EMGT 501
HW #1
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
Due Day: Sep 13
Ch. 2 – 18
For the linear program
Max
4 x1  1x2
s.t.
10 x1  2 x2  30
3x1  2 x2  12
2 x1  2 x2  10
x1 , x2  0
a. Write this linear program in standard form.
b. Find the optimal solution using the graphical solution
procedure.
c. What are the values of the three slack variables at the
optimal solution?
Ch. 2 – 20
Embassy Motorcycle (EM) manufactures two lightweight
motorcycles designed for easy handling and safety. The EZRider model has a new engine and a low profile that make it
easy to balance. The Lady-Sport model is slightly larger,
uses a more traditional engine, and is specifically designed
to appeal to women riders. Embassy produces the engines
for both models at its Des Moines, Iowa, plant. Each EZRider engine requires 6 hours of manufacturing time and
each Lady-Sport engine requires 3 hours of manufacturing
time. The Des Moines plant has 2100 hours of engine
manufacturing time available for the next production period.
Embassy’s motorcycle frame supplier can supply as many
EZ-Rider frames as needed.
However, the Lady-Sport frame is more complex and the
supplier can provide only up to 280 Lady-Sport frames for
the next production period. Final assembly and testing
requires 2 hours for each EZ-Rider model and 2.5 hours for
each Lady-Sport model. A maximum of 1000 hours of
assembly and testing time are available for the next
production period. The company’s accounting department
projects a profit contribution of $2400 for each EZ-Rider
produced and $1800 for each Lady-Sport produced.
a. Formulate a linear programming model that can be used
to determine the number of units of each model that
should be produced in order to maximize the total
contribution to profit.
b. Find the optimal solution using the graphical solution
procedure.
c. Which constraints are binding.
Ch. 3 – 28
National Insurance Associates carries an investment portfolio of
stocks, bonds, and other investment alternatives. Currently $200,000
of funds are available and must be considered for new investment
opportunities. The four stock options National is considering and the
relevant financial data are as follows:
Stock
A
B
C
D
Price per share
$100 $50
$80
$40
Annual rate of return
0.12 0.08 0.06 0.10
Risk measure per dollar invested
0.10 0.07 0.05 0.08
The risk measure indicates the relative uncertainty associated with the
stock in terms of its realizing the projected annual return; higher
values indicate greater risk. The risk measures are provided by the
firm’s top financial advisor.
National’s top management has stipulated the following
investment guidelines: the annual rate of return for the portfolio
must be at least 9% and no one stock can account for more than
50% of the total dollar investment.
a. Use linear programming to develop an investment portfolio that
minimizes risk.
b. If the firm ignores risk and uses a maximum return-on-investment
strategy, what is the investment portfolio?
c. What is the dollar difference between the portfolios in parts (a)
and (b)? Why might the company prefer the solution developed
in part (a)?
Ch. 4 – 3
The employee credit union at State University is planning
the allocation of funds for the coming year. The credit
union makes four types of loans to its members. In addition,
the credit union invests in risk-free securities to stabilize
income. The various revenue-producing investments
together with annual rates of return are as follows:
Type of Loan/Investment
Automobile loans
Furniture loans
Other secured loans
Signature loans
Risk-free securities
Annual Rate of Return (%)
8
10
11
12
9
The credit union will have $2,000,000 available for
investment during the coming year. State laws and credit
union policies impose the following restrictions on the
composition of the loans and investments.
• Risk-free securities may not exceed 30% of the total funds
available for investment.
• Signature loans may not exceed 10% of the funds invested
in all loans (automobile, furniture, other secured, and
signature loans).
• Furniture loans plus other secured loans may not exceed
the automobile loans
• Other secured loans plus signature loans may not exceed
the funds invested in risk-free securities.
How should the $2,000,000 be allocated to each of the
loan/investment alternatives to maximize total annual
return? What is the projected total annual return?
Ch. 5 – 6
Basis cB
x1
5
2
0
3
x2 x3
20 25
1
0
2
1
0 -1/2
s1
0
1
0
0
s2
0
0
1
0
s3
0
0
0
1
40
30
15
zj
cj  zj
a. Complete the initial tableau.
b. Write the problem in tableau form.
c. What is the initial basis? Does this basis correspond to
the origin? Explain.
d. What is the value of the objective function at this initial
solution?
e. For the next iteration, which variable should enter the
basis, and which variable should leave the basis?
f. How many units of the entering variable will be in the
next solution? Before making this first iteration, what do
you think will be the value of the objective function after
the first iteration?
g. Find the optimal solution using the simplex method.
Theory of Simplex Method
A two-variable linear programming problem
Max Z  3x1  5x2 ,
s.t.
x1
4
and
2x2  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
Z
s.t.
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 vector of basic variables
X N : a vector of non basic variables
c B : a vector of these correspond ing objective
coefficien ts (to X B )
c N : a vector of these correspond ing objective
coefficien ts (to X N )
B : a basic matrix
N : a nonbasic matrix
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
From Eq. (2),
 Z  1

  
 X B  0
1

0
(11)
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 
(12)
1
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
0
B  2 
0  1 1
Iteration BV
1
x3
x2
x5
1 0 0   4   4 
0 1 0 12  6,
1

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
1

Z1  c1  cB B a1  c1  0,5,00
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






1

1
B A  0 2 0  0 2   0 1 
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 13  13
 1 1 3  1 3   4   2
1   0 1 0 , 1   0 1 0  12  6,
2
2
B 
 B b 
   
 0  13 13 
 0  13 13  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
Z 4  c4
Z 5  c5
1 13  13 0
1
 cB B a4  c4  0,5,00 1 2 0 1  0  3 2
0  13 13 0
1 1  13 0
3
1

 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
2
1
0
0
0
3
0
0
1
1
0
1
0
1
1
0
0
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
2
1
0
0
0
3
0
0
1
1
0
1
0
1
1
0
0
3
2
1
3
1
3
0
1
3
Right
Side
36
2
6
2