Find the Optimal Solution Using the Graphical Solution Procedure
Think about this: you are traveling from Chandler, AZ, to San Diego, CA. Your hope is to get there in as little time as possible, hence aiming to minimize travel time. At the same time, you will be facing more or less traffic on certain stretches of the trip, you will need to stop for gas at least once (unless you are driving a hybrid vehicle), and, if you have kids, you'll definitely need to stop for a restroom break. While we have only mentioned a few, these are all
constraints—things that limit you in your goal to get to your destination in as little time as possible.
Solving Linear Programming Problems Graphically
A linear programming problem involves constraints that contain inequalities. An
inequality is denoted with familiar symbols, <, >, [latex]\le [/latex], and [latex]\ge [/latex]. Due to difficulties with strict inequalities (< and >), we will only focus on[latex]\le [/latex] and[latex]\ge [/latex].
In order to have a linear programming problem, we must have:
- Inequality constraints
- An objective function, that is, a function whose value we either want to be as large as possible (want to maximize it) or as small as possible (want to minimize it).
Example 1
Solution
The first step is to identify the unknown quantities. We are asked to find the number of each ticket that should be sold. Since there are coach and first-class tickets, we identify those as the unknowns. Let,
x = # of coach tickets
y = # of first-class tickets
Next, we need to identify the objective function. The question often helps us identify the objective function. Since the goal is the maximize profits, our objective is identified.
Profit for coach tickets is $225. If
x coach tickets are sold, the total profit for these tickets is 225x.
Profit for first-class tickets is $200. Similarly, if
y first class tickets are sold, the total profit for these tickets is 200y.
The total profit,P, is
P = 225x + 200y
We want to make the value of
as large as possible, provided the constraints are met. In this case, we have the following constraints:
- Sell at least 25 first-class tickets
- Sell at least 40 coach tickets
- No more than 150 tickets can be sold (no more than 150 people can fit on the plane)
We need to quantify these.
- At least 25 first-class tickets means that 25 or more should be sold. That is, y [latex]\ge [/latex] 25
- At least 40 coach tickets means that 40 or more should be sold. That is, x [latex]\ge [/latex] 40
- The sum of first-class and coach tickets should be 150 or fewer. That is x + y [latex]\le [/latex] 150
Thus, the objective function along with the three mathematical constraints is:
Objective Function: P = 225x + 200y
Constraints: y [latex]\ge [/latex] 25; x[latex]\ge [/latex] 40; x + y [latex]\le [/latex] 150
We will work to think about these constraints graphically and return to the objective function afterwards. We will thus deal with the following graph:
Note that we are only interested in the first quadrant, since we cannot have negative tickets.
We will first plot each of the inequalities as equations, and then worry about the inequality signs. That is, first plot,
x= 25
y = 40
x + y = 150
The first two equations are horizontal and vertical lines, respectively. To plot x + y= 150, it is preferable to find the horizontal and vertical intercepts.
To find the vertical intercept, we let
x = 0:
y= 150
Giving us the point (0,150)
To find the horizontal intercept, we let
y = 0:
x = 150
Giving us the point (150,0)
Plotting all three equations gives:
Our next task is to take into account the inequalities.
We first ask, when isy [latex]\ge [/latex] 25? Since this is a horizontal line running through a y-value of 25, anything above this line represents a value greater than 25. We denote this by shading above the line:
This tells us that any point in the green shaded region satisfies the constraint that
y [latex]\ge [/latex] 25.
Next, we deal with
x [latex]\ge [/latex] 40. We ask, when is the x-value larger than 40? Values to the left are smaller than 40, so we must shade to the right to get values larger than 40:
The blue area satisfies the second constraint, but since we must satisfy
all constraints, only the region that is green and blue will suffice.
We have one more constraint to consider:
x+ y [latex]\ge [/latex] 150. We have two options, either shade below or shade above. To help us better see that we will, in fact, need to shade below the line, let us consider an ordered pair in both regions. Selecting an ordered pair above the line, such as (64, 130) gives:
64 + 130[latex]\ge [/latex] 150
Which is a false statement since 64 + 130 = 194, a value larger than 150.
According to the graph, the point (64, 65) is one that falls below the graph. Putting this pair in yields the statement:
64 + 65 [latex]\ge [/latex] 150
Which is a true statement since 64+65 is 129, a value smaller than 150.
Therefore, we shade below the line:
The region in which the green, blue, and purple shadings intersect satisfies all three constraints. This region is known as thefeasible regions, since this set of points is feasible, given all constraints. We can verify that a point chosen in this region satisfies all three constraints. For example, choosing (64, 65) gives:
64 [latex]\ge [/latex] 40 TRUE
65 [latex]\ge [/latex] 25 TRUE
64 + 65 [latex]\ge [/latex] 150 TRUE
This gets us to a great point, but still does not answer the question:
which point maximizes profit? Fortunately, there is a theorem discovered by mathematicians that allows us to answer this question.
First off, we define a new term: acorner point is a point that falls along the corner of a feasible region. In our situation, we have three corner points, shown on the graph as the solid black dots:
The objective function along with the three corner points above forms abounded linear programming problem. That is, imagine you are looking at three fence posts connected by fencing (black point and lines, respectively). If you were to put your dog in the middle, you could be sure it would not escape (assuming the fence is tall enough). If this is the case, then you have a bounded linear programming problem. If the dog could walk infinitely in any one direction, then the problem is unbounded.
Fundamental Theorem of Linear Programming
- If a solution exists to a bounded linear programming problem, then it occurs at one of the corner points.
- If a feasible region is unbounded, then a maximum value for the objective function does not exist.
- If a feasible region is unbounded, and the objective function has only positive coefficients, then a minimum value exist
This means we have to choose among three corner points. To verify the "winner," we must see which of these three points maximizes the objective function. To find the corner points as ordered pairs, we must solve three systems of two equations each:
System 1
x= 40
x + y = 150
System 2
x= 40
y= 25
System 3
y = 25
x+ y = 150
We could decide to solve by using matrix equations, but these equations are all simple enough to solve by hand:
System 1
(40) + y = 150
y = 110
Point:(40,110)
System 2
Point already given
Point: (40,25)
System 3
x + 25 = 150
x = 125
Point: (125,25)
We test each of these three points in the objective function:
| Point | Profit |
| (40,110) | 225(40) + 200(110) = $31,000 |
| (40,25) | 225(40) + 200(25) = $14,000 |
| (125,25) | 225(125) + 200(25) = $33,125 |
The third point, (125,25) maximizes profit. Therefore, we conclude that the airline should sell 125 coach tickets and 25 first-class tickets in order to maximize profits.
The above example was rather long and had many steps to complete. We will summarize the procedure below:
Solving a Linear Programming Problem Graphically
- Define the variables to be optimized. The question asked is a good indicator as to what these will be.
- Write the objective function in words, then convert to mathematical equation
- Write the constraints in words, then convert to mathematical inequalities
- Graph the constraints as equations
- Shade feasible regions by taking into account the inequality sign and its direction. If,
a)A vertical line
[latex]\le [/latex], then shade to the left
[latex]\ge [/latex], then shade to the right
b) A horizontal line
[latex]\le [/latex], then shade below
[latex]\ge [/latex], then shade above
c) A line with a non-zero, defined slope
[latex]\le [/latex], shade below
[latex]\ge [/latex], shade above
6. Identify the corner points by solving systems of linear equations whose intersection represents a corner point.
7. Test all corner points in the objective function. The "winning" point is the point that optimizes the objective function (biggest if maximizing, smallest if minimizing)
There is one instance in which we must take great caution. First, consider the (true) inequality,
5 > 3
Suppose we were to divide both sides by –1. Would it still be true to say the following?
[latex]\displaystyle\frac{{5}}{{-{1}}}\gt\frac{{3}}{{-{1}}}[/latex]
[latex]\displaystyle-{5}\gt-{3}[/latex]
Clearly, –5 is not larger than –3! To keep the statement true, we should change the direction of the inequality sign so that,
–5 < –3
We can see by the number line below, that the two sets of numbers are symmetric about 0, except that the way in which we describe size is opposite. This justifies that we should also use the opposite sign when we reflect values to the other side of 0.
Changing the Inequality Sign
When multiplying/dividing any inequality by –1, the direction of the inequality should change
Example 2
A health-food business would like to create a high-potassium blend of dried fruit in the form of a box of 10 fruit bars. It decides to use dried apricots, which have 407 mg of potassium per serving, and dried dates, which have 271 mg of potassium per serving (SOURCE: www.thepotassiumrichfoods.com).
The company can purchase its fruit through
www.driedfruitbaskets.com in bulk for a reasonable price. Dried apricots cost $9.99/lb. (about 3 servings) and dried dates cost $7.99/lb. (about 4 servings). The company would like the box of bars to have at least the recommended daily potassium intake of about 4700 mg, but would like to keep it under twice the recommended daily intake. In order to minimize cost, how many servings of each dried fruit should go into the box of bars?
Solution
We begin by defining the variables. Let,
x = # of servings of dried apricots
y = # of servings of dried dates
We next work on the objective function.
For apricots, there are 3 servings in one pound. This means that the cost per serving is $9.99/3 = $3.33. The cost for
x servings would thus be 3.33x.
For dates, there are 4 servings per pound. This means that the cost per serving is $7.99/4
$2.00. The cost for y servings would thus be 2.00y.
The total cost for apricots and dates would be
C = 3.33x + 2.00y
We have two major constraints (in addition to the constraints that
x [latex]\ge [/latex] 0
and y [latex]\ge [/latex] 0, given that negative servings cannot be used):
- Product must contain at least 4700mg of potassium
- Product should contain no more than 4700 × 2 = 9400mg of potassium
Mathematically,
- There are 407x mg of potassium in x servings of apricots and 271y mg of potassium in y servings of dates. The sum should be greater than or equal to 4700mg of potassium, or 407x + 271y [latex]\ge [/latex] 4700
- The same sum should be less than or equal to 9400 mg of potassium, or 407x + 271y [latex]\le [/latex] 9400.
Thus we have,
Objective Function : C = 3.33x + 2.00y
Subject To Constraints:
407x + 271y [latex]\ge [/latex] 4700
407x + 271y [latex]\le [/latex] 9400
x [latex]\ge [/latex] 0
y [latex]\ge [/latex] 0
We graph the constraints as equations:
Since the first inequality has [latex]\ge [/latex], we must shade above and, since the second inequality has[latex]\le [/latex], we must shade below (This idea can be confirmed by selecting points above and below each line in order to verify.):
The feasible region is the green and blue shaded section between the two lines. We see that there are four corner points that form an upside-down trapezoid, as shown in the graph below:
We must solve the following systems to find the corner points (bottom-to-top, left-to-right)
System 1
x = 0
407x + 271y =4700
Solution:
0 + 271
y = 4700
y ≈ 17.3
Point: (0,17.3)
System 2
x = 0
407x + 271y = 9400
Solution:
0 + 271y = 9400
y ≈ 34.7
Point: (0,34.7)
System 3
y = 0
407x + 271y = 4700
Solution:
407x + 0 = 4700
x ≈ 11.5
Point: (11.5,0)
System 4
y = 0
407x + 271y = 9400
Solution:
407x + 0 = 9400
x ≈ 23.1
Point: (23.1,0)
Again, we could solve by using matrix equations, but the systems are straightforward to solve by substitution. Since the problem is bounded, we now check to see which one minimizes cost:
| Point | Cost |
| (0,17.3) | 33.3(0) = 2.00(17.3) = $34.60 |
| (0,34.7) | 33.3(0) = 2.00(34.7) = $69.40 |
| (11.5,0) | 33.3(11.5) = 2.00(0) = $38.30 |
| (23.1,0) | 33.3(23.1) = 2.00(0) = $76.92 |
The cheapest route for the company will be to create bars that contain no dried apricots and 17.3 servings of dried dates.
It is interesting to note that each of the corner points corresponds to either a horizontal or vertical intercept.
Why are we seeing what we're seeing? This is truly a case of real-world product creation! Of course, it doesn't make sense to increase the daily intake for the box, since this would mean increasing the amount of dried fruit, hence increasing cost. Since the cost of dried dates is cheaper ($2.00 per serving) and since for the price of one serving of apricots ($3.33 per serving) we can pay:
[latex]\displaystyle\frac{{{407}{m}{g}}}{{${3.33}}}\approx{122.2}[/latex]mg per dollar for apricots
and
[latex]\displaystyle\frac{{{271}{m}{g}}}{{${2.00}}}\approx{135.5}[/latex]mg per dollar for dates
It makes complete sense to buy dates, since the same dollar amount yields a higher content of potassium.
The question still remains: is it desirable to require a larger quantity of dates for a smaller price, or is it more desirable to require a smaller quantity of apricots for a larger price? This indeed depends on the constraints. The company might want to consider the amount of packaging/processing/etc. required in both instances. Perhaps the manufacturing and packaging costs could add constraints that alter the decision-making process. A similar problem will be left as a homework exercise for the reader to think about.
As a mathematical note, what we are seeing occurs as a result of having constraint lines that are parallel.
There are two terms we should be familiar with when dealing with inequalities:bounded and unbounded. A feasible region is said to be bounded if the constraints enclose the feasible region.
That is, if the shading does not continue to cover the entire plane, we are dealing with a bounded linear programming problem.
Both examples thus far have been examples of bounded linear programming problems, since the first feasible region was in the shape of a triangle and the second in the shape of a trapezoid.
If the feasible region cannot be enclosed among the lines formed by constraints, it is said to be unbounded. An example of an unbounded linear programming problem would be:
Example 3
A human resources office is working to implement an increase in starting salaries for new administrative secretaries and faculty at a community college. An administrative secretary starts at $28,000 and new faculty receive $40,000. The college would like to determine the percentage increase to allocate to each group, given that the college will be hiring 8 secretaries and 7 faculty in the upcoming academic year. The college has at most $5,000 to put towards raises. What should the percentage increase be for each group?
Solution
Our goal is to determine the percentage increase for administrative secretaries and faculty, so let
x = percentage increase for secretaries
y = percentage increase for faculty
The college would like to minimize its total expenditures, so the objective function must include the total amount of money outflows. Since the new secretaries will require a total budget of
$28,000 × 8 = $224,000 and the faculty a total budget of $40,000 × 7 = $280,000, the total cost will be the raise percentage for each group, multiplied by the total salaries:
C = 224x + 280y
There is one constraint given, which is that the total raises must be $5,000 or less. That is,
224x + 280y [latex]\le [/latex] 5
Of course, the college does not want to reduce the salaries, so
x [latex]\ge [/latex] 0 and y [latex]\ge [/latex] 0.
To visualize the situation, we graph the constraint as an equation. To help us find points, we first find the intercepts:
Horizontal Intercept:
224(0) + 280
y = 5
y ≈ .018
Vertical Intercept
224x + 280(0) = 5
x ≈ .022
We then plot the points and connect them with a straight line:
Since the inequality sign is [latex]\le [/latex], we shade below the line:
This gives us three corner points, as shown above. We test each to verify which of the pairs of percentages gives the minimum cost:
| Point | Cost |
| (0,0) | 224(0) + 280(0) = $0 |
| (0,.018) | 224(0) + 280(.018) = $5.04 |
| (.020,0) | 224(.020) + 280(0) = $4.48 |
Clearly, the first option gives the smallest cost; however, this combination of tells us to give a 0% raise to both groups, which, of course, is not practical, since the company's goal was to give a raise to each group.
Why did this happen, and what should we do to fix it? Well, when we think about the constraint of spending $5,000 or less and hoping to make expenditures as small as possible, wouldn't it make sense to say, "don't spend anything!"? This outcome will occur anytime we are minimizing, have constraints with the
le inequality sign, and when the origin is included in the feasible region. To fix the problem, the company should make additional specifications, such as, what is the minimum percentage raise to give to each group? Is it desirable for one of the raises to be larger than the other? These are questions the analyst should discuss with human resources and administration.
Practice Problems
1) Solve each of the following linear programming problems.
a) Maximize R = 2x + 3y
Subject to
–2x – y [latex]\ge [/latex] –10
x + 3y[latex]\ge [/latex] 6
x [latex]\ge [/latex] 0
y [latex]\ge [/latex] 0
Solution: R=30 at (0,10)
B) Minimize T = 3x + y
Subject to
x + 2y[latex]\ge [/latex] 4
x + 3y [latex]\ge [/latex] 6
x [latex]\ge [/latex]0
y [latex]\ge [/latex] 0
Solution: R=2 at (0,2)
2) A local school governing board approves a new math education program that is to be implemented at a series of elementary schools within the district. Money for the program will come from two different budgets: public expenditures budget and grade-school initiatives budget. The board is willing to pay at least half of what comes out of the initiatives budget from its public expenditures budget. Since this program is considered an initiative, the government mandates that at least $2,000 comes from the local initiatives budget. Both budgets are partially funded by federal emergency funding. For the public expenditures budget, the percentage is 55% and 23% for the grade-school initiatives budget. In order to properly use emergency funding, the district would like to minimize the use of federal dollars. How much should come from each budget?
Solution: x=amount from publix expenditures; y=amount from grade school initiatives
Minimize: C=0.55x+0.23y
(1/2)x≥y or {(1/2)x-y≥0}
y≥2000
x,y≥0
Solution: C=2660, x=4000, y=2000
3) A public relations director for a homeopathic is seeking to advertise her company's products on two different websites—one is a medical parts supplier and the other is a fitness e-zine (a web-based magazine). The medical parts supplier website receives, on average, about 1,200,000 hits per day per page, while the fitness e-zine receives about 2,000,000 hits per day per page. The daily cost to advertise is $1,100 per advertisement and $1,600 per advertisement, respectively. The director would like at least 15 ads and is able to allocate up to $50,000 for advertising. At least 3 ads should be placed on each website. How many adds should be placed on each website to maximize the potential number of readers (even if some viewers see the add on different pages of the website)?
x=number on medical website; y= number on fitness website.
Maximize: R=1200000x+2000000y
Subject to:
x+y≥15
1100x+1600y≤50000
x≥3
y≥3
x,y≥0
| Corner Points | R=1200000x+2000000y |
| (3,12) | 27,600,000 |
| (12,3) | 20,400,000 |
| (3,29.2) | 62,000,000 |
| (41.1,3) | 55,320,000 |
Optimal Solution
See more examples in next section.
Find the Optimal Solution Using the Graphical Solution Procedure
Source: https://courses.lumenlearning.com/sanjacinto-finitemath1/chapter/reading-meeting-demands-with-linear-programming/