Preparatory course for opre.315
Written by Yoosef Khadem
Excel, PH Stat 2 or SPSS in UB Computer Labs
These computer applications are available in all UB labs. Any UB student, staff, or faculty with
a net ID can also access these applications from home using Citrix. Instructions are at
http://www.ubalt.edu/about-ub/offices-and-services/technology-services/faqs/citrix_remote_access.cfm.
Below is the link that gives you the instructions on how to connect to install Citrix for Mac.
http://www.ubalt.edu/about-ub/offices-and-services/technology-services/faqs/citrix_remote_access.cfm
Dear students of opre.315,
The prerequisite of opre.315 is completion of Math111 (College Algebra) here at UB or
elsewhere. This preparatory class is neither perquisite of opre.315 (Decision sciences) nor will
guarantee your success in completion of oper.315. I will try to cover the following subjects to
boost your confidence level and brush up your rusty basic algebra skills for a good start in
Decision sciences class where you face more challenging problems. It will also increase your
problem solving skills for universal test such as GRE or GMAT. Wish you the best.
Contents covered in this handout are:
1) Linear equations and inequalities
2) Solving two equations and two unknowns
3) Slope and y-intercept of a line
4) Graph of the equation of a line
5) Graphing linier inequalities
6) Introduction to linear programming
7) Solving a basic linear programming problem by Excel
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Review for OPRE 315: Algebra with Applications
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Solving linear equality and inequalities:
EQUALITIES
Definition: An equality is two algebraic expressions linked by an equal sign =.
If a, b, and c are real numbers then the following properties can be used in solving
equalities.
 If a = b then a + c = b + c or a - c = b - c.
 If a = b then for any c  0, ac = bc and a  c = b  c.
Note: The second property is good for removing fractions in equalities.
 If a(b+c), use distributive property to remove the parentheses,
hence a(b+c) = ab + ac.
INEQUALITIES
Definition: When two algebraic expressions are linked by inequality signs such as > , < ,
<, and > they form an inequality. To solve an inequality use the following properties,
which in many cases are similar to the properties used in solving equalities.
If a, b, and c are real numbers then the following properties can be used in solving inequalities
 If a > b and b > c then a > c.
 If a > b then a+ c > b + c or a - c > b - c.
 If a > b and c > 0 (this means c is a positive number) then ac > bc or a / c > b / c.
Note: If you notice you see these properties so far are similar to the equations
properties.
 If a > b and c < 0 (this means c is negative) then ac < bc or a/c < b/c.
EXAMPLES
1. Solve the following equation for X.
2(3 - X) + 10 = 5 - 2(3X + 2)
Solution:
6 – 2X + 10 = 5 – 6X – 4
2X + 16 = -6X + 1
6X – 2X = -16 + 1
4X = -15 = -15/4 = -3.75
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2. Solve for y if 2x + 5y = 12
Note: When you solve for y in terms of you will find slope and y-intercept of the line. In
equation y =mx + b , m is the slope of the line and b is the y-intercept. Usually yintercept should be written as a pair point like (0, b).
3.
TWO EQUATIONS AND TWO UNKNOWNS
Our main focus in this section is to solve systems of equations with two unknown
variables. The solutions to these types of equations can be very tricky, especially in the
data sufficiency questions of the GMAT exam. Our objective is, not only to be able to
solve the equations and find answers, but to be able to convert verbal expressions into
equations and find the unknowns. The most popular way to solve two equations and
two unknowns is called the elimination method. In some books it is called the addition
or subtraction method. There are three possible answers for these types of equations.
The system may have a unique solution, infinite many solutions or no solutions at all.
My main objective is to show you how to find answers to application problems.
However, before solving application problems, it is appropriate to learn the mechanics
of solving systems of two equations and two unknowns using the elimination method.
Solve for X and Y
2X - Y = 3
X + 2Y =14
Solution: When X and Y in both equations are in the left side and the constant terms in
the right hand side, multiply first and second equations by numbers that leads to the
discovery of two opposite terms in the left side of the two equations. Add the two
equations together to eliminate one of the two variables. When this is done, the result
is one equation and one unknown. Solve this equation for the unknown variable. To find
the other unknown, simply substitute the unknown found in the previous step into one
of the original equations to find the other unknown. This is illustrated below.
2 {2X - Y = 3 }
X + 2Y = 14
4X – 2Y = 6
X +2Y = 14
5X
= 20
X= 4
Now substitute X = 4 in first equation and find the Y value.
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2(4) - Y = 3
8-Y=3
-Y = -8 + 3
-Y = -5
Y=5
Note: This system could have been given like following:
Solve for X1 and X2
2 X1 X2= 3
X1 + 2 X2 =14
Sustitution Method
Using the first equation solve for Y in terms of X. The answer is Y = 2 - X. Since one of the
two variables can be arbitrary, we let X be equal to any value and then calculate Y using
the arbitrary X value. For example if X = 5 then Y = 2 - 5 that is Y = -3. In this case we can
say a particular solution of this system is X = 5 and Y = -3.
Some more examples:
Solve the following systems of two equations and two unknowns
1) 3x+ 4y = 5
-x +2 y = 5
2) 2x = - y + 4
x- 3y = -5
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Solving Linear inequalities and finding feasible regions.
Examples:
Find the feasible region for the following linear inequalities:
1) X +Y ≤ 10
2X + 3Y ≥ 12
y
x
2) X -Y ≤ 20
-2X + 3Y ≥ 12
y
x
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Basic word problem
A daily diet consists of 300 grams of a mixture of two foods, low calorie food and high calorie
food. The low calorie and high calorie and high calorie foods each consists of 10% and 15%
protein, respectively. If the diet must provide exactly 38 grams of protein daily, how many
grams of low calorie food are in the mixture?
a) 100
b) 140
c) 150
d) 160
e) 200
Answer B
General questions regarding systems of equations and the equation of a straight line.
Slope of a line:
To find the slope of a line, solve the given line equation for dependent variable in term
of independent variable. Generally find standard equation of the line, Y = mX + b. Once
you find this equation refer to coefficient of X(independent variable). The coefficient of
independent variable is called the slope of the line. If dependent variable is X2 , solve
the given equation for X2 in terms of X1.
Examples
Find the slope of given lines:
a) Line 7 X1 + 10 X2 = 70
Solution:
Solve for X2 in terms of X1 , the coefficient of X1 is the slope of the line.
7 X1 + 10 X2 = 70  10 X2 = -7 X1 + 70  divide by 10 to isolate X2.
(10/10) X2 = ( -7/10) X1 + (70/10)  X2 = ( -7/10) X1 + 7 , hence the slope is
-7/10.
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Review for OPRE 315: Algebra with Applications
b)
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Line X1 = 10
Solution: The is a vertical line and the slope of it is undefined or infinity.
c) Line X2 = 5
Solution: this is a horizontal line and the slope is zero.
d) Line 1 x  2 y  3 .
2
5
Solution: to get rid of fractions, multiply all terms by LCD of all denominators. In
this case the LCD is 10. Hence
10( 1 x  2 y  3 ) that gives 5x  4 y  30 now isolate y, 4y = -5x + 30.
2
5
5
15
5
Dividing all terms by 4 we get y   x 
hence the slope is  .
4
2
4
e) Graph the following linear inequality and find feasible region.
x-2y ≤4
2x+4y ≤12
y
x
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Now let’s see how knowing these basic algebra concepts are going to help a student
with a good starts in Decision Sciences class.
Problem statement
Assume a department in a furniture company is only assigned to produce chairs and
tables. Past statistics show profit per chair is $7 while the profit per table is $10.
Manufacturing chairs and tables require labors and materials. The required labor and
material for each chair and table is given in the following table:
Labor
Material
A Chair
2
1
A Table
3
1
There are no more than 36 hours of labors and 16 units of materials available for this
department. Also because of space limitation the department can’t accommodate more
than 14 chairs per day. Formulate a linear program problem that allows you to produce
a daily output of chairs and tables for a maximum daily profit, using least amount of
labors and materials and not exceeding the space limitation.
SOLUTION
Step 1. Define decision variables. Decision variables tell how much or how many
of something to produce, invest, purchase, hire, etc.
In this problem, decision variables are:
X1 = daily number of chairs to produce.
X2 = daily number of tables to produce.
Note: X1 and X2 are similar to X and Y dimensions in a two dimensional systems.
The reason for choosing a letter followed by indices 1, 2, … is because in a
higher level formulations you may have to choose several variables at a
time in a linear program formulation. Choosing different indices limits
the number of letters used in a linear program.
Step 2. Find the objective function. In this problem, the goal is to maximize the
daily profit. Problem statement Indicate that profit per chair is $7 while the
profit per table is $10, hence the objective function in this is 7 X1 + 10 X2 or P
=7 X1 + 10 X2 .
Step 3. The most challenging part of a linear programming problem is to
translate all problem statements into linear equality or inequalities. There are
not many standard format to formulate a linear programming problem.
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Sometimes translation of problem statements is very simple while the other
times very complex and tedious. In order to be able to find answers by hand and
computer we must simplify inequalities in the form that all variables in simplest
form are in left side of equality or inequality signs and the constant is in right
hand side of equality or inequality signs. Constraints in our problem are:
2 X1 + 3X2 ≤ 36
X1 + X2 ≤ 16
X1
≤ 14
Labor Hours
Materials
Limitation for the number of chairs
So all together, we can say:
Maximize
7 X1 + 10 X2
Subjected to:
2 X1 + 3X2 ≤ 36 Labor Hours
X1 + X2 ≤ 16 Materials
X1
≤ 14
Limitation for number of chairs
Since number of chairs and tables can’t be negative, we should add to more
constraints. They are:
X1 ≥0 and X2 ≥0
We can rewrite all as:
Max 7 X1 + 10 X2
s.t.
2 X1 + 3X2 ≤ 36
X1 + X2 ≤ 16
X1
≤ 14
X1 , X2 ≥0
Solution by graphing method
To solve this problem, start with non-negativity constraints, X1 , X2 ≥0
X2
Quadrant II
X1<0
X2>0
Quadrant I
X1>0
X2>0
X1
Quadrant III
X1<0
X2<0
Quadrant IV
X1>0
X2<0
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X2
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As you see non-negativity constraints, X1 , X2 ≥0 indicate final answer should
fall in the Quadrant I.
X1
X2
0
12
18
0
X1
Notice: The equation of X2 axis (in basic algebra is known as y axis) is X1 = 0 and the
equation of X1 axis(in basic algebra is known as x axis) is X2 = 0. Line X1 = 0 is a vertical
line and its slope is undefined. Line X2 = 0 is a horizontal line and its slope is zero.
Now we should draw each constraint individually or simultaneously and identify
whether pair points below, above, left or right side of the line are feasible.
Graph 2 X1 + 3X2
≤ 36 , using x-intercept and y-intercept method:
X2
X1
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Graph
X 1 + X2
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≤ 16 , using x-intercept and y-intercept method:
X2
X1
Graph X1 ≤ 14.
X2
X1
0
16
14
X2
16
0
X1
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Now let’s combine three regions to find the final feasible region.
X2
x1 + x2 = 16
C
A
2x1 + 3x2 = 36
B
D
X1
As you see the shaded region is the set of all pair points that satisfy all constraints. Each
pair point represents the number of chairs and tables produce in one day. We have to
select a pair point that gives us the maximum profit. Obviously pair points closer to
origin (0,0) result to smaller profit so we do our best to evaluate the objective function
using pair points that are feasible but far away from the origin. Faraway pair points in
boundaries of feasible set should be tested in order to find the best answer for the
problem. Since there are so many pair points in the boundaries of feasible region, we
only test pair points that are found by intersecting any two lines in feasible region. In
this problem the pair points that we call VERTICES OF FEASIBLE REGION are: C, A, B
and D. Of course the origin (pair point (0,0) ) is a vertex of feasible region but since the
results when evaluating the objective function is always zero, we don’t care about it.
Once we clearly identify measurements of vertices of feasible region, we evaluate the
objective function for each vertex to see which one give the largest value in a
maximization problem and the smallest value in a minimization problem. Referring to
graph we see it is easy to find the measurement of some of these vertices but not all. A
vertex such as C is simply y-intercept of line 2 X1 + 3X2 = 36. So using this equation we
can simply let X1 = 0 and find X2 as shown below:
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Let’s set X1 = 0 in the equation 2 X1 + 3X2 = 36 and find X2 that is 2 (0 )+ 3X2 = 36 or
solving for X2 we get X2 = 12. Hence pair point C is (0 , 12). Similarly you can see
measurement of pair point D that is (14, 0). This is x-intercept of line X1 =14.
To fond pair points A and B, we have to solve to separate systems of two equations and
two unknowns.
To find A, solve:
2 X1 + 3X2 = 36
X1 + X2 = 16
Solution:
As I mentioned in review algebra part, usually you can use elimination method to
solve a system with two equations and two unknowns. In this problem as you
see if you multiply the second equation by -2 you can discover two opposite X1
terms in left sides of each equation.
2 X1 + 3X2 = 36
-2 X1 - 2X2 = -32
Add left sides and right sides in order we get X2 = 4. If we substitute X2 = 4 in the
first or the second equation, we find X1 value as following:
X2 = 4  2 X1 + 3X2 = 36 or 2 X1 + 3(4) = 36 that gives X1 =12. Pair point
A is (12,4).
To find B, solve system:
X1 + X 2
X1
= 16
= 14
Solution: As you see X1 = 14 so substituting this value in first equation we get
X2 = 2. Therefore pair point B is ( 14 ,2 ).
Now in a table we evaluate the objective function using these vertices.
Corner Points
origin
C
A
B
D
Pair point value
(0,0)
(0,12)
(12,4)
(14,2)
(14,0)
Objective function 7 X1 + 10 X2
7(0) + 10(0)
=0
7 (0) + 10 (12) =120
7 (12) + 10 (4) =124
7 (14) + 10 (2) =118
7 (14) + 10 (0) =98
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As you see the best answer occurs at (12,4) for a maximum of $124. This means the
department should produce 12 chairs and 4 tables for a maximum profit of $124.
The optimal value is X1 =12, X2 = 4 and optimal solution is $124.
Unused resources
The original problem:
Maximize 7 X1 + 10 X2
2 X1 + 3X2 ≤ 36
X1 + X2 ≤ 16
X1
≤ 14
X1 , X2 ≥0
Labor Hours
Materials
Limitation for the number of chairs
Could be converted to the standard form by adding a slack to constraints with ≤
or subtracting a surplus from constraint with ≥ signs. Hence the standard form
of our system is:
Maximize
s.t.
7 X1 + 10 X2
2 X1 + 3X2 + S1 = 36
X1 + X2 + S2 = 16
X1
+ S3 = 14
Labor Hours
Materials
Limitation for the number of chairs
In this standard format, S1 is the unused labor hour, S2 is the unused material
and S3 is the left over (if any) of our limitation constraint. To find these values,
we solve the system using our optimal value X1 =12, X2 = 4.
2 (12) + 3(4) + S1 = 36
12 + 4 + S2 = 16
12
+ S3 = 14
Labor Hours
Materials
Limitation for the number of chairs
S1= 0 , S2 = 0 and S3 =2. This means we used all of the labor hours and materials
and did not exceed our Limitation for the number of chairs. Constraints number
1 and 2 are called binding constraints and the third one is non-binding. A binding
constraint is also is a fully utilized constraint with slack of zero.
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Now let’s solve the same problem with Excel Solver.
Maximize
7 X1 + 10 X2
Subjected to:
2 X1 + 3X2 ≤ 36
X1 + X2 ≤ 16
X1
≤ 14
Number of Labor Hours
Number of units of Materials
Limitation for number of chairs
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Microsoft Excel 14.0 Answer Report
Worksheet: [Book1]Sheet1
Report Created: 1/15/2014 3:40:29 PM
Result: Solver found a solution. All Constraints and optimality conditions are satisfied.
Solver Engine
Engine: Simplex LP
Solution Time: 0.031 Seconds.
Iterations: 2 Subproblems: 0
Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative
Objective Cell (Max)
Cell
$D$1
Name
Original Value
Objective Function
0
Final Value
124
Variable Cells
Cell
$B$1
$B$2
Name
Original Value
0
0
Name
Cell Value
x1
x2
Final Value
Integer
12 Contin
4 Contin
Constraints
Cell
$A$4
$A$5
$A$6
x2
x2
x2
Formula
36 $A$4<=$C$4
16 $A$5<=$C$5
12 $A$6<=$C$6
Status
Binding
Binding
Not Binding
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Slack
0
0
2
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Microsoft Excel 14.0 Sensitivity Report
Worksheet: [Book1]Sheet1
Report Created: 1/15/2014 3:40:29 PM
Variable Cells
Cell
$B$1
$B$2
Final
Name Value
x1
12
x2
4
Reduced
Cost
0
0
Objective
Coefficient
7
10
Allowable
Increase
Final
Name Value
x2
36
x2
16
x2
12
Shadow
Price
3
1
0
Constraint
R.H. Side
36
16
14
Allowable
Increase
12
0.666666667
1E+30
3
0.5
Allowable
Decrease
0.333333333
3
Constraints
Cell
$A$4
$A$5
$A$6
Allowable
Decrease
2
4
2
Microsoft Excel 14.0 Limits Report
Worksheet: [Book1]Sheet1
Report Created: 1/15/2014 3:40:30 PM
Objective
Name
Cell
Value
$D$1 Objective Function
Cell
$B$1
$B$2
Variable
Name
x1
x2
124
Value
12
4
Lower
Limit
0
0
x1
x2
12
4
Objective Function
36
16
12
<=
<=
<=
36
16
14
Objective
Result
40
84
Upper Objective
Limit
Result
12
124
4
124
124
Based on this output, we can find the optimal solution and answer many different questions
related to range of optimality and range of feasibilities.
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