We found in the previous section that the graphical method of solving linear programming problems, while time-consuming, enables us to see solution regions and identify corner points. This, however, is not possible when there are multiple variables. We can visualize in up to three dimensions, but even this can be difficult when there are numerous constraints.
To handle linear programming problems that contain upwards of two variables, mathematicians developed what is now known as the
simplex method. It is an efficient algorithm (set of mechanical steps) that 'toggles' through corner points until it has located the one that maximizes the objective function. Although tempting, there are a few things we need to lookout for prior to using it.
Prior to providing the mathematical details, let's see an example of a linear programming problem that
would qualify for the simplex method:
Example 1
The following system can be solved by using the simplex method:
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Objective Function: P = 2x + 3y + z
Subject to Constraints:
3x + 2y ≤ 5
2x + y – z ≤ 13
z ≤ 4
x,y,z≥0
Standard Maximization Problem
Mathematically speaking, in order to use the simplex method to solve a linear programming problem, we need the standard maximization problem:
- an objective function, and
- one or more constraints of the form a1x1 + a2x2 + … anxn le V
- All of the anumber represent real-numbered coefficients and
- the xnumber represent the corresponding variables.
- V is a non-negative (0 or larger) real number
Focus Matrix 1 3 35 Cm
Having constraints that have upper limits should make sense, since when maximizing a quantity, we probably have caps on what we can do. If we had no caps, then we could continue to increase, say profit, infinitely! This contradicts what we know about the real world.
In order to use the simplex method, either by technology or by hand, we must set up an
initial simplex tableau, which is a matrix containing information about the linear programming problem we wish to solve.
First off, matrices don't do well with inequalities. For one, a matrix does not have a simple way of keeping track of the direction of an inequality. This alone discourages the use of inequalities in matrices. How, then, do we avoid this?
Consider the following linear programming problem
Maximize:
P = 7x + 12y
Subject to:
2x + 3y ≤ 6
3x + 7y ≤12
Because we know that the left sides of both inequalities will be quantities that are smaller than the corresponding values on the right, we can be sure that adding 'something' to the left-hand side will make them exactly equal. That is:
2x + 3y + s1 = 6
3x + 7y + s2 = 12
For instance, suppose that
x = 1, y = 1, Then
2+3+s1=6 or s1=1
3+7+s2=12 or s2=2
It is important to note that these two variables, s1 and s2, are not necessarily the same. They simply act on the inequality by picking up the 'slack' that keeps the left side from looking like the right side. Hence, we call them slack variables. This takes care of the inequalities for us. Since augmented matrices contain all variables on the left and constants on the right, we will rewrite the objective function to match this format:
–7x – 12y + P =0
We can now write an initial system of equations:
| 2x + 3y + s1 = 6 |
| 3x + 7y + s2 = 12 |
| –7x – 12y + P =0 |
This will require us to have a matrix that can handle x,y,s1,s2, and P. We will put it in this order. Finally, the simplex method requires that the objective function be listed as the bottom line in the matrix so that we have:
We have established the initial simplex tableau. Note that he horizontal and vertical lines are used simply to separate constraint coefficients from constants and objective function coefficients. Also notice that the slack variable columns, along with the objective function output, form the identity matrix.
We will present the algorithm for solving, however, note that it is not entirely intuitive. We will focus on this method for one example, and will then proceed to use technology to run through the process for us.
1. Select a Pivot Column
We first select a pivot column, which will be the column that contains the largest negative coefficient in the row containing the objective function. Note that the largest negative number belongs to the term that contributes most to the objective function. This is intentional since we want to focus on values that make the output as large as possible.
Our pivot is thus the y column.
2. Select a Pivot Row
Do this by computing the ratio of each constraint constant to its respective coefficient in the pivot column—this is called the test ratio. Select the row with the smallest test ratio.
We first calculate the test ratios:
[latex]displaystyle{left[matrix{{6÷3=2}{12÷7≈1.7}}right]}[/latex]
Since the test ratio is smaller for row 2, we select it as the pivot row. The boxed value is now called our pivot. To justify why we do this, observe that 2 and 1.7 are simply the vertical intercepts of the two inequalities. We select the smaller one to ensure we have a corner point that is in our feasible region:
3. Using Gaussian Elimination, Eliminate Rows 1 and 3
Multiply R2 by (1/7) to convert 7 to 1.
Then use the 1 to eliminate the 3 in R1: -3R2+R1→R1
And use the 1 to eliminate the -12 in R3: 12R2+R3→R3
We get the following matrix (you might prefer it as fractions)
What have we done? For one, we have maxed out the contribution of the 2-2 entry
y-value coefficient to the objective function. Have we optimized the function? Not quite, as we still see that there is a negative value in the first column. This tells us that can still contribute to the objective function. To eliminate this, we first find the pivot row by obtaining test ratios:
Wifi explorer pro 2 1 7 cr2 file. [latex]displaystyle{left[matrix{{5/7}&{0}&{1}&{-3/7}&{0}&{|}{6/7}{3/7}&{1}&{0}&{1/7}&{0}&{|}{12/7}{-13/7}&{0}&{0}&{12/7}&{1}&{|}{144/7}}right]}[/latex]
[latex]displaystyle{left[matrix{{(6/7)÷(5/7)≈1.2}{12÷3≈4}}right]}[/latex]
Since 1.2 < 4, R1 is our new pivot row.
Interestingly, this test ratio corresponds to the input value of the intersection of the two lines!
Similarly, we proceed to eliminate all non-pivot values.
Multiply R1 by (1/0.71) to convert 0.71 to 1.
Then use the 1 to eliminate the 3 in R3: -3R1+R2→R2
And use the 1 to eliminate the -12 in R3: 1.86R1+R3→R3
We get the following matrix
[latex]displaystyle{left[matrix{{1}&{0}&{7/5}&{-3/5}&{0}&{|}{6/5}{0}&{1}&{-3/5}&{2/5}&{0}&{|}{6/5}{0}&{0}&{13/5}&{3/5}&{1}&{|}{114/5}}right]}[/latex]
There remain no additional negative entries in the objective function row. We thus have the following matrix:
[latex]displaystyle{left[matrix{{1}&{0}&{7/5}&{-3/5}&{0}&{|}{6/5}{0}&{1}&{-3/5}&{2/5}&{0}&{|}{6/5}{0}&{0}&{13/5}&{3/5}&{1}&{|}{114/5}}right]}[/latex]
We are thus prepared to read the solutions.
4. Identify the Solution Set
To identify the solution set, we focus only on the columns with exactly one non-zero entry—these are called basic variables (columns with more than one non-zero entry are thus called nonbasic variables).
We notice that both the x any columns are basic variables. We really don't care about the slack variables, much like we ignore inequalities when we are finding intersections. We now see that,
[latex]displaystyle{left[matrix{{1}&{0}&{7/5}&{-3/5}&{0}&{|}{6/5}{0}&{1}&{-3/5}&{2/5}&{0}&{|}{6/5}{0}&{0}&{13/5}&{3/5}&{1}&{|}{114/5}}right]}[/latex]
Setting the slack variables to 0, gives:
x ≈ 1.2
y ≈ 1.2
P = 22.8
Thus, x=1.2, y=1.2, P=22.8 is the solution to the linear programming problem. That is, inputs of x=1.2 and y=1.2 will yield a maximum objective function value of 22.8
While somewhat intuitive, this process has more behind it than we are letting on. And, rather than going through these grueling steps ad nauseum, we will allow our technology to follow these steps. For this, we need a special program, which will be distributed in class,
To perform the simplex method with a graphing calculator, the following programs are needed:
- Pivot,
- Pivot1, and
- Simplex
Pivot and Pivot1 are not used directly. Instead, the Simplex program reaches into these two applications to assist it with some rather long and tedious code. How does the code work? Using instructions, it finds pivot columns, pivot rows, performs Gaussian elimination, checks for negatives in the objective function row, and repeats this process, as necessary until all negatives have been removed.
Now that we have illustrated that, in fact, the simplex method works for even two input variables, let's show a situation where this method is particularly useful.
Example 2
A new airline has decided to join the market. It is considering offering flights out of Phoenix, AZ, and would initially like to travel to three different locations: San Diego, San Francisco, and Las Vegas. The distances of each round-trip flight going out of Phoenix are (approximately): 720 miles, 1500 miles, and 1140 miles, respectively. The company would like to use the slogan, 'the average price per flight is never more than $200.' As for costs, it anticipates flights to San Diego will run about 10% of airfare. Similarly, San Francisco will run 12% and Las Vegas will run 14% of airfare. The company wants to ensure that the overall average cost is no more than 10% of earned airfare. Recent market research allows the company to conclude that it could probably sell about 1900 San Diego tickets, 700 San Francisco tickets, and 1000 Las Vegas ticket. Under these conditions and assuming that all tickets sold are round-trip flights, how much should the company charge per ticket in order to maximize its total revenue?
Solution
We want to know the price for airfare to each destination. Let,
x = price per round-trip ticket to San Diego
Fivenotes 2 2 2 0. y = price per round-trip ticket to San Francisco
z = price per round-trip ticket to Las Vegas
The company wants to maximize total revenue. This is based on the sum of number of tickets sold multiplied by the price per ticket, which is:
R = 1900x + 700y + 1000z
Subject to the constraints:
- Average price per flight is less than or equal to $200
- Average cost from airfare is no more than 10% of total
Mathematically,
- Add prices and divide by 3
[latex]displaystylefrac{{{x}+{y}+{z}}}{{3}}le{200}[/latex] - Or x+y+z ≤ 600
- Revenue from San Diego tickets will total and 10% of this amount is estimated to be cost. That is Cost = .10(1900x) = 190x. Similarly, we have .12(700y) = 84y and .14(1000z) = 140z. We want the sum of these costs to be less than or equal to 10% of total revenue, which is .10(1900x + 700y + 1000z) = 190x + 70y + 100z.
- 190x + 84y + 140z ≤ 190x + 70y + 100z
- or 14y + 40z ≤ 0
Since both constraints are of the correct form, we can proceed to set up the initial simplex tableau. As a note, be very cautious about when you use the simplex method, as unmet requirements invalidate the results obtained.
The rewritten objective function is:
–1900x – 700y – 1000z + R = 0
And simplified constraints are:
x + y + z ≤ 600 (Multiply both sides by 3)
14y + 40z ≤ 0
x,y≥0
Each of the two constraints will require a slack variable, and so we rewrite them as follows:
x + y + z + s1 = 600
14y + 40z + s2 = 0
We will have the following variable columns:
x,y,z,s1,s2,R, and the constant column, for a total of 7 columns. We have two constraints and one objective function for a total of three rows. We now write the initial simplex tableau:
[latex]displaystyle{left[matrix{{1}&{1}&{1}&{1}&{0}&{0}&{|}{600}{0}&{14}&{40}&{0}&{1}&{0}&{|}{0}{-1900}&{-700}&{-1000}&{0}&{0}&{1}&{|}{0}}right]}[/latex]
The tableau is now ready to be solved using Simplex.
Pivot on the 1st column and 1st row. (You are not allowed to divide by 0 to get the pivot row)
[latex]displaystyle{left[matrix{{1}&{1}&{1}&{1}&{0}&{0}&{|}{600}{0}&{14}&{40}&{0}&{1}&{0}&{|}{0}{0}&{1200}&{900}&{1900}&{0}&{1}&{|}{1140000}}right]}[/latex]
Since only the x column will be basic, we can see that x = 600 is a solution. The fact that y and z are nonbasic variables, we set y = z = 0. That is, they do not contribute to maximizing revenue. Additionally, R is an active variable, and so we see that R = $1,140,000 is the maximum revenue the company can earn, given the constraints. They should sell tickets to San Diego for $600 and market no flights to the other cities. As you might guess, the company is probably a bit stunned. We will explore this in the next example.
It is interesting that San Diego trips alone produce the highest revenue, based on the constraints given. Why is this? If we look at the constraints, we see that the company is fairly certain it can sell 1900 flights to San Diego. The company is also somewhat puzzled that it is expected to sell tickets at $600 each. At this point, it might decide to add some additional constraints to the model.
Example 3
Supposed the airliner in Example 2 decides that it can charge no more than $150 per ticket to San Diego in order to be competitive with other airliners that fly to the same destination. Assuming all other constraints will still be used, how are ticket prices and maximum revenue affected?
Solution
We use the same initial tableau, but we must deal with the following new constraint:
x ≤ 150
Adding a third slack variable, we have
x + s3 = 150
This adds one column and one row to our tableau:
[latex]displaystyle{left[matrix{{1}&{1}&{1}&{1}&{0}&{0}&{0}&{|}{600}{0}&{14}&{40}&{0}&{1}&{0}&{0}&{|}{0}{1}&{0}&{0}&{0}&{0}&{1}&{0}&{|}{150}{-1900}&{-700}&{-1000}&{0}&{0}&{0}&{1}&{|}{0}}right]}[/latex]
Solving this by Simplex yields
[latex]displaystyle{left[matrix{{0}&{0}&{-13/7}&{1}&{-1/14}&{-1}&{0}&{|}{450}{0}&{1}&{20/7}&{0}&{1/14}&{0}&{0}&{|}{0}{1}&{0}&{0}&{0}&{0}&{1}&{0}&{|}{150}{0}&{0}&{1000}&{0}&{50}&{1900}&{1}&{|}{285000}}right]}[/latex]
Solution: x=150, y=0, z=450 and R=285000
Example 4
A catering company is to make lunch for a business meeting. It will serve ham sandwiches, light ham sandwiches, and vegetarian sandwiches. A ham sandwich has 1 serving of vegetables, 4 slices of ham, 1 slice of cheese, and 2 slices of bread. A light ham sandwich has 2 serving of vegetables, 2 slices of ham, 1 slice of cheese and 2 slices of bread. A vegetarian sandwich has 3 servings of vegetables, 2 slices of cheese, and 2 slices of bread. A total of 10 bags of ham are available, each of which has 40 slices; 18 loaves of bread are available, each with 14 slices; 200 servings of vegetables are available, and 15 bags of cheese, each with 60 slices, are available. Given the resources, how many of each sandwich can be produced if the goal is to maximize the number of sandwiches?
Solution
We wish to maximize the number of sandwiches, so let:
x = # of ham sandwiches
y = # of light ham sandwiches
https://softwarecouturenova-drift-0-21-6-classic-arcade-space-combat.peatix.com. z = # of vegetarian sandwiches
The total number of sandwiches is given by
S = x + y + z
The constraints will be given by considering the total amount of ingredients available. That is, the company has a limited amount of ham, vegetables, cheese, and bread.
In total, the company has
10(40) = 400 slices of ham, 18(14) = 252 slices of bread, 200 servings of vegetables, and 15(60) = 900 slices of cheese available. At most, the company can use the above amounts.
There are two sandwiches that use ham—the first requires 4 slices of ham and the second requires only 2, per sandwich. That is, 4x + 2y ≤ 400
That is, the total number of slices of ham cannot exceed 400.
Each sandwich requires 2 slices of bread so 2x + 2y + 2z ≤ 252
The ham sandwiches have 1 and 2 servings of vegetables, respectively, and the vegetarian sandwich has 3 servings of vegetables. So, 1x + 2y + 3z ≤ 200
Both ham sandwiches require one slice of cheese, and the vegetarian sandwich requires two slices of cheese, so, 1x + 1y + 2z ≤ 900 Below is the completed linear programming model for this example.
| Maximize: S = x + y + z |
| Subject To: 4x + 2y ≤ 400 |
| 2x + 2y + 2z ≤ 252 |
| x + 2y + 3z ≤ 200 |
| 1x + 1y + 2z ≤ 900 |
| x, y, z≥0 |
These constraints satisfy the requirements for the simplex method, so we proceed.
Incorporating slack variables, we get:
4x + 2y + 0z + s1 = 400
2x + 2y + 2z + s2 = 252
x + 2y + 3z + s3 = 200
x + y + 2z + s4 = 900
–x – y – z + S = 0
The initial simplex tableau is:
[latex]displaystyle{left[matrix{{4}&{2}&{0}&{1}&{0}&{0}&{0}&{0}{|}&{400}{2}&{2}&{2}&{0}&{1}&{0}&{0}&{0}{|}&{252}{1}&{2}&{3}&{0}&{0}&{1}&{0}&{0}{|}&{200}{1}&{1}&{2}&{0}&{0}&{0}&{1}&{0}{|}&{900}{-1}&{-1}&{-1}&{0}&{0}&{0}&{0}&{1}{|}&{0}}right]}[/latex]
Focus Matrix 1 3 35 Equals
Since the most negative number on the bottom row is the same for the 3 columns, we can use either column. Might as well use the first column. The smallest quotient occurs by dividing 4 into 400 so row 1 is the pivot column. Pivoting on the '4' in R1C1 yields.
[latex]displaystyle{left[matrix{{1}&{1/2}&{0}&{1/4}&{0}&{0}&{0}&{0}{|}&{100}{0}&{1}&{2}&{-1/2}&{1}&{0}&{0}&{0}{|}&{52}{0}&{3/2}&{3}&{-1/4}&{0}&{1}&{0}&{0}{|}&{100}{0}&{1/2}&{2}&{-1/4}&{0}&{0}&{1}&{0}{|}&{800}{0}&{-1/2}&{-1}&{1/4}&{0}&{0}&{0}&{1}{|}&{100}}right]}[/latex]
Note: We have increased S from 0 to 200, but we still have a negative in the bottom row. Since '-1' is more negative than '-1/2' we will pivot on column 3. After dividing the positive numbers above '-1' into the constants, we get the smallest quotient in row 2. Pivoting on the '2' in R2C3 yields.
[latex]displaystyle{left[matrix{{x}&{y}&{z}&{s1}&{s2}&{s3}&{s4}&{S}{1}&{1/2}&{0}&{1/4}&{0}&{0}&{0}&{0}{|}&{100}{0}&{1/2}&{1}&{-1/4}&{1/2}&{0}&{0}&{0}{|}&{26}{0}&{0}&{0}&{1/2}&{-3/2}&{1}&{0}&{0}{|}&{22}{0}&{-1/2}&{0}&{1/4}&{-1}&{0}&{1}&{0}{|}&{748}{0}&{0}&{0}&{1/2}&{0}&{0}&{0}&{1}{|}&{126}}right]}[/latex]
We now have the optimal solution
Focus Matrix 1 3 35 Inches
- x=100 (basic variable in row 1)
- y=0 (nonbasic variable)
- z=26 (basic variable row 2)
- s1=0 (nonbasic variable)
- s2=0 (nonbasic variable)
- s3=22 (basic variable row 3)
- s4=748 (basic variable row 4)
- S=126 (basic variable row 5)
Of course we are really just interested in: x=100, y=0, z=26, S=126
Disk map analyzer clean your hard drive space 1 5. We find that 100 ham sandwiches, 26 vegetarian sandwiches, and 0 light ham sandwiches should be made to maximize the total number of sandwiches made.
The slack variables are not important in the solution. Just in reaching the solution.
See next section for more examples. Many problems use subscripted variables such as x1, x2, x3, etc. This is especially helpful if you have several variables. You will see this in later examples.
More resources are listed below:
Milos Podmanik, By the Numbers, 'Solving Standard Maximization Problems using the Simplex Method,' licensed under a CC BY-NC-SA 3.0 license.
MathIsGreatFun, 'MAT217 HW 2.2 #1,' licensed under a Standard YouTube license.
MathIsGreatFun, 'MAT217 2.2 #2,' licensed under a Standard YouTube license.
MathIsGreatFun, 'MAT217 2.2 #3,' licensed under a Standard YouTube license.
Focus Matrix 1 3 35 +
MathIsGreatFun, 'MATH217 2.2 #4,' licensed under a Standard YouTube license.
