Linear Programming and the

Simplex Method

By: Ron Aldad

“Money makes the world go round.” Throughout history, there has always

been a race for money and profit. As humans, we strive for the best, and we believe

that the best is where the most money is. The conclusion that mathematicians came

to was that we could minimize or maximize our profits using equations, inequalities,

and constraints. This process of using these tools to organize a problem into a clear

and accurate graph is called linear programming. We are able to transform word

problems and real world situations into these graphs. In economics, this is directly

applicable to find the amount of revenue needed to make the maximum profits.

Linear programming and the simplex method are invaluable instruments that

changed the investment and economic worlds.

During World War II, Leonid Kantorovich was the mathematician that

created linear programming. He won the Nobel Prize in Economics in 1975, and he

was the only winner to ever win this award from the USSR. He used linear

programming to decrease the expenses to the army and increase the loss of the

enemy. Through inequalities, he was able to devise graphs that indicated the

optimal revenue that fit his needs. Subsequently, George Dantzig created the

simplex method that transformed linear programming into tables, rather than

graphs, and made the problems easier and faster to solve.

The fundamental idea behind linear programming and the simplex method is

to take a real world dilemma in the form of a word problem and create an equation

along with a series of constraints. In linear programming, the simpler of the two, one

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takes the problem given to him/her and begins with the equation. Then, the

equation is put aside until the conclusion of the problem. The more important pieces

to a linear programming problem are the constraints. Usually written as

inequalities, these constraints are what help the mathematician or economist create

the graph. The graph or the inequalities depicts the points of intersection of the

inequalities, and these points tell the economist what values for (x,y) he/she would

need to plug into the original equation to find the minimal or maximum amount of

profit available. Linear programming is a graph format of answering these types of

problem and is more easily explicable through examples.

Ex. 1) “You need to buy some filing cabinets. You know that Cabinet X

costs $10 per unit, requires six square feet of floor space, and holds eight

cubic feet of files. Cabinet Y costs $20 per unit, requires eight square feet of

floor space, and holds twelve cubic feet of files. You have been given $140 for

this purchase, though you don't have to spend that much. The office has room

for no more than 72 square feet of cabinets. How many of which model

should you buy, in order to maximize storage volume” (Stapel)?

The first step is to find the two variables needed to make the equations

and inequalities.

X = the number of X cabinets bought

Y= the number of Y cabinets bought

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Then, create the equation. In this case, it would be for the maximum

storage volume, as shown below.

V=8 x+12 y

The next step is to create constraints in the form of inequalities that follow

the problem. When doing this problem, one is able to realize that the “x” and “y”

values cannot be negative, so two constraints are:

x≥ 0

y ≥0

There are two other constraints needed for this problem: one for the cost,

and one for the space.

Cost: 10 x+20 y ≤ 140

Space: 6 x+8 y≤ 72

In order to graph these inequalities, it is easier to make them into standard

“y=” form.

Cost: y ≤−12

x+7

Space: y ≤−34

x+9

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When graphing all four of these inequalities, one would receive a solution set

similar to one below.

(http://www.purplemath.com/modules/ineqsolv/linprog10.gif)

The final step in solving this linear programming problem is to take the three

corner points, (8, 3), (0, 7), and (12, 0), and plug them each into the original

equation that we developed.

V=8x+12 y

1) V=8 (8 )+12 (3 )

V=100

2) V=8 (0 )+12 (7 )

V=84

3) V=8 (12 )+12 (0 )

5

V=96

The final answer is that you would need to buy eight of the X cabinets

and three of the Y cabinets in order to receive the maximum storage volume

of one hundred. Example two is a simple sample problem that I created to

relate to investment business.

Ex. 2) There are two stocks that an investor wants to buy: Google and

Apple. For every Apple stock he buys, he wants to buy at least two Google

stocks. He has $400,000 to invest in Apple and $500,000 to invest in Google. If

each Apple stock at the time of purchase is $2.00 and each Google stock at the

time of purchase is $3.00, how much money should be invested in each stock

to maximize the revenue.

Just like in example one, the first step is to come up with variables and

the equation. In this case, they would be:

X= the number of Apple stocks purchased

Y= the number of Google stocks purchased

Revenue( R )=2 x+3 y

The following step is to devise the constraints. Since the investor

cannot buy a negative amount of the stock, we know that:

X ≥ 0

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Y ≥ 0

In addition, since the investor wants at least two Google stocks for

every Apple stock, we know that:

X ≥ 2 y

Or

Y ≤ 12

x

We also know the restrictions of the amount of money he has to spend.

X ≤ 400,000

Y ≤500,000

The final step is to graph all of the inequalities we just created and plug

the corner points into the equation.

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There are four corner intersecting points on this graph: (0,0), (0,5), (4,4), and

(4,5). When plugging each into the equation created for the optimization of

this problem, the point (4,5) would yield to the largest profit. This proves that

in order to gain the most profit, one must put in the most money, in this case,

$400,000 to Google and $500,000 to Apple. Stocks do have many other

components to them other than the money being put into them, in addition to

the fact that their prices fluctuate continuously, so this is simply a basic and

fundamental example to build on further.

Linear programming problems become more complex when there are

more than two variables. In this case, there must be a series of extra steps to

transform the inequalities to only two variables. Example three guides

through these steps.

0 1 2 3 4 5 60

1

2

3

4

5

6 Stock Investment

Y=(1/2)x

X=0

Y=0

X=4

Y=5

Amount of money for Google stock(in hundred thousands)

Am

oun

t of

mon

ey fo

r A

pp

le s

tock

(in

hu

nd

red

th

ousa

nd

s)

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Ex. 3) “A building supply has two locations in town. The office receives

orders from two customers, each requiring 3/4-inch plywood. Customer A needs

fifty sheets and Customer B needs seventy sheets. The warehouse on the east side of

town has eighty sheets in stock; the west-side warehouse has forty-five sheets in

stock. Delivery costs per sheet are as follows: $0.50 from the eastern warehouse to

Customer A, $0.60 from the eastern warehouse to Customer B, $0.40 from the

western warehouse to Customer A, and $0.55 from the western warehouse to

Customer B. Find the shipping arrangement which minimizes costs”(Stapel).

In this case, the variables are the east warehouse for Customer A, the west

warehouse for customer A, the east warehouse for Customer B, and the west

warehouse for customer B. These can be named Ae , Aw , Be , and Bw. To put these into

applicable equations, since Customer A needs fifty sheets and Customer B needs

seventy sheets, one would then receive:

Ae+ Aw=50

Be+Bw=70

These could then be solved for only one variable in order to make them

utilizable in the following inequalities.

Aw=50−Ae

Bw=70−B e

Because of the fact that the eastern warehouse can only ship up to eighty

sheets and will not ship less then zero, one would receive the following constraint.

0≤ A e+B e ≤80

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Similarly, one would receive the following constraint due to the facts that the

western warehouse ships more than zero sheets, but less than forty-five.

0≤ Aw+Bw≤ 45

The equation one would then create for this problem would be for the cost of

the shipment. The variable “C” represents the cost of shipment.

C=0.5 Ae+0.6 Be+0.4 Aw+0.55 Bw

Now that all of the necessary equations and inequalities created, one would

then take the inequality for the western warehouse and convert it into having the

same variables as the eastern warehouse. Using the equations on the top of this

page could do this. The first inequality does not change.

0≤ Aw+Bw≤ 45

0 ≤ (50−A e)+( 70−Be ) ≤ 45

This simplifies to:

0 ≤ 120−A e−Be ≤ 45

One could then multiply by negative one to receive:

0≥−120+ Ae+B e ≥−45

After adding one hundred and twenty to each side, the final two inequalities

are:

120≥ A e+Be ≥75

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0≤ A e+B e ≤80

These two inequalities can finally combine into one.

75 ≤ A e+Be ≤80

One could also simplify the equation created for the cost of the shipment

through the same process.

C=0.5 Ae+0.6 Be+0.4 Aw+0.55 Bw

C=0.5 Ae+0.6 Be+0.4 (50−A e)+0.55 (70−Be )

C=0.1 Ae+0.05 B e+58.50

Due to the needs of the two customers, two other constraints that are

necessary are:

0≤ A e ≤ 50

0 ≤ Be ≤70

To make these inequalities easier to graph, and because there are only two

variables, one could rename the variables.

x=A e

y=Be

In total, the inequalities one would need to graph and to minimize the

shipment cost are:

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x≥ 0

y ≥0

y ≥−x+75

y ≤−x+80

x≤ 50

y ≤70

The final graph should depict the solution set as shown on the following

page.

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The corner points are (5, 70), (10, 70), (50, 30), and (50, 25). Once plugged

into the equation, the point (5,70) would yield to the minimum cost. This means that

five sheets went to Customer A from the eastern warehouse, seventy sheets went to

Customer B from the eastern warehouse, forty five sheets went to Customer A from

the western warehouse, and zero sheets went to customer B from the western

warehouse.

Linear programming is a convenient way to solve these problems when there

are few variables, but it can get complex when there are more than two, as shown in

example three. In business and economics, there are usually more than two

variables in the problems faced. For this reason, George Dantzig devised the simplex

method. Through the use of this algorithm, he was able to solve problems like

example three and ones even more complex faster and more easily.

The simplex method is the algorithm that computer programs, like excel, use

to find the optimal amount of each variable to come to the highest profit. Millions of

investors and researchers use this process in order to determine how much money

to put in each stock they are interested in and how to maximize their income. The

“solver” utensil in excel uses this algorithmic process in seconds to come up with the

answers the investors are looking for, but it is crucial to also understand how to do

this by hand in order to truly understand what is happening with the money being

invested.

The basic key to the simplex method is the tableau that organizes all of the

data and allows the algorithm to work accurately. Similar to the linear

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programming method of solving these types of problems, the first three steps are to

find the variables, construct an equation, and create the constraints, or inequalities,

for the problem.

After these steps, the process becomes different. One would then need to

create the equations from the inequalities to be able to sync with the tableau. In

order to do this, one would need to add a “slack variable” to the “less than” side of

the inequalities. This variable represents any extra content that is left over to allow

the inequality to be transformed into an equation. Next, one would take the starting

equation that was made for this problem, not the constraints, and equal it to zero,

while keeping the income or profit variable positive. This allows the equation to be

easily applied to the tableau. The following step would be to create the tableau and

use the algorithm to come with a final answer. Examples four and five go through

the steps described above and the following.

Ex. 3) “Maximize: P = 3x + 4y subject to”(Some simplex method examples):

x+ y≤ 4

2 x+ y ≤ 5

x≥ 0

y ≥0

Since we already know the variables, equation, and inequalities, we are able

to skip to converting the inequalities to equations. For the first two inequities, we,

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using the slack variables “s” and “v.” We do not need to add slack variable to the last

two constraints because they are not used in the tableau. This yields to:

x+ y+s=4

2 x+ y+v=5

The equation used in the tableau would be equal to zero.

−3 x−4 y+P=0

The next step is to create the tableau. We do this by placing the variables on

the left sides of the equations on the top and the numbers on the right sides of the

equations on the right side of the tableau. The slack variables and profit variable go

on the right side of the tableau. The tableau would look like this:

X Y S V PS 4V 5P 0

In order to fill in the tableau, we write in how much of each variable we have

in each equation. The filled in tableau would be:

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X Y S V PS 1 1 1 0 0 4V 2 1 0 1 0 5P -3 -4 0 0 1 0

The most important element of the simplex method algorithm is the pivot

element. This number in the tableau is the crucial piece that we want to equal one.

Using the pivot number, we can have a series of row reductions to reach our goal: to

get every number in the final row to be positive. The way we find this pivot number

is by first finding the pivot column. It is the column with the lowest negative

number; in this case it is the negative four column. Once we find this column, we

need to take the constant, or the number in the farthest right column, and divide it

by the number in the same in the pivot column of that same row.

4 ÷1=4

5 ÷ 1=5

The row with the smaller number becomes the pivot row, and the number

that is in both the pivot column and row is the pivot number. In this case, 1 is the

pivot number. The objective in this step is to make a row reduction to make the

pivot number equal to 1. Coincidentally, the pivot number in this case is already 1;

therefore, there is no work that we need to do to change that. Now that we found the

pivot number, the next step is to get the rest of the pivot column equal to zero. This

is accomplished with a couple of row reduction steps. For the second row,

−1R1+R2→ R2

And for the third row,

4 R1+R3→ R3

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Doing these processes will give us the following tableau with the pivot

column having two zeros and the pivot number of one. In addition, another rule to

this algorithm is that we must swap the top variable of the pivot column with the

left side variable of the row with the pivot number in it.

We would repeat these steps until all of the numbers in the bottom row are

positive. In this case, after one round, all of the numbers are already positive.

The final step to the simplex method algorithm is to interpret the results. The

variables on the left side of the tableau are equal to the corresponding numbers on

the right side of the tableau, and the other variables that are left over are set equal

to zero.

Y=4

V=1

P=16

X=0

S=0

So, the final answer to this problem is that we would get the maximum profit

of sixteen when:

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X Y S V PY 1 1 (Pivot number) 1 0 0 4

V 1 0 -1 1 0 1P 1 0 4 0 1 16

X=0

and

Y=4

The “V” variable is the slack variable that indicates that the point does not lie

on the equations that we created from the inequalities, but it does sync with the

inequalities themselves; therefore, we are able to disregard the slack variable in this

case. The slack variable shows us that we went below one of our constraints.

Example five is an example that I created that explains how this relates to the

economic and investment worlds in addition to giving an example of a more

complex problem.

Ex. 5) An investor wants to maximize his profits when buying three stocks:

Facebook, Disney, and Yahoo (A, B, C). Stock A costs $3.00 each and has a $2.00

trading fee; stock B costs $4.00 and has a $2.00 trading fee; and stock C costs $2.00

and has a $3.00 trading fee. He has $1,000 to invest and only wants to spend $900

on trading fees. If stocks A, B, and C come up with a profit of $4, $5, and $6

respectfully, how many of each would give him the largest profit.

Again, the first steps are to identify the variables, come up with an equation,

and create the constraints. In this example, we know that:

X=¿ Number of Stock A bought

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Y=¿ Number of Stock B bought

Z=¿ Number of Stock C bought

The equation, using the profits coming from each stock, would be:

Maximize: P=4 x+5 y+6 z

The constraints for this problem are:

X , Y ,Z ≥ 0

3 x+4 y+2 z≤ 1000

2 x+2 y+3 z≤ 900

Now, we must make the equation equal to zero and the inequities into

equations using the slack variables.

−4 x−5 y−6 z+P=0

3 x+4 y+2 z+s≤ 1000

2 x+2 y+3 z+v ≤ 900

Then, we construct the tableau.

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X Y Z S V PS 3 4 2 1 0 0 1000V 2 2 3 0 1 0 900P -4 -4 -6 0 0 1 0

The pivot column would be the Z column because negative six is the lowest

number in the last row. Since 900 ÷ 3=300 and 1000 ÷ 2=500, the second row has the

pivot number, which is the number three. In order to get that number to equal one,

we need to divide every number in that row by three. The tableau would then look

like:

Next, we create row reductions to get the other two numbers in the pivot column

equal to zero.

−2 R2+ R1→ R1

6 R2+R3 →R3

The new tableau, with the Z variable transferred, looks like:

Since there are no negative numbers in the final row after this round, we are done.

The answer to this problem is:

X=0

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X Y Z S V PS 3 4 2 1 0 0 1000V 2

323

1 0 13

0 300

P -4 -4 -6 0 0 1 0

X Y Z S V PS 5

383

0 1 −23

0 400

Z 23

23

1 0 13

0 300

P 0 0 6 0 2 1 1800

Y=0

Z=300

S=400

P=1800

So, the maximum profit of $1800 is reached when he buys zero of stock A, zero

of Stock B, and three hundred of stock C, but the slack variables show us that the answer

is far less than the maximum for the inequalities.

Linear programming and the simplex method are two highly important tools used

every day in the investment and economic worlds. They both take complex problems

with more than one variable, and transform them into easily solved graphs and tables. As

shown in examples two and five, both of these mathematical apparatuses are directly

applicable to business and investments.

Currently, we have even more complex mathematical algorithms and

processes to solve longer, more challenging problems. One of these, to do further

research into, is the research done at the University of Cambridge on solving semi-

infinite linear programming problems. There are still new, faster ways to solve

everyday problems we face, but without the original ideas of Leonid Kantorovich

and George Dantzig, economics and investment research would not be the same

today.

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References

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Mifflin Canada.

Example (part 1): Simplex method. (n.d.). Retrieved December 26, 2013, from

PHPsimplex website:

http://www.phpsimplex.com/en/simplex_method_example.htm

Reeb, J., & Leavengood, S. (n.d.). Using the simplex method to solve linear programming

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http://ir.library.oregonstate.edu/xmlui/bitstream/handle/1957/20086/em8720-e.pdf

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from

http://www.ms.uky.edu/~rwalker/Class%20Work%20Solutions/class%20work

%208%20solutions.pdf

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http://www.purplemath.com/modules/linprog2.htm

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Stapel, E. (n.d.). Linear programming: A word problem with four variables. Retrieved

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http://www.purplemath.com/modules/linprog5.htm

Stapel, E. (n.d.). Linear programming: Introduction. Retrieved December 26, 2013, from

Purplemath website: http://www.purplemath.com/modules/linprog.htm

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http://www.purplemath.com/modules/linprog4.htm

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from Purplemath website: http://www.purplemath.com/modules/linprog3.htm

Steps in the simplex method [Image]. (n.d.). Retrieved from

http://i.stack.imgur.com/WTm0f.png

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