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Q 2-31
Min 3A + 4B
s.t.
1A + 3B ≧ 6
B = - 1/3A + 2
1A + 1B ≧ 4
B=-A+4
A, B ≧ 0
B
B=-A+4
Feasible
Region
Objective Function = 3A + 4B
Optimal Solution A = 3, B = 1
Objective
Function Value
= 3(3) + 4(1) = 13
B = - 1/3A + 2
A
Q 2-38 a.
Let
S = yards of the standard grade material / frame
P = yards of the professional grade material / frame
Min 7.50S + 9.00P
s.t.
0.10S + 0.30P ≧ 6 carbon fiber
(at least 20% of 30 yards)
0.06S + 0.12P ≦ 3 kevlar
(no more than 10% of 30 yards)
S+
P = 30 total
S, P ≧ 0
P
Total
Extreme Point
S = 10, P = 20
Feasible region is
the line segment
Kevlar
Carbon fiber
Extreme Point
S = 15, P = 15
S
Q 2-38 c.
Extreme Point
Cost
(15, 15)
7.50(15) + 9.00(15) = 247.50
(10, 20)
7.50(10) + 9.00(20) = 255.00
255.00 > 247.50
Therefore, the optimal solution is
S = 15, P = 15
Q 2-38 d.
Changing is only this value
Min 7.50S + 8.00P
(From 9.00 to 8.00)
s.t.
0.10S + 0.30P ≧ 6 carbon fiber
(at least 20% of 30 yards)
0.06S + 0.12P ≦ 3 kevlar
(no more than 10% of 30 yards)
S+
P = 30 total
S, P ≧ 0
Therefore, optimal solution does not change:
S = 15, and P =15
New optimal solution value =
7.50(15) + 8(15) = $232.50
Q 2-38 e.
At $7.40 per yard, the optimal solution is S = 10,
P = 20.
So, the optimal solution is changed.
The value of the optimal solution is reduced to
7.50 (10) + 7.40 (20) = 223.00.
A lower price for the professional grade will
change the optimal solution.
Q 3-7 a.
From figure 3.14,
Optimal solution: U = 800, H = 1200
Estimated Annual Return
3(800) + 5(1200) = $8400
Q 3-7 b.
Constraints 1 and 2 because of 0 at
Slack/Surplus column. All funds available
are being fully utilized and the maximum
risk is being incurred.
Q 3-7 c.
Constraint
Dual Prices
Funds Avail
0.09
Risk Max
1.33
U.S. Oil Max
0
A unit increase in the RHS of Funds Avail makes 0.09
improvement in the value of the objective function. Risk Max
also makes 1.33 improvement in the value of the objective
function. A unit increase in the RHS of U.S. Oil Max makes
no improvement in the objective function value.
Q 3-7 d.
NO
A unit increase in the RHS of U.S. Oil Max
makes no improvement in the objective
function value.
Q 3-10 a.
From figure 3.16,
Optimal solution: S = 4,000, M = 10,000
Estimated total risk =
8(4,000) + 3(10,000) = 62,000
Q 3-10 b.
Variable
Range of Optimality
S
3.75 to No Upper Limit
M
No Lower Limit to 6.4
Q 3-10 c, d, e, f
(c) 5(4,000) + 4(10,000) = $60,000
(d) 60,000/1,200,000 = 0.05
(e) 0.057
(f) 0.057(100) = 5.7
Theory of Simplex Method
A two-variable linear programming problem
Max Z  3x1  5x2 ,
s.t.
x1
4
and
2 x2  12
3x1  2 x2  18
x1  0, x2  0
For any LP problem with n decision variables, each
CPF (Corner Point Feasible) solution lies at the
intersection of n constraint boundaries; i.e., the
simultaneous solution of a system of n constraint
boundary equations.
 x1 
 b1 
0 
x 
b 
0 
2
2


c  c1 , c2 ,  , cn , x 
,b 
,0    ,



 
 
 
0 
 xn 
bn 
Max
s.t.
cx
Ax  b
x0
 a11
a
21

A


am1
a12
a22

am 2
 a1n 

 a2 n 
 

 amn 
Original Form
Augmented Form
Max Z
s.t. 1Z  cX  0 X S  0
Max cX
s.t. AX  b
x0
0Z  AX  IX S  b (1)
X  0, X S  0
Matrix Form
1
0

c
A
Z 
0    0 
(2)
X




I    b 
 X S 
Matrix Form (2) is
Max
s.t.
Z
Z  cX  0 X S  0
AX  IX S  b
or
Max
Z
s.t.
Z  cˆXˆ  0
Aˆ Xˆ  b
where
cˆ  c, 0
X


Xˆ   
XS 
(3)
(4)
Aˆ  A, I 
X B : a vectorof basic variables
X N : a vectorof non basic variables
c B : a vectorof thesecorresponding objective
coefficient s (to X B )
c N : a vectorof thesecorresponding objective
coefficient s (to X N )
B : a basic matrix
N : a nonbasicmatrix
Then, we have
Max Z
s.t.
where
Z  cB X B  c N X N  0
(5)
BX B  NX N  b
(6)
X


B
Xˆ    cˆ  cB , cN  Aˆ  B, N 
X N 
Eq. (6) becomes
1
1
X B  B NX N  B b
(7)
Putting Eq. (7) into (5), we have
1
1
Z  cB ( B b  B NX N )  cN X N  0
So,
1
(8)
1
Z  0 X B  (cB B N  cN ) X N  cB B b (9)
Currently,X N  0, Eq. (7) and Eq. (9) become
1
1
X B  B b, Z  cB B b
(10)
Eq. (10) can be expressed by
1

 Z  1 cB B  0 cB B b

  

 1 



1
 X B  0 B  b  B b 
1
(11)
From Eq. (2),
 Z  1
   
 X B  0
1

0
Z 
1
cB B  1  c 0   cB B b
X

 1 

1  


B  0 A I 
B b 


 X S 
Z 
1
1 
1


cB B A  c cB B   cB B b 
X


 1 
1
1


B A
B 
B b 


 X S 
1
(12)
Thus, initial and later simplex tableau are
Iteration
0
BV
Z
Z
1
0
XB
Iteration BV
Any
Z
Z
1
XB
0
Original
Variables
-c
A
Original
Variables
1
cB B A  c
1
B A
Slack
RHS
Variables
0
0
I
b
Slack
Variables
cB B
B
1
1
RHS
1
cB B b
1
B b
The Overall Procedure
1. Initialization:
Same as for the original simplex method.
2. Iteration:
Step 1
Determine the entering basic variable:
Same as for the Simplex method.
Step 2
Determine the leaving basic variable:
Same as for the original simplex method,
except calculate only the numbers required to
do this [the coefficients of the entering basic
variable in every equation but Eq. (0), and
then, for each strictly positive coefficient, the
right-hand side of that equation].
Step 3
Determine the new BF solution:
Derive
B
1
and set
1
xB  B b.
3. Optimality test:
Same as for the original simplex method, except
calculate only the numbers required to do this test,
i.e., the coefficients of the nonbasic variables in
Eq. (0).
Fundamental Insight
Z
Z
1
XB 0
X
XS
RHS
1
1
c
B
c
B
b Row0
cB B A  c B
B
1
1
B A
B
1
1
B b
Row1~N
1 0
1 0 0
4






1
1
A  0 2, I ( B )  0 1 0, b( B b)  12
3 2
0 0 1
18
 x3 
xB   x4 , c  3,5, cS  0,0,0,
 x5 
Iteration BV
x1
x3
x4
x5
-3
1
0
3
0
Coefficient of:
x2
x3
x4
x5
-5
0
2
2
0
1
0
0
0
0
1
0
0
0
0
1
Right
Side
0
4
12
18
cB
 0,5,0,
1 0 0

 1  0 1
B  2 0 ,
0  1 1
Iteration BV
1
x3
x2
x5
1 0 0   4   4 
0 1
 12  6,
1

0
B b 
2
   
0  1 1 18 6
Coefficient of:
x5
Right
Side
x2
x3
0
0
1
0
1
0
0
4
0
1
0
1
2
0
6
3
0
0
-1
1
6
x1
x4
0
1
0
1



0
,
5
,
0
Z1  c1  cB B a1  c1

0
1
1

Z 4  c4  cB B a4  c4  0,5,00
0
Iteration BV
1
x3
x2
x5
0  1 
0 0  3  3
1 3
0  0 



0 1  0  5 2
1 0
0
1
2
1
0
1
2
1
Coefficient of:
x2
x3
x4
x5
Right
Side
-3
1
0
0
0
1
5
0
0
0
4
0
1
0
1
2
0
6
3
0
0
-1
1
6
x1
2
1 0 0 1 0 1 0
 0 1 0  0 2   0 1 

1

B A 
2

 

0  1 1 3 2 3 0
Iteration BV
1
x3
x2
x5
Coefficient of:
x2
x3
x4
x5
Right
Side
-3
1
0
0
0
1
5
0
0
0
4
0
1
0
1
2
0
6
3
0
0
-1
1
6
x1
2
so
Iteration BV
1
x3
x2
x5
 4


1
cB B b  0,5,06  30
6
Coefficient of:
x2
x3
x4
x5
Right
Side
-3
1
0
0
0
1
5
0
0
0
30
4
0
1
0
1
2
0
6
3
0
0
-1
1
6
x1
2
4
6
4
4
1
6
2
3
minimum
The most negative coefficient
Iteration BV
1
x3
x2
x5
Coefficient of:
x2
x3
x4
x5
Right
Side
-3
1
0
0
0
1
5
0
0
0
30
4
0
1
0
1
2
0
6
3
0
0
-1
1
6
x1
2
cB  0,5,3,
 1 1 3  1 3   4  2
 1 1 3  1 3
1   0 1 0  , 1   0 1 0  12  6 ,
2
2
B

 B b 
   
 0  1 3 1 3 
 0  1 3 1 3  18 2
Iteration BV
2
x3
x2
x1
Coefficient of:
x2
x3
0
0
0
0
0
1
1
0
1
x1
0
1
1
0
0
x4
3
2
1
3
x5
1
3
0
1
3
Right
Side
2
6
2
1 1 3  1 3 0
1
Z4  c4  cB B a4  c4  0,5,00 1 2 0 1  0  3 2
0  1 3 1 3 0
1 1  1 3 0
3
1

Z5  c5  cB B a5  c5 0,5,30 1 2 0 0  0  1
0  1 1  1
3 3
Iteration BV
2
x3
x2
x1
x1
Coefficient of:
x2
x3
x4
x5
1
0
0
0
3
0
0
1
1
0
1
0
1
1
0
0
2
3
2
1
3
1
3
0
1
3
Right
Side
2
6
2
so
 2


1
cB B b  0,5,36  36
2
Iteration BV
2
x3
x2
x1
x1
Coefficient of:
x2
x3
x4
x5
1
0
0
0
3
0
0
1
1
0
1
0
1
1
0
0
2
3
2
1
3
1
3
0
1
3
Right
Side
36
2
6
2
Duality Theory
One of the most important discoveries in the
early development of linear programming was
the concept of duality.
Every linear programming problem is associated
with another linear programming problem called
the dual.
The relationships between the dual problem and
the original problem (called the primal) prove
to be extremely useful in a variety of ways.
Primal and Dual Problems
Primal Problem
Dual Problem
m
n
Max
s.t.
Z  cjxj,
Min
n
s.t.
j 1
 a ijx j  bi ,
j1
W   bi yi ,
i 1
m
a
i 1
ij
yi  c j ,
for i  1,2,, m.
for j  1,2,, n.
x j  0, for j  1,2,, n.
yi  0, for i  1,2,, m.
The dual problem uses exactly the same parameters
as the primal problem, but in different location.
In matrix notation
Primal Problem
Maximize
Dual Problem
Z  cx,
W  yb,
Minimize
subject to
subject to
yA  c
Ax  b
y  0.
x  0.
y1, y2 ,, ym 
Where c and y 
vectors but b and
x
are row
are column vectors.
Example
Dual Problem
Primal Problem
Max Z  3x1  5x2 ,
s.t.
x1
4
2 x2  12
3x1  2 x2  18
x1  0, x 2  0
Min
W  4 y1  12y2  18y3 ,
s.t.
y1
 3 y3  3
2 y2  2 y3  5
y1  0, y2  0, y3  0
Primal Problem
in Matrix Form
Max
s.t.
 x1 
Z  3,5 ,
 x2 
1 0
4
0 2  x1 ,  12

x   
3 2  2  18
 x1  0
 x   0.
 2  
Dual Problem
in Matrix Form
4
W   y1 , y2 , y3 12
s.t.
18
1 0 
 y1 , y2 , y3 0 2  3,5
3 2
Min
y1, y2 , y3   0,0,0.
Dual Problem
Coefficient
of:
Right
Side
Primal Problem
Coefficient of:
Right
x1 x2  xn Side
y1 a11
y 2 a21
a12  a1n
a22  a2 n
 
am 2  amn
y m am1
VI
VI
c1
c2


VI
cn
Coefficients for
Objective Function
(Maximize)
 b1
 b2

 bm
Coefficients for
Objective Function
(Minimize)
Primal-dual table for linear programming
Relationships between Primal and Dual Problems
One Problem
Constraint i
Objective function
Minimization
0
Variables
0
Unrestricted
Constraints



Other Problem
Variable i
Right sides
Maximization



Constraints
0
Variables
0
Unrestricted
The feasible solutions for a dual problem are
those that satisfy the condition of optimality for
its primal problem.
A maximum value of Z in a primal problem
equals the minimum value of W in the dual
problem.
Rationale: Primal to Dual Reformulation
Max cx
s.t. Ax  b
x 0
Min yb
s.t. yA  c
y 0
Lagrangian Function [ L( X , Y )]
L(X,Y) = cx - y(Ax - b)
= yb + (c - yA) x
[ L( X , Y )]
= c-yA
X
The following relation is always maintained
 yb (from Primal: Ax  b) (1)
yAx  cx (from Dual : yA  c) (2)
yAx
From (1) and (2), we have
cx
 yAx  yb
(3)
At optimality
cx* = y*Ax* = y*b
is always maintained.
(4)
“Complementary slackness Conditions” are
obtained from (4)
( c - y*A ) x* = 0
(5)
y*( b - Ax* ) = 0
(6)
xj* > 0
y*aj = cj , y*aj > cj
xj* = 0
yi* > 0
aix* = bi , ai x* < bi
yi* = 0
Any pair of primal and dual problems can be
converted to each other.
The dual of a dual problem always is the primal
problem.
Dual Problem
Min W = yb,
s.t.
yA  c
y  0.
Converted to
Standard Form
Max (-W) = -yb,
s.t.
-yA  -c
y  0.
Converted to
Standard Form
Max Z = cx,
s.t.
Ax  b
x  0.
Its Dual Problem
Min (-Z) = -cx,
s.t.
-Ax -b
x  0.
Min
s.t.
0.4 x1  0.5 x2
0.3x1  0.1x2  2.7
0.5x1  0.5x2  6
0.6 x1  0.4 x2  6
x1  0, x2  0
Min
s.t.
0.4 x1  0.5 x2
 0.3x1  0.1x2  2.7 [y1 ]
0.5 x1  0.5 x2  6
[y2 ]
 0.5 x1  0.5 x2  6
[y-2 ]
0.6 x1  0.4 x2  6
x1  0, x2  0
[y3 ]

2

2
Max  2.7 y1  6( y  y )  6 y3
s.t.  0.3 y1  0.5( y2  y2 )  0.6 y3  0.4
 0.1y1  0.5( y2  y2 )  0.4 y3  0.5
y1  0, y2  0, y2  0, y3  0.
Max  2.7 y1  6 y2  6 y3
s.t.  0.3 y1  0.5 y2  0.6 y3  0.4
 0.1y1  0.5 y2  0.4 y3  0.5
y1  0, y2 : URS, y3  0.
Home Work
•
•
•
•
Ch 4: Problem 1
Ch 4: Problem 10
Additional Problems: A-1 and A-2
(See the proceeding PPS)
• Due Date: September 16
A–1
Theory of Simplex Method: Consider the following problem.
Maximize
Z  4 x1  3x 2  x 3  2 x 4 ,
subject to
4 x1  2 x 2  x 3  x 4  5
3x1  x 2  2 x 3  x 4  4
x1  0, x 2  0, x 3  0 & x 4  0
Let x5 and x6 denote slack variables for the two constraints.
After you apply the simplex method, a portion of the final
simplex tableau is as follows:
A – 1 cont’d
Coefficient of:
Basic
Variable
Eq.
Z
Z
(0)
x2
x4
x1
x2
x3
x4
x5
x6
1
1
1
(1)
0
1
-1
(2)
0
-1
2
•
Solve the problem.
•
What is B-1 ? How about B-1b and CBB-1b ?
•
If the right hand side is changed from (5, 4) to (5, 5),
how is and optimal solution changed? How about an
optimal objective value?
Right
Side
A–2
Linear Programming: Slim-Down Manufacturing makes
a line of nutritionally complete, weight-reduction
beverages. One of their products is a strawberry shake
which is designed to be a complete meal. The
strawberry shake which is designed to be a complete
meal. The strawberry shake consists of several
ingredients. Some information about each of these
ingredients is given below.
A – 2 cont’d
Ingredient
Calories
from Fat
(per tbsp)
Total
Calories
(per tbsp)
Vitamin
Content
(mg/tbsp)
Thickeners
(mg/tbsp)
Cost
(¢/tbsp)
Strawberry
flavoring
1
50
20
3
10
Cream
75
100
0
8
8
Vitamin
supplement
0
0
50
1
25
Artificial
sweetener
0
120
0
2
15
Thickening
agent
30
80
2
25
6
A – 2 cont’d
The nutritional requirements are as follows. The beverage must
total between 380 and 420 calories (inclusive). No more than
20 percent of the total calories should come from fat. There
must be at least 50 milligrams (mg) of vitamin content. For
taste reasons, there must be at least 2 tablespoons (tbsp) of
strawberry flavoring for each tablespoon of artificial sweetener.
Finally, to maintain proper thickness, there must be exactly 15
mg of thickeners in the beverage.
Management would like to select the quantity of each
ingredient for the beverage which would minimize cost while
meeting the above requirements.
A – 2 cont’d
•
Formulate a linear programming model for this problem.
•
Show its dual formulation.
•
Solve this model by your computer and confirm
complementary slackness condition.