ø
2
7
1
2
æ3 5 ö
Minimum value occurs at ç , ÷.
è 28 28 ø
B)
Corner points
(0, 0)
æ1 ö
çè , 0 ÷
6 ø
æ1 1 ö
çè , ÷
7 7ø
æ 1ö
çè0, ÷
5ø
y = z1 + z2
0
1
6
2
7
1
5
æ1 1 ö
Maximum value occurs at ç , ÷.
è7 7 ø
10-21
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 4:
v1 =
1 1 7
=
=
y 27 2
éæ7 öæ 3 ö æ7 öæ 5 öù é 3
P* = [v1x1 v1x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ëè 2 øè 28 ø è 2 øè 28 øû ë8
5ù
8ú
û
éæ7 öæ1 öù é 1 ù
ç ÷ç ÷
é v1z1 ù êè 2 øè 7 øú ê 2 ú
ê
ú= ê ú
Q* = ê
=
ú
ëv1z2 û êæ7 öæ1 öú ê 1 ú
÷ú
êçè 2 ÷ç
û
ë øè 7 øû êë 2 ú
v = v1 - k =
7
3
- 2=
2
2
Therefore the best strategy is for Colt Gunner to spend 37.5% of her time on speeding issues
and 62.5% of her time on teenagers staying out late at night. Reminton Rifler should spend
his time equally on the issues. However, no matter what Reminton Rifler does, Colt Gunner
gains at least 1.5 units by employing her best strategy.
10-22
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Name ________________________________ Date ______________ Class ____________
Section 10-4 Linear Programming and m×n
Games: Simplex Method and the Dual Problem
Goal: To solve problems using the simplex and dual problem methods
Given the nonstrictly determined matrix game, free of recessive rows and columns:
é q1 ù
ér1 r2 r3 ù
*
* ê ú
M =ê
ú to find P = [ p1 p2 ], Q = êq2 ú, and v,
ës1 s2 s3 û
êëq3 ú
û
proceed as follows:
1. Find matrix M1 , (if M does not have all positive entries, the add a
number k such that all entries of M are positive.
éa1 a2
M1 = ê
ëb1 b2
a3 ù
b3 ú
û
This new matrix game, M1 , has the same optimal strategies
P* and Q* as M . However, if v1 is the value of the game M1 , then
v = v1 - k
is the value of the original game M.
2. Set up the two corresponding linear programming problems:
A) Minimize y = x1 + x2
B) Maximize y = z1 + z2
ì a1x1 + b1x2 ³ 1
ï a x +b x ³ 1
ï 2 1 2 2
subject to í
ï a3 x1 + b3 x2 ³ 1
ïî x1, x2 ³ 0
ì a1z1 + a2 z2 + a3 z3 £ 1
ï
subject to í b1z1 + b2 z2 + b3 z3 £ 1
ï
z1, z2 , z3 ³ 0
î
3. Solve the maximization problem, part B, the dual of part A, using the simplex
method as modified in Chapter 6 Section 2. [You will automatically get the
solution of the minimization problem, part A as well, by following this process.]
10-23
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
4. Use the solutions in step 3 to find the value v1 for the game M1 and
the optimal strategies and value v for the original game M:
1
1
1
1
v1 = =
or v1 = =
y z1 + z2 + z3
y x1 + x2
*
P = [ p1
p2 ] = [v1x1 v1x2 ]
é q1 ù é v1z1 ù
êv z ú
Q = êêq2 ú
ú= ê 1 2ú
êëq3 ú
û êëv1z3 ú
û
*
v = v1 - k
5. A further check of the solution is provided by showing that
P*MQ* = v
Solve each matrix game using the simplex approach:
é1 5 - 1ù
1. ê
ú
ë3 - 1 4 û
Step 1: Since there is a negative value in the matrix, we will add 2 to all values. Therefore
the new matrix will be written as follows:
é3 7 1ù
M1 = ê
ú and k = 2
ë5 1 6û
Step 2: Set up the two linear programming problems:
ì 3 x1 + 5 x2 ³ 1
ï 7x + x ³ 1
ï 1 2
A) Minimize y = x1 + x2 subject to í
ï x1 + 6 x2 ³ 1
ïî x1, x2 ³ 0
ì 3z1 + 7 z2 + z3 £ 1
ï
B) Maximize y = z1 + z2 + z3 subject to í 5 z1 + z2 + 6 z3 £ 1
ï
z1, z2 ³ 0
î
10-24
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the maximization problem. (Pivot elements are underlined)
z1
z2
z3
x1 x2 y
x1 é 3 7 1 1 0 0 1ù
ê
ú
x2 ê 5 1 6 0 1 0 1ú
y êë- 1 - 1 - 1 0 0 1 0 úû
z1
z2 z3
x1 x2 y
z3 é 3
7 1 1 0 0
1ù
ê
ú
x2 ê- 13 - 41 0 - 6 1 0 - 5ú
y êë 2
6 0
1 0 1 1úû
z1
z2
z3
z3 é 3
ê
x2 ê13
41
yê2
ëê
7
1
z1
é 32
z3 ê 41
z2 ê 13
ê 41
y ê4
ëê 41
x1
x2 y
1ù
ú
6 - 1 0 5
41
41
41 ú
ú
1
0 1 1û
ú
1
0 0
z2 z3
x1
x2 y
0
1
41
6
41
5
41
7
41
1
41
6
41
1 0
6 0
1 -
1 0
0 0
-
0
0
1
6ù
41 ú
5ú
41 ú
11 ú
41 û
ú
Based on the above matrix: z1 = 0, z2 =
5 , z
41 3
6 , x = 5 , x = 6 , and y = 11 .
= 41
1 41
2
41
41
Step 4:
v1 =
1
1
41
=
=
y 1141 11
éæ41 öæ 5 ö æ41 öæ 6 öù é 5 6 ù
P* = [v1x1 v2 x2 ] = êç ÷ç ÷ ç ÷ç ÷ú = ê
ú
ëè 11 øè 41 ø è 11 øè 41 øû ë11 11û
10-25
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
é æ41 ö
ù
0) ú é ù
(
ê çè 11 ÷
ø
ú êê 0 ú
é v1z1 ù ê
êæ41 öæ 5 öú ê 5 ú
ú
* ê
ú
Q = êv1z2 ú = êç ÷ç ÷ú =
è 11 øè 41 ø ê 41ú
ê
ú ê ú
êëv1z3 ú
û ê
æ41 öæ 6 öú ê 6 ú
÷ú
êçè ÷ç
ë 11 øè 41 øû ë 41û
v = v1 - k =
2.
41
19
- 2=
11
11
é- 2 2 4 ù
ê 3 1 - 1ú
ë
û
Step 1: Since there is a negative value in the matrix, we will add 3 to all values. Therefore
the new matrix will be written as follows:
é1 5 7 ù
M1 = ê
ú and k = 3
ë6 4 2û
Step 2: Set up the two linear programming problems:
ì x1 + 6 x2 ³ 1
ï 5x + 4 x ³ 1
ï 1
2
A) Minimize y = x1 + x2 subject to í
7
x
+
2
x
2 ³ 1
ï 1
ïî x1, x2 ³ 0
ì z1 + 5 z2 + 7 z3 £ 1
ï
B) Maximize y = z1 + z2 + z3 subject to í 6 z1 + 4 z2 + 2 z3 £ 1
ï
z1, z2 ³ 0
î
10-26
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the maximization problem. (Pivot elements are underlined)
z1
z2
z3
x1 x2 y
x1 é 1 5 7 1 0 0 1ù
ê
ú
x2 ê 6 4 2 0 1 0 1ú
y êë- 1 - 1 - 1 0 0 1 0 úû
z1
z2
z3
x1 x2 y
z1 é 1
5
7
1 0 0
1ù
ê
ú
x2 ê0 - 26 - 40 - 6 1 0 - 5ú
y êë0
4
6
1 0 1 1ú
û
z1
z1 é 1
ê
x2 ê0
y ê0
ëê
z2
z3
x1
x2 y
1ù
ú
13 1 3
1
1
0
20
20
40
8ú
ú
4 6 1
0 1 1û
ú
5 7
z1
z2
z3
é
z1 ê 1
z3 ê0
ê
y ê0
êë
9
20
13
20
1
10
0 1
0
1
0 0
x1
1
20
3
20
1
10
x2
-
7
40
1
40
3
20
y
0
0
0
1ù
8ú
1ú
8ú
1ú
4ú
û
1 , x =
Based on the above matrix: z1 = 18 , z2 = 0, z3 = 18 , x1 = 10
2
Step 4:
v1 =
1 1
=
=4
y 14
é æ1 ö
P* = [v1x1 v2 x2 ] = ê(4) ç ÷
ë è10 ø
3 öù é 2
÷ =ê
20 øú
û ë5
(4) æ
çè
3ù
5ú
û
10-27
Copyright © 2015 Pearson Education, Inc.
3 ,
20
and y = 14 .
Finite Mathematics
Chapter 10
é æ1 öù é 1 ù
(4) ç ÷
é v1z1 ù ê è 8 øú ê 2 ú
ê
ú ê ú
Q* = êêv1z2 ú
=
4
0
ê
ú = ê0 ú
(
)(
)
ú
ê
ú
êëv1z3 ú
û ê 4 æ1 öú ê 1 ú
( ) çè ÷ø êë 2 úû
êë
8 ú
û
v = v1 - k = 4 - 3 = 1
é 5 - 1ù
3. ê- 1 4 ú
ê
ú
êë 1 3 ú
û
Step 1: Since there is a negative value in the matrix, we will add 2 to all values. Therefore
the new matrix will be written as follows:
é7 1 ù
M1 = ê1 6ú and k = 2
ê
ú
ëê3 5úû
Step 2: Set up the two linear programming problems:
ì 7 x1 + x2 + 3x3 ³ 1
ï
A) Minimize y = x1 + x2 + x3 subject to í x1 + 6 x2 + 5 x3 ³ 1
ï x ,x ,x ³ 0
î
1 2 3
ì 7 z1 + z2 £ 1
ï z + 6z £ 1
ï 1
2
B) Maximize y = z1 + z2 subject to í
ï 3 z1 + 5 z2 £ 1
ïî z1, z2 ³ 0
10-28
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the maximization problem. (The pivot elements are underlined)
z1 z2 x1 x2 x3 y
x1 é 7 1
ê
x2 ê 1 6
x3 ê 3 5
ê
y êë–1 –1
z1
z2 é 7
ê
x2 ê - 41
x3 ê- 32
ê
y ëê 6
z2
z1
é0
z2 ê
x2 ê0
ê
z1 ê 1
ê
yê
0
ë
1ù
0 - 6 1 0 0 - 5ú
ú
0 - 5 0 1 0 - 4ú
ú
0
1 0 0 1 1û
ú
0
x1 x2
x3 y
1ù
0 -6 1
0 0 - 5ú
ú
5
1
0 32
0 - 32
0 18 ú
ú
0
1 0
0 1 1ú
û
1 -
0
1 0 0 0
1
z2
0
x1 x2 x3 y
1
z1 z2
z2 é 7
ê
x2 ê- 41
x3 ê 1
ê
yê 6
ë
1ù
0 1 0 0 1ú
ú
0 0 1 0 1ú
ú
0 0 0 1 0ú
û
1 0 0 0
1 0
x1
3
32
13
32
5
32
1
16
0 0
x2
x3
y
0
7
32
41
32
1
32
3
16
0
1 0 0
0
0
1
1ù
8ú
1ú
8
ú
1ú
8
ú
1ú
4û
1 , x = 0, x = 3 , and y = 1 .
Based on the above matrix: z1 = 18 , z2 = 18 , x1 = 16
2
3 16
4
Step 4:
v1 =
1 1
=
=4
y 14
P* = [v1x1 v1x2
é æ1 ö
v1x3 ] = ê(4) ç ÷
ë è16 ø
3 öù é 1
0
÷ =ê
16 øú
û ë4
(4)(0) (4) æ
çè
10-29
Copyright © 2015 Pearson Education, Inc.
3ù
4ú
û
Finite Mathematics
Chapter 10
é æ1 öù é 1 ù
ê(4) çè 8 ÷
øú ê 2 ú
* é v1z1 ù ê
ú= ê ú
Q =ê
ú=
v
z
ê
ú ê1 ú
1
æ
ö
ë 1 2û
ê(4) çè 8 ÷
ú
øú
ë
û ëê 2 û
v = v1 - k = 4 - 2 = 2
é- 4 0 ù
4. ê- 3 - 4ú
ê
ú
êë 0 - 5úû
Step 1: Since there is a negative value in the matrix, we will add 6 to all values. Therefore
the new matrix will be written as follows:
é2 6 ù
M1 = ê3 2úand k = 6
ê
ú
ëê6 1 úû
Step 2: Set up the two linear programming problems:
ì 2 x1 + 3 x2 + 6 x3 ³ 1
ï
A) Minimize y = x1 + x2 + x3 subject to í 6 x1 + 2 x2 + x3 ³ 1
ï x ,x ,x ³ 0
î
1 2 3
ì 2 z1 + 6 z2 £ 1
ï 3z + 2 z £ 1
ï 1
2
B) Maximize y = z1 + z2 subject to í
ï 6 z1 + z2 £ 1
ïî z1, z2 ³ 0
10-30
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
Step 3: Solve the maximization problem (The pivot elements are underlined)
z1
z2 x1 x2 x3 y
x1 é 2 6
ê
x2 ê 3 2
x3 ê 6 1
ê
y êë–1 –1
z1
z2 x1 x2
x1 é- 34
ê
x2 ê - 9
z2 ê 6
ê
y êë 5
1 0 - 6 0 - 5ù
0 0 1 - 2 0 - 1ú
ú
1 0 0
1 0
1ú
ú
0 0 0
1 1 1ú
û
z2
x1 é 1
ê
x2 ê- 9
z2 ê 6
ê
yê 5
ë
é1
z1 ê
x2 ê0
ê
z2 ê0
ê
yê
0
ë
x1 x2
0 -
1
34
1
0
z2
x1
0 -
1
34
9
34
3
17
5
34
0 1
0
x3 y
3
17
5ù
34 ú
0
0 1 - 2 0 - 1ú
0 0
1 0 1ú
ú
0 0
1 1 1ú
û
0
0
x3 y
0
z1
z1
1ù
0 1 0 0 1ú
ú
0 0 1 0 1ú
ú
0 0 0 1 0ú
û
1 0 0 0
x2
x3
y
0
3
17
7
- 317
1
- 17
2
17
0
0
0
0
0
0
1
5ù
34 ú
11 ú
34
ú
2ú
17
ú
9ú
34 û
5 , z = 2 , x = 5 , x = 0, x = 2 , and y = 9 .
Based on the above matrix: z1 = 34
2 17 1 34
2
3 17
34
Step 4:
v1 =
1
1
34
=9 =
y
9
34
10-31
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 10
éæ34 öæ 5 ö æ34 ö
4ù
æ34 öæ 2 öù é 5
v1x3 ] = êç ÷ç ÷ ç ÷(0) ç ÷ç ÷ú = ê
0
è 9 øè17 øû ë9
9ú
û
ëè 9 øè 34 ø è 9 ø
éæ34 öæ 5 öù é 5 ù
ç ÷ç ÷
é v1z1 ù êè 9 øè 34 øú ê 9 ú
ê
ú= ê ú
Q* = ê
=
ú
ëv1z2 û êæ34 öæ 2 öú ê 4 ú
êçè 9 ÷ç
øè17 ÷
øú
û
ë
û êë 9 ú
P* = [v1x1 v1x2
v = v1 - k =
34
20
- 6=9
9
10-32
Copyright © 2015 Pearson Education, Inc.