Chapter 1 Linear Equations and Graphs

Chapter 1 Linear Equations and Graphs
Section 1 Linear Equations and Inequalities

Linear Equations, Standard Form
In general, a first-degree, or linear, equation in one variable is any equation that can be written in the form where a is not equal to zero. This is called the standard form of the linear equation. For example, the equation is a linear equation because it can be converted to standard form by clearing of fractions and simplifying.

Equivalent Equations Two equations are equivalent if one can be transformed into the other by performing a series of operations which are one of two types: 1. The same quantity is added to or subtracted from each side of a given equation. 2. Each side of a given equation is multiplied by or divided by the same nonzero quantity. To solve a linear equation, we perform these operations on the equation to obtain simpler equivalent forms, until we obtain an equation with an obvious solution.

Example of Solving a Linear Equation
Example: Solve

Example of Solving a Linear Equation
Example: Solve Solution: Since the LCD of 2 and 3 is 6, we multiply both sides of the equation by 6 to clear of fractions. Cancel the 6 with the 2 to obtain a factor of 3, and cancel the 6 with the 3 to obtain a factor of 2. Distribute the 3. Combine like terms.

Solving a Formula for a Particular Variable
Example: Solve M =Nt +Nr for N.

Solving a Formula for a Particular Variable
Example: Solve M=Nt+Nr for N. Factor out N: Divide both sides by (t + r):

Linear Inequalities If the equality symbol = in a linear equation is replaced by an inequality symbol (<, >, ≤, or ≥), the resulting expression is called a first-degree, or linear, inequality. For example is a linear inequality.

Solving Linear Inequalities
We can perform the same operations on inequalities that we perform on equations, except that the sense of the inequality reverses if we multiply or divide both sides by a negative number. For example, if we start with the true statement –2 > –9 and multiply both sides by 3, we obtain –6 > –27. The sense of the inequality remains the same. If we multiply both sides by -3 instead, we must write 6 < 27 to have a true statement. The sense of the inequality reverses.

Example for Solving a Linear Inequality
Solve the inequality 3(x – 1) < 5(x + 2) – 5

Example for Solving a Linear Inequality
Solve the inequality 3(x – 1) < 5(x + 2) – 5 Solution: 3(x –1) < 5(x + 2) – 5 3x – 3 < 5x + 10 – Distribute the 3 and the 5 3x – 3 < 5x + 5 Combine like terms. –2x < 8 Subtract 5x from both sides, and add 3 to both sides x > -4 Notice that the sense of the inequality reverses when we divide both sides by -2.

Interval and Inequality Notation
If a < b, the double inequality a < x < b means that a < x and x < b. That is, x is between a and b. Interval notation is also used to describe sets defined by single or double inequalities, as shown in the following table. Interval Inequality [a,b] a ≤ x ≤ b (–∞,a] x ≤ a [a,b) a ≤ x < b (–∞,a) x < a (a,b] a < x ≤ b [b,∞) x ≥ b (a,b) a < x < b (b,∞) x > b

Interval and Inequality Notation and Line Graphs
(A) Write [–5, 2) as a double inequality and graph . (B) Write x ≥ –2 in interval notation and graph.

Interval and Inequality Notation and Line Graphs
(A) Write [–5, 2) as a double inequality and graph . (B) Write x ≥ –2 in interval notation and graph. (A) [–5, 2) is equivalent to –5 ≤ x < 2 [ ) x -5 2 (B) x ≥ –2 is equivalent to [–2, ∞) [ x -2

Procedure for Solving Word Problems
Read the problem carefully and introduce a variable to represent an unknown quantity in the problem. Identify other quantities in the problem (known or unknown) and express unknown quantities in terms of the variable you introduced in the first step. Write a verbal statement using the conditions stated in the problem and then write an equivalent mathematical statement (equation or inequality.) Solve the equation or inequality and answer the questions posed in the problem. Check the solutions in the original problem.

Example: Break-Even Analysis
A recording company produces compact disk (CDs). One-time fixed costs for a particular CD are $24,000; this includes costs such as recording, album design, and promotion. Variable costs amount to $6.20 per CD and include the manufacturing, distribution, and royalty costs for each disk actually manufactured and sold to a retailer. The CD is sold to retail outlets at $8.70 each. How many CDs must be manufactured and sold for the company to break even?

Break-Even Analysis (continued)
Solution Step Let x = the number of CDs manufactured and sold. Step Fixed costs = $24,000 Variable costs = $6.20x C = cost of producing x CDs = fixed costs + variable costs = $24,000 + $6.20x R = revenue (return) on sales of x CDs = $8.70x

Step The company breaks even if R = C, that is if $8.70x = $24,000 + $6.20x Step x = 24, x Subtract 6.2x from both sides 2.5x = 24, Divide both sides by 2.5 x = 9,600 The company must make and sell 9,600 CDs to break even.

Step 5. Check: Costs = $24,000 + $6.2 ∙ 9,600 = $83,520 Revenue = $8.7 ∙ 9,600 = $83,520

Graphing an Equation To sketch the graph an equation in x and y, we need to find ordered pairs that solve the equation and plot the ordered pairs on a grid. This process is called point-by-point plotting. For example, let’s plot the graph of the equation

Graphing an Equation: Making a Table of Ordered Pairs
Make a table of ordered pairs that satisfy the equation x y –3 (–3)2+2 = 11 –2 (–2)2+2 = 6 –1 (–1)2+2 = 6 (0)2+2 = 2 1 (1)2+2 = 3 2 (2)2+2 = 6

Graphing an Equation: Plotting the points
Next, plot the points and connect them with a smooth curve. You may need to plot additional points to see the pattern formed.

Functions The previous graph is the graph of a function. The idea of a function is this: a correspondence between two sets D and R such that to each element of the first set, D, there corresponds one and only one element of the second set, R. The first set is called the domain, and the set of corresponding elements in the second set is called the range. For example, the cost of a pizza (C) is related to the size of the pizza. A 10 inch diameter pizza costs $9.00, while a 16 inch diameter pizza costs $12.00.

Function Definition You can visualize a function by the following diagram which shows a correspondence between two sets: D, the domain of the function, gives the diameter of pizzas, and R, the range of the function gives the cost of the pizza. 10 9.00 12 10.00 16 12.00 domain D range R

Functions Specified by Equations
If in an equation in two variables, we get exactly one output (value for the dependent variable) for each input (value for the independent variable), then the equation specifies a function. The graph of such a function is just the graph of the specifying equation. If we get more than one output for a given input, the equation does not specify a function.

Functions Specified by Equations
Consider the equation that was graphed on a previous slide Input: x = –2 –2 Process: square (–2), then subtract 2 (–2,2) is an ordered pair of the function. Output: result is 2 2

Vertical Line Test for a Function
If you have the graph of an equation, there is an easy way to determine if it is the graph of an function. It is called the vertical line test which states that: An equation specifies a function if each vertical line in the coordinate system passes through at most one point on the graph of the equation. If any vertical line passes through two or more points on the graph of an equation, then the equation does not specify a function.

Vertical Line Test for a Function (continued)
This graph is not the graph of a function because you can draw a vertical line which crosses it twice. This is the graph of a function because any vertical line crosses only once.

Function Notation The following notation is used to describe functions. The variable y will now be called f (x). This is read as “ f of x” and simply means the y coordinate of the function corresponding to a given x value. Our previous equation can now be expressed as

Function Evaluation Consider our function What does f (–3) mean?

Function Evaluation Consider our function
What does f (–3) mean? Replace x with the value –3 and evaluate the expression The result is 11 . This means that the point (–3,11) is on the graph of the function.

Some Examples 1.

Domain of a Function Consider which is not a real number.
Question: for what values of x is the function defined?

Domain of a Function Answer:
is defined only when the radicand (3x – 2) is equal to or greater than zero. This implies that

Domain of a Function (continued)
Therefore, the domain of our function is the set of real numbers that are greater than or equal to 2/3. Example: Find the domain of the function

Domain of a Function (continued)
Therefore, the domain of our function is the set of real numbers that are greater than or equal to 2/3. Example: Find the domain of the function Answer:

Domain of a Function: Another Example
Find the domain of

Domain of a Function: Another Example
Find the domain of In this case, the function is defined for all values of x except where the denominator of the fraction is zero. This means all real numbers x except 5/3.

Mathematical Modeling
The price-demand function for a company is given by where x represents the number of items and P(x) represents the price of the item. Determine the revenue function and find the revenue generated if 50 items are sold.

Solution Revenue = Price ∙ Quantity, so
R(x)= p(x) ∙ x = (1000 – 5x) ∙ x When 50 items are sold, x = 50, so we will evaluate the revenue function at x = 50: The domain of the function has already been specified. We are told that

Break-Even and Profit-Loss Analysis
Any manufacturing company has costs C and revenues R. The company will have a loss if R < C, will break even if R = C, and will have a profit if R > C. Costs include fixed costs such as plant overhead, etc. and variable costs, which are dependent on the number of items produced C = a + bx (x is the number of items produced)

Break-Even and Profit-Loss Analysis (continued)
Price-demand functions, usually determined by financial departments, play an important role in profit-loss analysis p = m – nx (x is the number of items than can be sold at $p per item.) The revenue function is R = (number of items sold) ∙ (price per item) = xp = x(m – nx) The profit function is P = R – C = x(m – nx) – (a + bx)

Example of Profit-Loss Analysis
A company manufactures notebook computers. Its marketing research department has determined that the data is modeled by the price-demand function p(x) = 2,000 – 60x, when 1 < x < 25, (x is in thousands). What is the company’s revenue function and what is its domain?

Answer to Revenue Problem
Since Revenue = Price ∙ Quantity, The domain of this function is the same as the domain of the price-demand function, which is 1 ≤ x ≤ 25 (in thousands.)

Profit Problem The financial department for the company in the preceding problem has established the following cost function for producing and selling x thousand notebook computers: C(x) = 4, x (x is in thousand dollars). Write a profit function for producing and selling x thousand notebook computers, and indicate the domain of this function.

Answer to Profit Problem
Since Profit = Revenue – Cost, and our revenue function from the preceding problem was R(x) = 2000x – 60x2, P(x) = R(x) – C(x) = 2000x – 60x2 – ( x) = –60x x – 4000. The domain of this function is the same as the domain of the original price-demand function, 1< x < 25 (in thousands.) 5000 Thousand dollars 25 Thousand cameras

Chapter 3 Mathematics of Finance
Section 1 Simple Interest

3.1 Simple Interest Definition: I = Prt where I = interest earned
P = principal (amount invested) r = interest rate (as a decimal) t = time

An Example Find the interest on a boat loan of $5,000 at 16% for 8 months.

Example Find the interest on a boat loan of $5,000 at 16% for 8 months. Solution: Use I = Prt I = 5,000(0.16)(0.6667) (8 months = 8/12 of one year years) I = $533.33

Total Amount to Be Paid Back
The total amount to be paid back for the boat loan would be $5000 plus the interest of $533.33, for a total of $5,

Present Value and Future Value
A = P + Prt = P (1 + rt) where A = amount, or future value P = principal, or present value r = annual simple interest rate (as a decimal) t = time in years

Another Example Find the total amount due on a loan of $600 at 16% interest at the end of 15 months.

Another Example Find the total amount due on a loan of $600 at 16% interest at the end of 15 months. Solution: A = P (1 + rt) A = 600(1+0.16(1.25)) A = $720.00

Interest Rate Earned on a Note
What is the annual interest rate earned by a 33-day T-bill with a maturity value of $1,000 that sells for $996.16?

Interest Rate Earned on a Note Solution
Solution: Use the equation A = P (1 + rt) ,000 = We normally use 360 days for a financial year

Another Application A department store charges 18.6% interest (annual) for overdue accounts. How much interest will be owed on a $1080 account that is 3 months overdue?

Another Application solution
A department store charges 18.6% interest (annual) for overdue accounts. How much interest will be owed on a $1080 account that is 3 months overdue? Solution: I = Prt I = 1080(0.186)(0.25) I = $50.22

Chapter 4 Systems of Linear Equations; Matrices
Section 1 Review: Systems of Linear Equations in Two Variables

Opening Example A restaurant serves two types of fish dinners- small for $5.99 and large for $ One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $1, How many small fish dinners and how many large fish dinners were ordered?

Systems of Two Equations in Two Variables
We are given the linear system ax + by = h cx + dy = k A solution is an ordered pair (x0, y0) that will satisfy each equation (make a true equation when substituted into that equation). The solution set is the set of all ordered pairs that satisfy both equations. In this section, we wish to find the solution set of a system of linear equations. To solve a system is to find its solution set.

Solve by Graphing One method to find the solution of a system of linear equations is to graph each equation on a coordinate plane and to determine the point of intersection (if it exists). The drawback of this method is that it is not very accurate in most cases, but does give a general location of the point of intersection. Lets take a look at an example: Solve the following system by graphing: 3x + 5y = –9 x+ 4y = –10

Solve by Graphing Solution
3x + 5y = –9 x+ 4y = –10 First line (intercept method) If x = 0, y= –9/5 If y = 0 , x = –3 Plot points and draw line (2, –3) Second line: Intercepts are (0, –5/2) , ( –10,0) From the graph we estimate that the point of intersection is (2,–3). Check: 3(2) + 5(–3)= –9 and 2 + 4(–3)= –10. Both check.

Another Example Now, you try one: Solve the system by graphing:
2x + 3 = y x + 2y = –4

Another Example Solution
Now, you try one: Solve the system by graphing: 2x + 3 = y x + 2y = –4 We can do this on a graphing calculator by first solving the second equation for y to get y = –0.5x – 2 If we enter both of these equations into the calculator and find the intersection point we get the screen shown. The solution is (–2,–1).

Method of Substitution
Although the method of graphing is intuitive, it is not very accurate in most cases, unless done by calculator. There is another method that is 100% accurate. It is called the method of substitution. This method is an algebraic one. It works well when the coefficients of x or y are either 1 or –1. For example, let’s solve the previous system 2x + 3 = y x + 2y = -4 using the method of substitution.

Method of Substitution (continued)
The steps for this method are as follows: Solve one of the equations for either x or y. Substitute that result into the other equation to obtain an equation in a single variable (either x or y). Solve the equation for that variable. Substitute this value into any convenient equation to obtain the value of the remaining variable.

Example (continued) 2x + 3 = y x + 2y = –4 x + 2(2x + 3) = –4
Substitute y from first equation into second equation Solve the resulting equation After we find x = –2, then from the first equation, we have 2(–2) + 3 = y or y = –1. Our solution is (–2, –1)

Another Example Solve the system using substitution: 3x – 2y = 7
y = 2x – 3

Another Example Solution
Solve the system using substitution: 3x – 2y = 7 y = 2x – 3 Solution:

Basic Terms A consistent linear system is one that has one or more solutions. If a consistent system has exactly one solution (unique solution), it is said to be independent. An independent system will occur when the two lines have different slopes. If a consistent system has more than one solution, it is said to be dependent. A dependent system will occur when the two lines have the same slope and the same y intercept. In other words, the graphs of the lines will coincide. There will be an infinite number of points of intersection. Two systems of equations are equivalent if they have the same solution set.

Basic Terms An inconsistent linear system is one that has no solutions. This will occur when two lines have the same slope but different y intercepts. In this case, the lines will be parallel and will never intersect. Example: Determine if the system is consistent, independent, dependent or inconsistent: 2x – 5y = –4x + 10y = –1

Solution of Example Solve each equation for y to obtain the slope intercept form of the equation: Since each equation has the same slope but different y intercepts, they will not intersect. This is an inconsistent system.

Elimination by Addition
The method of substitution is not preferable if none of the coefficients of x and y are 1 or –1. For example, substitution is not the preferred method for the system below: x – 7y = 3 –5x + 3y = 7 A better method is elimination by addition. The following operations can be used to produce equivalent systems: Two equations can be interchanged. An equation can be multiplied by a non-zero constant. A constant multiple of one equation and is added to another equation.

Elimination by Addition Example
For our system, we will seek to eliminate the x variable. The coefficients are 2 and –5. Our goal is to obtain coefficients of x that are additive inverses of each other. We can accomplish this by multiplying the first equation by 5, and the second equation by 2. Next, we can add the two equations to eliminate the x-variable. Solve for y Substitute y value into original equation and solve for x Write solution as an ordered pair

Elimination by Addition Another Example
Solve 2x – 5y = 6 –4x + 10y = –1

Elimination by Addition Another Example
Solve 2x – 5y = 6 –4x + 10y = –1 Solution: 1. Eliminate x by multiplying equation 1 by 2 . 2. Add two equations 3. Upon adding the equations, both variables are eliminated producing the false equation 0 = 11 4. Conclusion: If a false equation arises, the system is inconsistent and there is no solution.

Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog?

Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog? Solution: Let x represent the amount of time spent walking and y represent the amount of time spent jogging. Since the total time spent walking and jogging is 0.9 hours, we have the equation x + y = We are given the total distance traveled as 3.5 miles. Since Distance = Rate x Time, distance walking = 3x and distance jogging = 5y. Then total distance is 3x + 5y= 3.5.

Application (continued)
We can solve the system using substitution. 1. Solve the first equation for y 2. Substitute this expression into the second equation. 3. Solve second equation for x 4. Find the y value by substituting this x value back into the first equation. 5. Answer the question: Time spent jogging is 0.4 hours. Solution:

Supply and Demand The quantity of a product that people are willing to buy during some period of time depends on its price. Generally, the higher the price, the less the demand; the lower the price, the greater the demand. Similarly, the quantity of a product that a supplier is willing to sell during some period of time also depends on the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices. The simplest supply and demand model is a linear model where the graphs of a demand equation and a supply equation are straight lines.

Supply and Demand (continued)
In supply and demand problems we are usually interested in finding the price at which supply will equal demand. This is called the equilibrium price, and the quantity sold at that price is called the equilibrium quantity. If we graph the the supply equation and the demand equation on the same axis, the point where the two lines intersect is called the equilibrium point. Its horizontal coordinate is the value of the equilibrium quantity, and its vertical coordinate is the value of the equilibrium price.

Supply and Demand Example
Example: Suppose that the supply equation for long-life light bulbs is given by p = 1.04 q – 7.03, and that the demand equation for the bulbs is p = –0.81q + 7.5 where q is in thousands of cases. Find the equilibrium price and quantity, and graph the two equations in the same coordinate system.

Supply and Demand (Example continued)
If we graph the two equations on a graphing calculator and find the intersection point, we see the graph below. Demand Curve Supply Curve Thus the equilibrium point is (7.854, 1.14), the equilibrium price is $1.14 per bulb, and the equilibrium quantity is 7,854 cases.

Now, Solve the Opening Example
A restaurant serves two types of fish dinners- small for $5.99 each and large for $ One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $ How many small dinners and how many large dinners were ordered?

Solution A restaurant serves two types of fish dinners- small for $5.99 each and large for $ One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $ How many small dinners and how many large dinners were ordered? Answer: 60 small orders and 74 large orders

Chapter 5 Linear Inequalities and Linear Programming
Section 1 Linear Inequalities in Two Variables

Systems of Linear Inequalities in Two Variables
In this section, we will learn how to graph linear inequalities in two variables and then apply this procedure to practical application problems.

Half-Planes A line divides the plane into two regions called half planes. A vertical line divides it into left and right half planes. A nonvertical line divides it into upper and lower half planes. In either case, the dividing line is called the boundary line of each half plane, as indicated in the figure. Boundary Line Upper Half-plane Left half-plane Right half-plane Lower Half-plane Boundary Line

Graphs of Linear Inequalities
The graph of the linear inequality Ax + By < C or Ax + By > C with B ≠ 0 is either the upper half-plane or the lower half-plane (but not both) determined by the line Ax + By = C. If B = 0 and A ≠ 0, the graph of Ax < C or Ax > C is either the left half-plane or the right half-plane (but not both) determined by the line Ax = C.

Procedure for Graphing Linear Inequalities
Step 1. First graph Ax + By = C as a dashed line if equality is not included in the original statement, or as a solid line if equality is included. Step 2. Choose a test point anywhere in the plane not on the line (the origin (0,0) usually requires the least computation) and substitute the coordinates into the inequality. Step 3. Does the test point satisfy the original inequality? If so, shade the half-plane that contains the test point. If not, shade the opposite half-plane.

Graphing a Linear Inequality Example 1
Our first example is to graph the linear equality

Graphing a Linear Inequality Example 1
Our first example is to graph the linear equality Solution: Replace the inequality symbol with an equal sign 2. Graph the line. If the original inequality is a > or < sign, the graph of the line should be dotted, otherwise solid.

Example 1 (continued) In this example, since the original problem contained the inequality symbol (<) the line that is graphed should be dotted. For our problem, the equation of our line is already in slope-intercept form, (y = mx + b) so we easily sketch the line by first starting at the y intercept of –1, then moving up 3 units and to the right 4 units, corresponding to our slope of ¾. After locating the second point, we sketch the dotted line passing through these two points.

Example 1 (continued) 3. Now, we have to decide which half plane to shade. The solution set will either be (a) the half plane above the line, or (b) the half plane below the graph of the line. To determine which half-plane to shade, we choose a test point that is not on the line. Usually, a good test point to pick is the origin (0,0), unless the origin happens to lie on the line. In our case we can choose the origin as a test point. Substituting the origin in the inequality produces the statement 0 < 0 – 1, or 0 < –1.

Example 1 Graph Since this is a false statement, we shade the region on the side of the line not containing the origin. Had the origin satisfied the inequality, we would have shaded the region on the side of the line containing the origin. Here is the complete graph of the first inequality:

Example 1 Calculator Graph
We can also draw the graph on a graphing calculator, but we won’t be able to graph the dotted boundary line.

Example 2 For our second example, we will graph the inequality 3x – 5y ≥ 15.

Example 2 For our second example, we will graph the inequality 3x – 5y ≥ 15. Step 1. Replace inequality symbol with equal sign: 3x – 5y = 15 Step 2. Graph the line 3x – 5y = 15. We will graph the line using the x and y intercepts: When x = 0, y = -3 and when y = 0, x = 5. Plot these points and draw a solid line. The original inequality symbol is ≥, which means that the graph of the line itself is included. Graph is as shown.

Example 2 (continued) Step 3. Choose a point not on the line. Again, the origin is a good test point since it is not part of the line itself. We have the following statement which is clearly false. Therefore, we shade the region on the side of the line that does not include the origin.

Example 2 (continued)

Example 3 Our third example is unusual in that there is no y variable present. The inequality 2x > 8 is equivalent to the inequality x > 4. How shall we proceed to graph this inequality?

Example 3 Our third example is unusual in that there is no y variable present. The inequality 2x > 8 is equivalent to the inequality x > 4. How shall we proceed to graph this inequality? The answer is: the same way we graphed previous inequalities: Step 1: Replace the inequality symbol with an equals sign: x = 4. Step 2: Graph the line x = 4. Is the line solid or dotted? The original inequality is >. Therefore, the line is dotted. Step 3. Choose the origin as a test point. Is 2(0) > 8? Clearly not. Shade the side of the line that does not include the origin. The graph is displayed on the next slide.

Example 3 Graph

Example 4 Graph y ≤ –2. This example illustrates the type of problem in which the x variable is missing.

Example 4 Solution Graph y ≤ –2.
This example illustrates the type of problem in which the x variable is missing. We will proceed the same way. Step 1. Replace the inequality symbol with an equal sign: y = –2 Step 2. Graph the equation y = –2 . The line is solid since the original inequality symbol is ≤. Step 3. Shade the appropriate region. Choosing again the origin as the test point, we find that 0 ≤ -2 is a false statement so we shade the side of the line that does not include the origin. Graph is shown in next slide.

Example 4 Graph

Application A concert promoter wants to book a rock group for a stadium concert. A ticket for admission to the stadium playing field will cost $125, and a ticket for a seat in the stands will cost $175. The group wants to be guaranteed total ticket sales of at least $700,000. How many tickets of each type must be sold to satisfy the group’s guarantee? Express the answer as a linear inequality and draw its graph.

Application Solution x = number of tickets sold for the playing field
y = number of tickets sold for seats in the stands Total tickets sale must be at least $700,000. 1. Graph: 2. Test point (0, 0) and it’s false. 3. The graph is the upper half-plane including the boundary line.

Application Solution It’s not possible to sell a negative number of tickets, we must restrict both x and y to the first quadrant.

Chapter 6 Linear Programming: The Simplex Method
Section 1 A Geometric Introduction to the Simplex Method

6.1 Geometric Introduction to the Simplex Method
The geometric method of linear programming from the previous section is limited in that it is only useful for problems involving two decision variables and cannot be used for applications involving three or more decision variables. It is for this reason that a more sophisticated method must be developed.

George Dantzig George B. Dantzig developed such a method in 1947 while being assigned to the U.S. military. Ideally suited to computer use, the method is used routinely on applied problems involving hundreds and even thousands of variables and problem constraints.

An Interview with George Dantzig, Inventor of the Simplex Method
IRV How do you explain optimization to people who haven't heard of it? GEORGE I would illustrate the concept using simple examples such as the diet problem or the blending of crude oils to make high-octane gasoline. IRV What do you think has held optimization back from becoming more popular? GEORGE It is a technical idea that needs to be demonstrated over and over again. We need to show that firms that use it make more money than those who don't. IRV Can you recall when optimization started to become used as a word in the field? GEORGE From the very beginning of linear programming in 1947, terms like maximizing, minimizing, extremizing, optimizing a linear form and optimizing a linear program were used.

An Interview with George Dantzig (continued)
GEORGE The whole idea of objective function, which of course optimization applies, was not known prior to linear programming. In other words, the idea of optimizing something was something that nobody could do, because nobody tried to optimize. So while you are very happy with it and say it's a very familiar term, optimization just meant doing it better than somebody else. And the whole concept of getting the optimum solution just didn't exist. So my introducing the whole idea of optimization in the early days was novel. IRV I understand that while programming the war effort in World War II was done on a vast scale, the term optimization as a word was never used. What was used instead? GEORGE A program can be thought of as a set of blocks, or activities, of different shapes that can to be fitted together according to certain rules, or mass balance constraints. Usually these can be combined in many different ways, some more, some less desirable than other combinations. Before linear programming and the simplex method were invented, it was not possible to computationally determine the best combination such as finding the program that maximizes the number of sorties flown. Instead, all kinds of ground rules were invented deemed by those in charge to be desirable characteristics for a program to have.

A typical example of a ground rule that might have been used was: "Go ask General Arnold which alternative he prefers." A big program might contain hundreds of such highly subjective rules. And I said to myself: "Well, we can't work with all these rules." Because what it meant was that you set up a plan. Then you have so many rules that you have to get some resolution of these rules and statements of what they were. To do this, you had to be running to the general, and to his assistants and asking them all kinds of questions. IRV Name some of your most important early contributions. GEORGE The first was the recognition that most practical planning problems could be reformulated mathematically as finding a solution to a system of linear inequalities. My second contribution was recognizing that the plethora of ground rules could be eliminated and replaced by a general objective function to be optimized. My third contribution was the invention of the simplex method of solution.

IRV And these were great ideas that worked and still do. GEORGE Yes, I was very lucky. IRV What would you say is the most invalid criticism of optimization? GEORGE Saying: "It's a waste of time to optimize because one does not really know what are the exact values of the input data for the program." IRV Ok, let's turn this around. What would you say is the greatest potential of optimization? GEORGE It has the potential to change the world.

Standard Maximization Problem in Standard Form
A linear programming problem is said to be a standard maximization problem in standard form if its mathematical model is of the following form: Maximize P = c1x1 + c2x2 + … + cnxn subject to problem constraints of the form a1x1 + a2x2 + … + anxn < b, b > 0 with nonnegative constraints x1, x2, …, xn > 0

Example We will use a modified form of a previous example. Consider the linear programming problem of maximizing z under the given constraints. This is a standard maximization problem.

Example (continued) Any two of them intersect in a point. There are a total of six such points. The coordinates of the point can be found by solving the system of two equations in two unknowns created by the equations of the two lines. Some of the points are corner points of the feasible region, and some are outside. There are four lines bordering the feasible region:

Example (continued) (0,0) feasible (0,15) (0,20) not feasible
(7.5,12.5) feasible (optimal) (20,0) (45,0) grid spacing = 5 units

Slack Variables To use the simplex method, the constraint inequalities must be converted to equalities. Consider the two constraint inequalities To make this into a system of equations, we introduce slack variables They are called slack variables because they take up the slack between the left and right hand sides of the inequalities. Note that the slack variables must be nonnegative.

Slack Variables (continued)
We now have two equations in four unknowns x, y, s1, s2 The system has an infinite number of solutions since there are more unknowns than equations. We can make the system have a unique solution by assigning two of the variables a value of zero and then solving for the remaining two variables. This is called a basic solution. We select any two of the four variables as basic variables. The remaining two variables automatically become non-basic variables. The non-basic variables are always assigned a value of zero. Then we solve the equations for the two basic variables.

Example (continued) The equations are
Select x = 0 and y = 0 as the non-basic variables. Then the first equation reads 8(0) + 8(0) + s1 = 160, so s1 = The second equation becomes 4(0) + 12(0) + s2 = 180, so s2 = 180. This corresponds to the point (x, y) = (0,0). According to the graph we saw earlier, this is a feasible point. The other 5 choices of non-basic variables are done similarly. The table on the next slide summarizes the results. x y s1 s2 point feasible? 160 180 (0,0) yes

Example (continued) Basic Solutions x y s1 s2 point feasible? 160 180
160 180 (0,0) yes 20 -60 (0,20) no 15 40 (0,15) 100 (20,0) 45 -200 (45,0) 7.5 12.5 (7.5,12.5)

Discovery! Each basic solution corresponds to an intersection point of the boundary lines of the feasible region. A feasible basic solution corresponds to a corner point (vertex) of the feasible region. This includes the optimum solution of the linear programming problem. A basic solution that is not feasible includes at least one negative value; a basic feasible solution does not include any negative values. We can determine the feasibility of a basic solution simply by examining the signs of all the variables in the solution.

Generalization In a linear programming problem with slack variables there will always be more variables than equations. Basic variables are selected arbitrarily (as many basic variables as there are equations). The remaining variables are called non-basic variables. We obtain a basic solution by assigning the non-basic variables a value of zero and solving for the basic variables. If a basic solution has no negative values, it is basic feasible solution. Theorem: If the optimal value of the objective function in a linear programming problem exists, then that value must occur at one (or more) of the basic feasible solutions.

Conclusion This is simply the first step in developing a procedure for solving a more complicated linear programming problem. But it is an important step in that we have been able to identify all the corner points (vertices) of the feasible set without a having three or more variables.

Section 2 The Simplex Method: Maximization with Problem Constraints of the Form ≤

Example 1 We will solve the same problem that was presented earlier, but this time we will use the Simplex Method. We wish to maximize the profit function subject to the constraints below. The method introduced here can be used to solve larger systems that are more complicated.

Introduce Slack Variables; Rewrite Objective Function
The first step is to rewrite the system without the inequality symbols and to introduce slack variables. We also rewrite the objective function in a form that matches the other equations.

Set up the Initial Simplex Tableau
A simplex tableau is essentially the same as an augmented matrix from chapter 4. The manipulations on a tableau are also the same as in the Gauss-Jordan method before. We rename the variable y to x2, for consistency of notation.

The Initial Simplex Tableau (continued)
The variables s1, s2, P on the left are the basic variables. A basic variable always has one 1 in its column, otherwise 0. We can read off is value directly from the rightmost column. x1 and x2 are 0 (the non-basic variables), so s1=160, s2=180, P=0. This is the initial basic feasible solution that corresponds to the origin, since (x1,x2) = (0,0).

Determine the Pivot Element
160/8 = /12 = 15 The most negative element in the last row is –10. Therefore, the x2-column containing –10 is the pivot column. To determine the pivot row, we divide the coefficients above the -10 into the numbers in the rightmost column and determine the smallest quotient. That happens in the second row, labeled s2. x2 is the entering variable, and s2 is the exiting variable. x2 will become a basic variable, and s2 will become non-basic.

The Pivot Step The goal is to use elementary row operations to produce a 1 in the column of the entering variable (in the row corresponding to the exiting variable), and 0 otherwise. The elementary row operations are applied the same way as in chapter 4. Step 1: Divide row s2 by 12, to get a 1 in pivot position.

The Pivot Step (continued)
Step 2. Obtain zeros in the other positions of the pivot column by subtracting multiples of the pivot row from the other rows. Relabel the rows: x2 is the entering variable, and s2 exits.

Find the Next Pivot Element
The process must continue since the last row of the matrix still contains a negative value (–5/3). The next pivot column is column x1.

Find the Next Pivot Element (continued)
Divide the entries in the last column by the entries in the pivot column, and pick the smallest value. The pivot row is row s1. The entering variable is x1, the exiting variable is s1. 40/(16/3) = /(1/3) = 45

Do the Pivot Step Step 2: Multiply row 1 by 5/3 and add the result to row 3. Multiply row 1 by -1/3 and add result to row 2. Step 1: Divide row 1 by 16/3, to get a 1 in pivot position.

Read Off the Solution Since the last row of the matrix contains no negative numbers, we can stop the procedure. Assign 0 to the non-basic variables and read off x1 = x = 7.5 and x2 = y = This is the same solution we obtained geometrically. The profit is maximized at $162.50

Interpreting the Simplex Process Geometrically
We can now interpret the simplex process just completed geometrically, in terms of the feasible region graphed in the preceding section. The following table lists the three basic feasible solutions we just found using the simplex method, in the order they were found. The table also includes the corresponding corner points of the feasible region. x1 x2 s1 s2 P Corner Point 160 180 (0, 0) 15 40 150 (0, 15) 7.5 12.5 162.50 (7,5, 12.5)

Interpreting the Simplex Process Geometrically (continued)
Looking at the table on the previous slide, we see that the simplex process started at the origin, moved to the adjacent corner point (0, 15) and then to the optimal solution (7.5, 12.5). This is typical of the simplex process. 2 3 1

Selecting Basic and Nonbasic Variables for the Simplex Method
Given a simplex tableau, 1. Numbers of variables. Determine the number of basic variables and the number of nonbasic variables. These numbers do not change during the simplex process. 2. Selecting basic variables. A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element (usually 1) and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. This procedure always selects P as a basic variable, since the P column never changes during the simplex process.

Selecting Basic and Nonbasic Variables for the Simplex Method
3. Selecting nonbasic variables. After the basic variables are selected in step 2, the remaining variables are selected as the nonbasic variables. The tableau columns under the nonbasic variables usually contain more than one nonzero element.

Selecting the Pivot Element
Locate the most negative indicator in the bottom row of the tableau to the left of the P column (the negative number with the largest absolute value). The column containing this element is the pivot column. If there is a tie for the most negative indicator, choose either column. Divide each positive element in the pivot column above the dashed line into the corresponding element in the last column. The pivot row is the row corresponding to the smallest quotient obtained. If there is a tie for the smallest quotient, choose either row. If the pivot column above the dashed line has no positive elements, there is no solution, and we stop.

Selecting the Pivot Element
The pivot (or pivot element) is the element in the intersection of the pivot column and pivot row. Note: The pivot element is always positive and never appears in the bottom row. Remember: The entering variable is at the top of the pivot column, and the exiting variable is at the left of the pivot row.

Performing a Pivot Operation
A pivot operation, or pivoting, consists of performing row operations as follows: 1. Multiply the pivot row by the reciprocal of the pivot element to transform the pivot element into a 1. (If the pivot element is already a 1, omit this step.) 2. Add multiples of the pivot row to other rows in the tableau to transform all other nonzero elements in the pivot column into 0’s.

Simplex Method Summarized
Step 1: Write the standard maximization problem in standard form with slack variables. Yes Step 2: Are there any negative indicators in the bottom row? Stop. The optimal solution has been found. Step 3: Select the pivot column No Step 5: Select the pivot element and perform the pivot operation. Step 4: Are there any positive elements in the pivot column above the dashed line? Yes No Stop. The linear programming problem has no optimal solution.

Example 2 Agriculture A farmer owns a 100-acre farm and plans to plant at most three crops. The seed for crops A, B and C costs $40, $20, and $30 per acre, respectively. A maximum of $3,200 can be spent on seed. Crops A, B, and C require 1, 2, and 1 work days per acre, respectively, and there are a maximum of 160 work days available. If the farmer can make a profit of $100 per acre on crop A, $300 per acre on crop B, and $200 per acre on crop C, how many acres of each crop should be planted to maximize profit?

Example 2 (continued) The farmer must decide on the number of acres of each crop that should be planted. Thus, the decision variables are x1 = number of acres of crop A x2 = number of acres of crop B x3 = number of acres of crop C The farmer’s objective is to maximize profit: P = 100x x x3

Example 2 (continued) The farmer is constrained by the number of acres available for planting, the money available for seed, the the available work days. These lead to the following constraints: x1 + x2 + x3 < Acreage constraint 40x1 + 20x2 + 30x3 < 3,200 Monetary constraint x x2 + x3 < 160 Labor constraint

Example 2 (continued) Adding the nonnegative constraints, we have the following model for a linear programming problem: Maximize P = 100x x x3 Subject to x1 + x2 + x3 < 100 40x1 + 20x2 + 30x3 < 3,200 x x2 + x3 < 160 x1, x2, x3 > 0

Example 2 (continued) Next, we introduce slack variables and form the initial system: x x x3 + s = 100 40x1 + 20x2 + 30x s = 3200 x x x s = 160 -100x x2 -200x P = 0 x1, x2, x3, s1, s2, s3 > 0 The initial system has 7 – 4 = 3 nonbasic variables and 4 basic variables.

Example 2 (continued) Now we form the simplex tableau and solve by the simplex method: Enter x x x s1 s2 s3 P s1 s2 s3 P Exit 0.5R3 -> R3

Example 2 (continued) x1 x2 x3 s1 s2 s3 P s1 s2 s3 P
(–1)R3 + R1 -> R1 (–20)R3 +R2->R2 300R3 + R4 ->R4

Example 2 (continued) x1 x2 x3 s1 s2 s3 P s1 s2 x2 P Enter Exit
2R1 -> R1

Example 2 (continued) x1 x2 x3 s1 s2 s3 P s1 s2 x2 P
(–20)R1 + R2 -> R2 (–0.5)R1 + R3 -> R3 50R1 + R4 -> R4

Example 2 (continued) x1 x2 x3 s1 s2 s3 P x3 s2 x2 P
All indicators in the bottom row are nonnegative, and we can now read off the optimal solution: x1 = 0, x2 = 60, x3 = 40, s1 = 0, s2 = 800, s3 = 0, P = $26,000

Example 2 (continued) Thus, if the farmer plants 60 acres in crop B, 40 acres in crop C and no crop A, the maximum profit of $26,000 will be realized. The fact that s2 = 800 tells us that this max profit is reached by using only $2400 of the $3200 available for seed; that is, we have a slack of $800 that can be used for some other purpose.

Section 3 The Dual Problem: Minimization with Problem Constraints of the Form ≥

Learning Objectives for Section 6.3
Dual Problem: Minimization with Problem Constraints of the Form > The student will be able to formulate the dual problem. The student will be able to solve minimization problems. The student will be able to solve applications of the dual such as the transportation problem. The student will be able to summarize problem types and solution methods.

Dual Problem: Minimization With Problem Constraints of the Form >
Associated with each minimization problem with > constraints is a maximization problem called the dual problem. The dual problem will be illustrated through an example. We wish to minimize the objective function C subject to certain constraints:

Initial Matrix We start with an initial matrix A which corresponds to the problem constraints:

Transpose of Matrix A To find the transpose of matrix A, interchange the rows and columns so that the first row of A is now the first column of A transpose. The transpose is used in formulating the dual problem to follow.

Dual of the Minimization Problem
The dual of the minimization problem is the following maximization problem: Maximize P under the constraints:

Formation of the Dual Problem
Given a minimization problem with ≥ problem constraints, 1. Use the coefficients and constants in the problem constraints and the objective function to form a matrix A with the coefficients of the objective function in the last row. 2. Interchange the rows and columns of matrix A to form the matrix AT, the transpose of A. 3. Use the rows of AT to form a maximization problem with ≤ problem constraints.

Theorem 1: Fundamental Principle of Duality
A minimization problem has a solution if and only if its dual problem has a solution. If a solution exists, then the optimal value of the minimization problem is the same as the optimum value of the dual problem.

Forming the Dual Problem
We transform the inequalities into equalities by adding the slack variables x1, x2, x3:

Form the Simplex Tableau for the Dual Problem
The first pivot element is 2 (in red) because it is located in the column with the smallest negative number at the bottom (-16), and when divided into the rightmost constants yields the smallest quotient (16/2=8)

Simplex Process Divide row 1 by the pivot element (2) and change the entering variable to y2

Simplex Process (continued)
Perform row operations to get zeros in the column containing the pivot element. Result is shown below. Identify the next pivot element (0.5) (in red) New pivot element Pivot element located in this column because of negative indicator

Variable y1 becomes new entering variable Divide row 2 by 0.5 to obtain a 1 in the pivot position (circled.)

Get zeros in the column containing the new pivot element. We have now obtained the optimal solution since none of the bottom row indicators are negative.

Solution of the Linear Programming Problem
Solution: An optimal solution to a minimization problem can always be obtained from the bottom row of the final simplex tableau for the dual problem. Minimum of P is 136, which is also the maximum of the dual problem. It occurs at x1 = 4, x2 = 8, x3 = 0

Solution of a Minimization Problem
Given a minimization problem with non-negative coefficients in the objective function, Write all problem constraints as ≥ inequalities. (This may introduce negative numbers on the right side of some problem constraints.) Form the dual problem. Write the initial system of the dual problem, using the variables from the minimization problem as slack variables.

Solution of a Minimization Problem
4. Use the simplex method to solve the dual problem. Read the solution of the minimization problem from the bottom row of the final simplex tableau in step 4. Note: If the dual problem has no optimal solution, the minimization problem has no optimal solution.

Application: Transportation Problem
One of the first applications of linear programming was to the problem of minimizing the cost of transporting materials. Problems of this type are referred to as transportation problems. Example: A computer manufacturing company has two assembly plants, plant A and plant B, and two distribution outlets, outlet I and outlet II. Plant A can assemble at most 700 computers a month, and plant B can assemble at most 900 computers a month. Outlet I must have at least 500 computers a month, and outlet II must have at least 1,000 computers a month.

Transportation Problem (continued)
Transportation costs for shipping one computer from each plant to each outlet are as follows: $6 from plant A to outlet I; $5 from plant A to outlet II: $4 from plant B to outlet I; $8 from plant B to outlet II. Find a shipping schedule that will minimize the total cost of shipping the computers from the assembly plants to the distribution outlets. What is the minimum cost?

Solution: To form a shipping schedule, we must decide how many computers to ship from either plant to either outlet. This will involve 4 decision variables: x1 = number of computers shipped from plant A to outlet I x2 = number of computers shipped from plant A to outlet II x3 = number of computers shipped from plant B to outlet I x4 = number of computers shipped from plant B to outlet II

Constraints are as follows: x1 + x2 < 700 Available from A x3 + x4 < 900 Available from B x1 + x3 > 500 Required at I x2 + x4 > 1,000 Required at II Total shipping charges are: C = 6x1 + 5x2 + 4x3 + 8x4

Thus, we must solve the following linear programming problem: Minimize C = 6x1 + 5x2 + 4x3 + 8x4 subject to x1 + x < 700 Available from A x3 + x4 < 900 Available from B x x > 500 Required at I x x4 > 1,000 Required at II Before we can solve this problem, we must multiply the first two constraints by -1 so that all are of the > type.

The problem can now be stated as: Minimize C = 6x1 + 5x2 + 4x3 + 8x4 subject to -x1 - x > -700 - x3 - x4 > -900 x x > 500 x x4 > 1,000 x1, x2, x3, x4 > 0

The dual problem is; Maximize P = -700y y2 +500y3 + 1,000y4 subject to -y y < 6 -y y4 < 5 -y2 + y < 4 -y y4 < 8 y1, y2, y3, y4 > 0

Introduce slack variables x1, x2, x3, and x4 to form the initial system for the dual: -y y x = 6 -y y x = 5 -y2 + y x = 4 -y y x = 8 -700y y2 +500y3 + 1,000y P = 0

Transportation Problem Solution
If we form the simplex tableau for this initial system and solve, we find that the shipping schedule that minimizes the shipping charges is 0 from plant A to outlet I, 700 from plant A to outlet II, 500 from plant B to outlet I, and 300 from plant B to outlet II. The total shipping cost is $7,900.

Summary of Problem Types and Simplex Solution Methods
Constraints of Type Right-Side Constants Coefficients of Objective Function Method of Solution Maximization < Nonnegative Any real number Simplex Method Minimization > Form dual and solve by simplex method

Section 4 Maximization and Minimization with Problem Constraints

Introduction to the Big M Method
In this section, we will present a generalized version of the simplex method that will solve both maximization and minimization problems with any combination of ≤, ≥, = constraints M

Example Maximize P = 2x1 + x2 subject to x1 + x2 < 10
To form an equation out of the first inequality, we introduce a slack variable s1, as before, and write x1 + x2 + s1 = 10.

Example (continued) To form an equation out of the second inequality we introduce a second variable s2 and subtract it from the left side so that we can write –x1 + x2 – s2 = 2. The variable s2 is called a surplus variable, because it is the amount (surplus) by which the left side of the inequality exceeds the right side.

Example (continued) We now express the linear programming problem as a system of equations: x1 + x2 + s = 10 –x1 + x – s = 2 –2x1 – x P = 0 x1, x2, s1, s2 > 0

Example (continued) It can be shown that a basic solution of a system is not feasible if any of the variables (excluding P) are negative. Thus a surplus variable is required to satisfy the nonnegative constraint. An initial basic solution is found by setting the nonbasic variables x1 and x2 equal to 0. That is, x1 = 0, x2,= 0,, s1= 10, s2 = -2, P = 0. This solution is not feasible because the surplus variable s2 is negative.

Artificial Variables In order to use the simplex method on problems with mixed constraints, we turn to a device called an artificial variable. This variable has no physical meaning in the original problem and is introduced solely for the purpose of obtaining a basic feasible solution so that we can apply the simplex method. An artificial variable is a variable introduced into each equation that has a surplus variable. To ensure that we consider only basic feasible solutions, an artificial variable is required to satisfy the nonnegative constraint.

Example (continued) Returning to our example, we introduce an artificial variable a1 into the equation involving the surplus variable s2: x1 + x2 – s2 + a1 = 2 To prevent an artificial variable from becoming part of an optimal solution to the original problem, a very large “penalty” is introduced into the objective function. This penalty is created by choosing a positive constant M so large that the artificial variable is forced to be 0 in any final optimal solution of the original problem.

Example (continued) We then add the term –Ma1 to the objective function: P = 2x1 + x2 – Ma1 We now have a new problem, called the modified problem: Maximize P = 2x1 + x2 - Ma1 subject to x1 + x2 + s = 10 x1 + x2 – s2 + a1 = 2 x1, x2, s1, s2, a1 > 0

Big M Method: Form the Modified Problem
If any problem constraints have negative constants on the right side, multiply both sides by -1 to obtain a constraint with a nonnegative constant. Remember to reverse the direction of the inequality if the constraint is an inequality. Introduce a slack variable for each constraint of the form ≤. Introduce a surplus variable and an artificial variable in each ≥ constraint. Introduce an artificial variable in each = constraint. For each artificial variable a, add –Ma to the objective function. Use the same constant M for all artificial variables.

Key Steps for Solving a Problem Using the Big M Method
Now that we have learned the steps for finding the modified problem for a linear programming problem, we will turn our attention to the procedure for actually solving such problems. The procedure is called the Big M Method.

Example (continued) The initial system for the modified problem is
x1 + x2 + s = 10 –x1 + x2 – s2 + a1 = 2 –2x1 – x2 + Ma1 + P = 0 x1, x2, s1, s2, a1 > 0 We next write the augmented coefficient matrix for this system, which we call the preliminary simplex tableau for the modified problem.

Example (continued) x x2 s s a P To start the simplex process we require an initial simplex tableau, described on the next slide. The preliminary simplex tableau should either meet these requirements, or it needs to be transformed into one that does.

Definition: Initial Simplex Tableau
For a system tableau to be considered an initial simplex tableau, it must satisfy the following two requirements: 1. The requisite number of basic variables must be selectable. Each basic variable must correspond to a column in the tableau with exactly one nonzero element. Different basic variables must have the nonzero entries in different rows. The remaining variables are then selected as non-basic variables. 2. The basic solution found by setting the non-basic variables equal to zero is feasible.

Example (continued) The preliminary simplex tableau from our example satisfies the first requirement, since s1, s2, and P can be selected as basic variables according to the criterion stated. However, it does not satisfy the second requirement since the basic solution is not feasible (s2 = -2.) To use the simplex method, we must first use row operations to transform the tableau into an equivalent matrix that satisfies all initial simplex tableau requirements. This transformation is not a pivot operation.

Example (continued) x x2 s1 s a1 P If you inspect the preliminary tableau, you realize that the problem is that s2 has a negative coefficient in its column. We need to replace s2 as a basic variable by something else with a positive coefficient. We choose a1.

Example (continued) We want to use a1 as a basic variable instead of s2. We proceed to eliminate M from the a1 column using row operations: x1 x2 s1 s2 a1 P (-M)R2 + R3 ->R3

Example (continued) From the last matrix we see that the basic variables are s1, a1, and P because those are the columns that meet the requirements for basic variables. The basic solution found by setting the nonbasic variables x1, x2, and s2 equal to 0 is x1 = 0, x2 = 0, s1 = 10, s2 = 0, a1 =2, P = –2M. The basic solution is feasible (P can be negative) and all requirements are met.

Example (continued) We now continue with the usual simplex process, using pivot operations. When selecting the pivot columns, keep in mind that M is unspecified, but we know it is a very large positive number. In this example, M – 2 and M are positive, and –M – 1 is negative. The first pivot column is column 2.

Example (continued) If we pivot on the second row, second column, and then on the first row, first column, we obtain: x1 x2 s1 s2 a1 P x1 x2 P Since all the indicators in the last row are nonnegative, we have the optimal solution: Max P = 14 at x1 = 4, x2 = 6, s1 = 0, s2 = 0, a1 = 0.

Big M Method: Summary To summarize:
1. Form the preliminary simplex tableau for the modified problem. 2. Use row operations to eliminate the Ms in the bottom row of the preliminary simplex tableau in the columns corresponding to the artificial variables. The resulting tableau is the initial simplex tableau. 3. Solve the modified problem by applying the simplex method to the initial simplex tableau found in the second step.

Big M Method: Summary (continued)
4. Relate the optimal solution of the modified problem to the original problem. A) if the modified problem has no optimal solution, the original problem has no optimal solution. B) if all artificial variables are 0 in the optimal solution to the modified problem, delete the artificial variables to find an optimal solution to the original problem C) if any artificial variables are nonzero in the optimal solution, the original problem has no optimal solution.

Chapter 7 Logic, Sets, and Counting
Section 1 Logic

Logic Logic is not only the foundation of mathematics, but also is important in numerous fields including law, medicine, and science. Although the study of logic originated in antiquity, it was rebuilt and formalized in the 19th and early 20th century. George Boole (Boolean algebra) introduced mathematical methods to logic in 1847, while Georg Cantor did theoretical work on sets and discovered that there are many different sizes of infinite sets.

Statements or Propositions
A proposition or statement is a declaration which is either true or false. Some examples: “2 + 2 = 5” is a statement because it is a false declaration. “Orange juice contains vitamin C” is a statement that is true. “Open the door.” This is not considered a statement since we cannot assign a true or false value to this sentence. It is a command, but not a statement or proposition.

Negation The negation of a statement p is “not p”, denoted by ¬ p
Truth table: If p is true, then its negation is false. If p is false, then its negation is true. p ¬ p T F

Disjunction A disjunction is of the form p  q and is read “p or q.”
Truth table for disjunction: A disjunction is true in all cases except when both p and q are false. p q p  q T F

Conjunction A conjunction is of the form p  q and is read “p and q.”
Truth table for conjunction: A conjunction is only true when both p and q are true. p q p  q T F

Conditional A conditional is of the form p  q and is read “if p then q.” Truth table for conditional: A conditional is only false if p is true and q is false, otherwise it is true. p q p  q T F

Conditional (continued)
To understand the logic behind the truth table for the conditional statement, consider the following statement. “If you get an A in the class, I will give you five bucks.” Let p be the statement “You get an A in the class” Let q be the statement “I will give you five bucks.” Think of the statement as a contract. The statement is T if the contract is satisfied, and F if the contract is broken. Now, if p is true (you got an A) and q is true (I give you the five bucks), the truth value of p  q is T. The contract was satisfied and both parties fulfilled the agreement.

Conditional (continued)
Now, suppose p is true (you got the A) and q is false (you did not get the five bucks). You fulfilled your part of the bargain, but weren’t rewarded with the five bucks. p  q is false since the contract was broken by the other party. Suppose p is false and q is true (you did not get an A, but received five bucks anyway.) No contract was broken. There was no obligation to receive five bucks, but is was not forbidden, either. The truth value of p  q is T. Finally, if both p and q are false, the contract was not broken. You did not receive the A and you did not receive the five bucks. The statement is again T.

Tautologies, Contradictions and Contingencies
The disjunction, conjunction and conditional statements introduced on previous slides are examples of contingencies because their truth values depend on the truth of its components. Some of the entries in the last column of the truth tables are true, and some are false. A proposition is a tautology if each entry in its column of the truth table is T, and a contradiction if each entry is F.

Tautologies and Contradictions (continued)
p  p p   p p   p T F For example, p   p is a tautology because it is always true, and p   p is a contradiction because it is always false.

Variations of the Conditional
The converse of p  q is q  p. The contrapositive of p  q is  q   p.

Example Let p = “your score is 90%” Let q = “your letter grade is A”
Conditional: p → q means “If your score is 90%, then your letter grade will be an A.” Let’s assume this is true. Converse: q → p means “If your letter grade is A, then your score is 90%.” Is the statement true? No. A student with a score of 95% also gets an A.

Example (continued) Contrapositive:  q   p means “If your letter grade is not an A, then your score was not 90%.” Is this true? Yes. If the original statement is true, the contrapositive will also be true because a statement and its contrapositive are logically equivalent. We will explain what that means on the next slide.

Logically Equivalent Statements
Two statements are logically equivalent if they have the same truth tables. is Example: Show that p  q is logically equivalent to  p  q.

Logically Equivalent Statements
Two statements are logically equivalent if they have the same truth tables. is Example: Show that p  q is logically equivalent to  p  q. We will construct the truth tables for both sides and determine that the truth values for each statement are identical.

Logically Equivalent Statements (continued)
p q p  q  p  q T F These two columns represent logically equivalent statements, so we can say that p  q   p  q.

Logical Implications P  Q
Consider the compound propositions P and Q. If whenever P is true, Q is also true, we say that P logically implies Q, or that P  Q is a logical implication, and write P  Q We can determine if a logical implication exists by examining a truth table.

Logical Equivalences Example
To verify  P  Q  P  Q construct the following truth table: Note that whenever  P  Q is true (which is only in the third row), P  Q is also true. P Q  P  Q P  Q T F

Chapter 1 Linear Equations and Graphs

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Chapter 1 Linear Equations and Graphs

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Presentation on theme: "Chapter 1 Linear Equations and Graphs"— Presentation transcript:

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