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Q 2-31
Min 3A + 4B
s.t.
1A + 3B ≧ 6
1A + 1B ≧ 4
A, B ≧ 0
B = - 1/3A + 2
B = - A + 4
Feasible Region
A
B
B = - 1/3A + 2
B = - A + 4
Objective Function = 3A + 4BOptimal Solution A = 3, B = 1
Objective Function Value = 3(3) + 4(1) = 13
Q 2-38 a.
Let
S = yards of the standard grade material / frame
P = yards of the professional grade material / frameMin 7.50S + 9.00Ps.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 totalS, P ≧ 0
Total
Carbon fiberKevlar
Feasible region is the line segment
Extreme PointS = 10, P = 20
S
P
Extreme PointS = 15, P = 15
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.
Min 7.50S + 8.00Ps.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 totalS, P ≧ 0
Changing is only this value(From 9.00 to 8.00)
Therefore, optimal solution does not change: S = 15, and P =15New 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
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.
,53 21 xxZ
1x
0,0
1823
21
21
xx
xx122 2 x4
and
Max
s.t.
A two-variable linear programming problem
0
x
bAx
cx
,
0
0
0
0,,,,,, 2
1
2
1
21
nn
n
b
b
b
b
x
x
x
xcccc
Max
s.t.
mnmm
n
n
aaa
aaa
aaa
A
21
22221
11211
Original Form Augmented Form
Max s.t. bAX
0x
Maxs.t.
cX Z
0,0
0
001
S
S
S
XX
bIXAXZ
XcXZ
Matrix Form
b
X
X
Z
IA
c
S
00
0
1
(1)
(2)
Matrix Form (2) is
Max
s.t.
bIXAX
XcXZ
Z
S
S
00
(3)
orMax
s.t.
bXA
XcZ
Z
ˆˆ
0ˆˆ(4)
where
IAAX
XXcc
S
,ˆˆ0,ˆ
matrix nonbasic a :N
matrix basic a :B
)X (to tscoefficien
objective ingcorrespond theseof vector a :c
)X (to tscoefficien
objective ingcorrespond theseof vector a :c
variablesbasicnon of vector a :X
variablesbasic of vector a :X
N
N
B
B
N
B
Max
s.t.
bNXBX
XcXcZ
Z
NB
NNBB
0
Then, we have
(5)
(6)
where
NBAcccX
XX NB
N
B ,ˆ,ˆˆ
Eq. (6) becomes
bBNXBX NB11
Putting Eq. (7) into (5), we have
0)( 11 NNNB XcNXBbBcZ
(7)
(8)So,
bBcXcNBcXZ BNNBB11 )(0 (9)
become (9) Eq. and (7) Eq. ,0 Currently, NX
bBcZbBX BB11 , (10)
bB
bBc
bB
Bc
X
Z BB
B1
1
1
1 0
0
1
Eq. (10) can be expressed by
(11)
From Eq. (2),
bB
bBc
X
X
Z
BAB
BccABc
bB
bBc
X
X
Z
IA
c
B
Bc
X
Z
B
S
BB
B
S
B
B
1
1
11
11
1
1
1
1
0
1
0
01
0
1
(12)
Thus, initial and later simplex tableau are
IterationBVZOriginalVariables
SlackVariables
RHS
Z1 -c 0 00
BX0 A I b
IterationBVZOriginalVariables
SlackVariables
RHS
Z 1 cABcB 1 1BcB bBcB1
AnyBX 0 AB1 1B bB1
1. Initialization:
Same as for the original simplex method.
2. Iteration:
Step 1
Determine the entering basic variable:
Same as for the Simplex method.
The Overall Procedure
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 and set
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).
1B .1bBxB
Fundamental Insight
Z
Z
RHS
Row0
Row1~N
1BcB
1B
bBcB1
bB 1
X
BX
SX
1
0 AB 1
cABcB 1
0
5x4x3x
RightSideBVIteration 3x 4x 5x2x1x
18
12
4
)(,
100
010
001
)(,
2
2
0
3
0
111 bBbBIA
-3 -5 0 0 0 0 1 0 1 0 0 4 0 2 0 1 0 12 3 2 0 0 1 18
Coefficient of:
,0,0,0,5,3,
5
4
3
SB cc
x
x
x
x
BVIteration
1
5x2x3x 1 0 1 0 0 4
0 1 0 0 6
3 0 0 -1 1 6
,
110
00
001
,0,5,0
21 ,
6
6
4
18
12
4
110
00
001
1B bB 1
Bc
21
21
RightSide3x 4x 5x2x1x
Coefficient of:
0 0 0
BVIteration
25
1
5x2x3x 1 0 1 0 0 4
0 1 0 0 6
3 0 0 -1 1 62
1
RightSide3x 4x 5x2x1x
Coefficient of:
-3 0 0 0
0
0
1
0
110
00
001
0,5,0
33
3
0
1
110
00
001
0,5,0
2
1
21
11 cZ 111 caBcB
44 cZ 441 caBcB 2
5
BVIteration
25
1
5x2x3x 1 0 1 0 0 4
0 1 0 0 6
3 0 0 -1 1 62
1
RightSide3x 4x 5x2x1x
Coefficient of:
-3 0 0 0
0
1
0
3
0
1
2
2
0
3
0
1
110
00
001
AB 12
1
BVIteration
25
1
5x2x3x 1 0 1 0 0 4
0 1 0 0 6
3 0 0 -1 1 62
1
RightSide3x 4x 5x2x1x
Coefficient of:
-3 0 0 0 30
30
6
6
4
0,5,01
bBcB
so
BVIteration
25
1
5x2x3x 1 0 1 0 0 4
0 1 0 0 6
3 0 0 -1 1 62
1
RightSide3x 4x 5x2x1x
Coefficient of:
-3 0 0 0 30
minimum
The most negative coefficient
23
6
41
4
4
6
BVIteration
2
1x2x3x 0 0 1 2
0 1 0 0 6
1 0 0 2
21
RightSide3x 4x 5x2x1x
Coefficient of:
0 0 0
,
0
00
1
,3,5,0Bc
21 ,
2
6
2
18
12
4
0
00
1
2
13
13
1
31
31
31
31
31
31
1B bB 1
31
31
31
31
BVIteration
2
1x2x3x 0 0 1 2
0 1 0 0 6
1 0 0 2
21
RightSide3x 4x 5x2x1x
Coefficient of:
0 0 0 1
31
31
31
31
10
1
0
0
0
0
0
1
3,5,0
230
0
1
0
0
0
0
1
0,5,0
44
144 caBccZ B
551
55 caBccZ B
23
21
31
31
31
31
21
31
31
31
31
BVIteration
2
1x2x3x 0 0 1 2
0 1 0 0 6
1 0 0 2
21
RightSide3x 4x 5x2x1x
Coefficient of:
0 0 0 1 36
31
31
31
31
36
2
6
2
3,5,01
bBcB
so
23
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.
The dual problem uses exactly the same parameters as the primal problem, but in different location.
Primal and Dual Problems
Primal Problem Dual Problem
Max
s.t.
Min
s.t.
n
jjj xcZ
1
,
m
iii ybW
1
,
n
1jijij ,bxa
m
ijiij cya
1,
for for.,,2,1 mi .,,2,1 nj
for .,,2,1 mi for .,,2,1 nj ,0jx ,0iy
In matrix notation
Primal Problem Dual Problem
Maximize
subject to
.0x .0y
Minimize
subject tobAx cyA
,cxZ ,ybW
Where and are row vectors but and are column vectors.
c myyyy ,,, 21 b x
Example
Maxs.t.
Min
s.t.
Primal Problem Dual Problem
,53 21 xxZ ,18124 321 yyyW
1823 21 xx
122 2 x41x
0x,0x 21
522 32 yy
33 3 y1y
0y,0y,0y 321
Max
s.t.
Primal Problem in Matrix Form
Dual Problem in Matrix Form
Min
s.t.
,5,32
1
x
xZ
18
12
4
,
2
2
0
3
0
1
2
1
x
x
.0
0
2
1
x
x .0,0,0,, 321 yyy
5,3
2
2
0
3
0
1
,, 321
yyy
18
12
4
,, 321 yyyW
Primal-dual table for linear programmingPrimal Problem
Coefficient of: RightSide
Rig
ht
Sid
eDu
al P
rob
lem
Co
effi
cien
to
f:
my
y
y
2
1
21
11
a
a
22
12
a
a
n
n
a
a
2
1
1x 2x nx
1c 2c ncVI VI VI
Coefficients forObjective Function
(Maximize)
1b
mna2ma1ma
2b
mb
Coe
ffic
ient
s fo
r O
bjec
tive
Fun
ctio
n(M
inim
ize)
One Problem Other Problem
Constraint Variable
Objective function Right sides
i i
Relationships between Primal and Dual Problems
Minimization Maximization
Variables
Variables
Constraints
Constraints
0
0
0
0
Unrestricted
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 cxs.t. Ax b x 0
L(X,Y) = cx - y(Ax - b) = yb + (c - yA) x
Min yb
s.t. yA c
y 0
Lagrangian Function )],([ YXL
X
YXL
)],([
= c-yA
The following relation is always maintained
yAx yb (from Primal: Ax b)
yAx cx (from Dual : yA c)
From (1) and (2), we have
cx yAx yb
At optimality
cx* = y*Ax* = y*b
is always maintained.
(1)
(2)
(3)
(4)
“Complementary slackness Conditions” are
obtained from (4)
( c - y*A ) x* = 0
y*( b - Ax* ) = 0
xj* > 0 y*aj = cj , y*aj > cj xj* = 0
yi* > 0 aix* = bi , ai x* < bi yi* = 0
(5)
(6)
Any pair of primal and dual problems can be
converted to each other.
The dual of a dual problem always is the primal
problem.
Min W = yb,
s.t. yA c
y 0.
Dual ProblemMax (-W) = -yb,
s.t. -yA -c
y 0.
Converted to Standard Form
Min (-Z) = -cx,
s.t. -Ax -b
x 0.
Its Dual Problem
Max Z = cx,
s.t. Ax b
x 0.
Converted toStandard Form
Mins.t.
64.06.0
65.05.0
7.21.03.0
21
21
21
xx
xx
xx
0,0 21 xx
21 5.04.0 xx
Mins.t.
][y 64.06.0
][y 65.05.0
][y 65.05.0
][y 7.21.03.0
321
-221
221
121
xx
xx
xx
xx
0,0 21 xx
21 5.04.0 xx
Maxs.t.
.0,0,0,0
5.04.0)(5.01.0
4.06.0)(5.03.0
6)(67.2
3221
3221
3221
3221
yyyy
yyyy
yyyy
yyyy
Maxs.t.
.0, URS:,0
5.04.05.01.0
4.06.05.03.0
667.2
321
321
321
321
yyy
yyy
yyy
yyy
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
Theory of Simplex Method: Consider the following
problem.
0x&0x,0x,0x
4xx2xx3
5xxx2x4
,x2xx3x4Z
4321
4321
4321
4321
to subject
Maximize
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
A – 1 cont’d
Basic Variable
Coefficient of:Right SideEq. Z x1 x2 x3 x4 x5 x6
Z (0) 1 1 1
x2 (1) 0 1 -1
x4 (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?
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
A – 2 cont’d
IngredientCalories 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
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.
A – 2 cont’d