Transcript Document

EMGT 501
HW #1 Solutions
Chapter 2 - SELF TEST 18
Chapter 2 - SELF TEST 20
Chapter 3 - SELF TEST 28
Chapter 4 - SELF TEST 3
Chapter 5 - SELF TEST 6
Ch. 2 – 18
(a)
Max
4x1
+ 1x2
+ 0s1
+ 2x2
+ 1s1
+ 0s2
+ 0s3
s.t.
10x1
3x1
+ 2x2
2x1
+ 2x2
x1 ,
= 30
+ 1s2
= 12
+ 1s3
x2,
s1,
s2,
s3
0
= 10
x2
Ch. 2 – 18
(b)
14
12
10
8
6
Optimal Solution
x1 = 18/7, x2 = 15/7, Value = 87/7
4
2
(c)
s1 = 0, s2 = 0, s3 = 4/7
x1
0
2
4
6
8
10
Ch. 2 – 20
(a)
Let E = number of units of the EZ-Rider produced
L = number of units of the Lady-Sport produced
Max
2400E
+
1800L
6E
+
3L
s.t.
L
2E
+
2.5L
 2100
Engine time

280
Lady-Sport maximum
 1000
Assembly and testing
E, L  0
L
Ch. 2 – 20
(b)
700
Number of EZ-Rider Produced
600
Engine
Manufacturing Time
500
400
Frames for Lady-Sport
300
Optimal Solution E = 250, L = 200
Profit = $960,000
200
100
Assembly and Testing
0
E
100
200
300
400
Number of Lady-Sport Produced
500
Ch. 2 – 20
(c)
The binding constraints are the manufacturing time and the
assembly and testing time.
Ch. 3 – 28
(a)
Let A = number of shares of stock A
B = number of shares of stock B
C = number of shares of stock C
D = number of shares of stock D
To get data on a per share basis multiply price by rate of
return or risk measure value.
Min
10A
+ 3.5B
+
4C
+ 3.2D
100A
+
50B
+
80C
+ 40D
=
200,000
12A
+
4B
+
4.8C
+

18,000

100,000

100,000

100,000

100,000
s.t.
4D
100A
50B
80C
40D
A,
B,
C,
D
(9% of 200,00)
0
Solution:
A = 333.3, B = 0, C = 833.3, D = 2500
Risk: 14,666.7
Return: 18,000 (9%) from constraint 2
Ch. 3 – 28
Max
(b)
12A
+
4B
+
4.8C
+
4D
s.t.
100A
+
50B
+
80C
+
=
200,000

100,000

100,000

100,000

100,000
40D
100A
50B
80C
40D
A,
Solution:
Risk:
Return:
B,
C,
D0
A = 1000, B = 0, C = 0, D = 2500
10A + 3.5B + 4C + 3.2D = 18,000
22,000 (11%)
Ch. 3 – 28
(c)
The return in part (b) is $4,000 or 2% greater, but the risk
index has increased by 3,333.
Obtaining a reasonable return with a lower risk is a preferred
strategy in many financial firms. The more speculative,
higher return investments are not always preferred because
of their associated higher risk.
Ch. 4 – 3
0.08x1
Max
x1 = $ automobile loans
x2 = $ furniture loans
x3 = $ other secured loans
x4 = $ signature loans
x5 = $ "risk free" securities
+
0.10x2
+
0.11x3
+
0.12x4
+
0.09x5
s.t.
x5  600,000
x4
or
or
-0.10x1
-
x1
-
+
-
0.10x3
x2
+
x3
 x1
x2
+
x3
 0
or
x1
+
 0.10(x1 + x2 + x3 + x4)
0.10x2
x2
+
[1]
0.90x4
 0
[2]
[3]
x3
+
x4
+
x3
+
x4
-
x5
 0
[4]
+
x3
+
x4
+
x5
= 2,000,000
[5]
x1 ,
x2,
x3,
 x5
x4,
x5
0
Solution
Automobile Loans
(x1)
=
$630,000
Furniture Loans
(x2)
=
$170,000
Other Secured Loans
(x3)
=
$460,000
Signature Loans
(x4)
=
$140,000
Risk Free Loans
(x5)
=
$600,000
Annual Return $188,800 (9.44%)
Ch. 5 – 6
(a)
x1
x2
x3
s1
s2
s3
Basis
cB
5
20
25
0
0
0
s1
0
2
1
0
1
0
0
40
s2
0
0
2
1
0
1
0
30
s3
0
3
0
-1/2
0
0
1
15
Z
-cj +zj
0
-5
-20 -25
0
0
0
Ch. 5 – 6
(b)
Max
5x1
+ 20x2
2x1
+
+
25x3
+ 0s1
+ 0s2
+ 0s3
s.t.
1x2
2x2
3x1
+ 1s1
+
1x3
= 40
+ 1s2
-
= 30
+ 1s3
1/2x3
x1,
x 2,
x3,
s1,
s2,
s3,
 0.
= 15
Ch. 5 – 6
(c) The original basis consists of s1, s2, and s3. It is the
origin since the nonbasic variables are x1, x2, and x3
and are all zero.
(d) 0
x3 enters because it has the largest negative zj - cj
(e)
and s2 will leave because row 2 has the only positive
coefficient.
(f) 30; objective function value is 30 times 25 or 750.
(g) Optimal Solution:
x1 = 10 s1 = 20
x2 = 0 s2 = 0
x3 = 30 s3 = 0
z = 800.
EMGT 501
HW #2
Chapter 6 - SELF TEST 21
Chapter 6 - SELF TEST 22
Due Day: Sep 27
Ch. 6 – 21
Consider the following linear programming problem:
Max
s.t.
4 x1  3x2  6 x3
1x1  0.5 x2  1x3  15
2 x2  1x3  30
1x1  1x2  2 x3  20
x1 , x2 , x3  0
a. Write the dual problem.
b. Solve the dual.
c. Use the dual solution to identify the optimal solution to
the original primal problem.
d. Verify that the optimal values for the primal and dual
problems are equal.
Ch. 6 – 22
A sales representative who sells two products is trying to
determine the number of sales calls that should be made
during the next month to promote each product. Based on
past experience, representatives earn an average $10
commission for every call on product 1 and a $5
commission for every call on product 2. The company
requires at least 20 calls per month for each product and
not more than 100 calls per month on any one product. In
addition, the sales representative spends about 3 hours on
each call for product 1 and 1 hour on each call for
product 2. If 175 selling hours are available next month,
how many calls should be made for each of the two
products to maximize the commission?
a. Formulate a linear program for this problem.
b. Formulate and solve the dual problem.
c. Use the final simplex tableau for the dual problem to
determine the optimal number of calls for the products.
What is the maximum commission?
d. Interpret the values of the dual variables.
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
Primal Problem
in Algebraic Form
Max Z  3x1  5x2 ,
s.t.
x1
4
2 x2  12
3x1  2 x2  18
x1  0, x 2  0
Dual Problem
in Algebraic Form
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.
Question 1: Consider the following problem.
Maximize Z  2x1  4x2  3x3 ,
subject to
3x1  4 x2  2 x3  60
2 x1  x2  2 x3  40
x1  3x2  2 x3  80
and
x1  0, x2  0, x3  0.
(b) Work through the simplex method step by step in tabular form.
(c) Use a software package based on the simplex method to solve
the problem.
Question 2:
For each of the following linear programming models, give your recommendation on
which is the more efficient way (probably) to obtain an optimal solution: by applying the
simplex method directly to this primal problem or by applying the simplex method directly
to the dual problem instead. Explain.
(a) Maximize
Z  10x1  4x2  7 x3 ,
subject to
3 x1  x2  2 x3  25
x1  2 x2  3 x3  25
5 x1  x2  2 x3  40
x1  x2  x3  90
2 x1  x2  x3  20
and
x1  0, x2  0, x3  0.
(b) Maximize
Z  2x1  5x2  3x3  4x4  x5 ,
subject to
x1  3x2  2 x3  3x4  x5  6
4 x1  6 x2  5x3  7 x4  x5  15
and
x j  0,
for j  1, 2, 3, 4, 5.
Question 3:
Consider the following problem.
Maximize
Z   x1  2x2  x3 ,
subject to
x1  x2  2 x3  12
x1  x2  x3  1
and
x1  0, x2  0, x3  0.
(a) Construct the dual problem.
(b) Use duality theory to show that the optimal solution
for the primal problem has
Z  0.