Midterm Review PPT

ADM2302 DGD - THURSDAY 7 - 8:30 Midterm Review

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Agenda

• Forming Linear Programming Problems

• Plotting Lines and Graphing

• Using Graphs to Solve Linear Problems

• Problems you Might Encounter

▫ Product mix, transportation, production, etc.

• Spreadsheets & Sensitivity Analysis

• Practice Midterm (if time permits)

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A linear problem is any problem which consists of a series of variables

multiplied by a series of coefficients. If variables are multiplied by themselves

or by other variables, the problem becomes non-linear (and is not covered in

this course).

Forming linear problems always involves three steps:

1) Defining the decision variables

2) Creating the objective function

3) Stating the constraints

Forming Linear Programming

Problems

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Defining the Decision Variables

The first thing to be done is to let the reader know what each variable in your problem

represents. Typically we use x1, x2, etc. to define our problems, but you don’t have to –

in fact, it might be more sensible for you to use letters which make sense with the

problem!

Do not be afraid to define variables if you think you need to. Some models have many,

many variables. Don’t think that you necessarily have “too many variables.” If you start a

problem and finish it and realize that there are variables you haven’t used, you can

simply remove them later!

Sometimes variables are identified by two or more subscripts:

For example:

Let xij represent the amount of ingredient i to put in cake j

(where for i, 1 = flour, 2 = salt, 3 = sugar and for j, 1 = wedding and 2 = birthday)

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Creating the Objective Function

Typically in this class, you will only deal with two situations: maximizing profit or

minimizing cost.

If you are given information on both cost and revenue, this becomes a maximization

problem (because you are trying to maximize profit, which is revenue minus cost).

Here, it is important to make use of the variables you just defined.

For example:

How much revenue does each cake bring in? How much does each ingredient cost?

Take these coefficients (the word we use to describe numbers multiplied by variables)

and multiply them by their respective variable. The amount of each cake sold multiplied

by the amount of revenue each cake brings in is total revenue. The amount of each

ingredient used (in both cakes) multiplied by the cost of each ingredient is total cost. The

difference is profit – the thing you want to maximize!

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Stating the Constraints

Arguably the hardest part is to think about how the details given to you in the question

can be translated numerically.

Here, it is important to think logically and translate whatever constraints the question

gives you verbally into a mathematical expression.

Don’t forget about non-negativity constraints…they’re easy marks!

For example:

If I only had 5 pounds of flour, I would have the following constraint:

The amount of flour used in birthday cakes plus the amount of flour used in wedding

cakes must be less than or equal to five pounds total.

X11 + X12 ≤ 5

(the key here is to think logically – does it make sense!?)

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In ADM2302, you will be required to solve (by-hand) a linear system with two

variables (i.e. find the intersection of two straight lines)

Plotting Lines and Graphing

Recall that the equation of a line can be expressed in two ways:

ax + by = c

by = -ax + c

y = (-a/b)x + (c/b)

y = mx + b

Slope

Slope

Y-intercept

Y-intercept

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In this class, we use x1 instead of x, and we use x2 instead of y.


The math doesn’t change (just the names of the variables)

ax1 + bx2 = c x2 = mx1 + b

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There are three ways of finding the intersection of two straight lines:

• Graphing

• Substitution

• Elimination


Practice Problem: Find the intersection of the following two lines:

4x1 + 2x2 = 16 -x1 + 3x2 = 3

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Practice Problem: Find the intercept of the following two lines:

4x1 + 2x2 = 16 and -x1 + 3x2 = 3

Method One: Graph

One of the easiest methods to graph a line is to find the x1-intercept (the x-intercept) and

the x2-intercept (the y-intercept), plot them, and then connect the two. In fact, you can

graph by finding any two points on the line and connecting them!

4x1 + 2x2 = 16

4(x1) + 2(0) = 16

4x1 = 16

x1 = 4

The x1 intercept is (4,0)

4x1 + 2x2 = 16

4(0) + 2(x2) = 16

2x2 = 16

x2 = 8


(x1,0)

(0,x2)

-x1 + 3x2 = 3

-x1 + 2(0) = 3

-x1 = 3

x1 = -3

The x1 intercept is (-3,0)

-x1 + 3x2 = 3

-(0) + 3x2 = 3

3x2 = 3

x2 = 1


(x1,0)

(0,x2)

Note: you can only graph constraints like this; NOT the objective function!

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4x1 + 2x2 = 16 and -x1 + 3x2 = 3

Method One: Graph

Our intercepts are (4,0) and (0,8) for the first line and (-3,0) and (0,1) for the second line.

If we plot these points and draw the lines by connecting them, we can see where they

intersect.

5

10

10 5 -5 -10 (4,0)

(0,8)

(-3,0)

(0,1) (3,2)

Therefore they intersect at (3,2)

Problem: what do you do if they intersect at a decimal?

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4x1 + 2x2 = 16 and -x1 + 3x2 = 3

Method Two: Substitution

This method is much faster and more precise than graphing; it involves solving each

equation for either x1 or x2, then setting them equal and solving.

4x1 + 2x2 = 16

2x2 = 16 - 4x1

x2 = 8 - 2x1

-x1 + 3x2 = 3

3x2 = 3 + x1

x2 = 1 + (1/3)x1

8 – 2x1 = 1 + (1/3)x1

8 – 1 = (1/3)x1 + 2x1

7 = (7/3)x1

x1 = 3

x2 = 8 – 2x1

x2 = 8 – 2(3)

x2 = 2

Therefore they intersect at (3,2)

You should always try to find the

intersection algebraically (sometimes

your eyes can trick you)

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4x1 + 2x2 = 16 and -x1 + 3x2 = 3

Method Three: Elimination

This method is the fastest of them all; it involves modifying the coefficients of one (or

both) variables so that they are the same in each equation (so they cancel out), adding

(or subtracting) the equations, and then solving.

4x1 + 2x2 = 16

-x1 + 3x2 = 3

Therefore they

intersect at (3,2)

Let’s make the x1 coefficients cancel by multiplying the second equation by four.

4x1 + 2x2 = 16

-4x1 + 12x2 = 12 +

14x2 = 28

x2 = 2

4x1 + 2x2 = 16

4x1 + 2(2) = 16

4x1 = 12

x1 = 3

x4

If we add these two equations together,

the 4x1 and -4x1 cancel each other out

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Practice Problem: Graph the following constraint

4x1 + 2x2 ≥ 16

Graphing Inequalities

Most of the time, you’ll be graphing inequalities (problems with a ≤ or ≥ sign). The

common area forms the feasible region. Sometimes it can be tricky knowing which side

of the line the feasible region is on. Luckily, there’s an easy trick!

5

10

10 5 (4,0)

(0,8)

1) Sub the point (0,0) (or any point) into the

inequality.

4(0) + 2(0) ≥ 16

0 ≥ 16 X

2) Determine if the inequality holds (does it

make sense?)

0 is not greater than 16; this doesn’t make sense!

3) If the inequality holds, the side of the line

containing that point (in this case 0,0) is the

side with the FR. If not, the other side of the

line holds the FR

The inequality doesn’t hold, so we are on the far

side of the line (the side without the point 0,0) Don’t assume anything!

Some equations are

misleading

(0,0)

Feasible Region

• This side doesn’t

contain (0,0)

Note: if a constraint has a = sign, the FR

is on the line, not above/below it

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After plotting your constraints and finding the feasible region, the next step is

to find the optimal solution. There are two ways of doing this:

• Testing each vertex (corner) of the feasible region by plugging it into the

objective function

• Graphing the objective function

Using Graphs to Solve Problems

(0,10)

(10,0)

(3.33,3.33)

Feasible

Region

(5,0) (0,0)

20

(0,5)

10

5 10

Practice Problem: Find the optimal

solution of the graph to the right,

assuming the objective function is

MAX z = 2x1 + 7x2

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Practice Problem: Find the optimal solution of the graph assuming the objective function is:

MAX z = 2x1 + 7x2

Method One: Test each vertex

A famous scientist named George Dantzig proved that the optimal solution will always lie

on a boundary (usually a corner) of the feasible region (i.e. never “in the middle” of the

feasible region). Therefore, to find the optimal solution, all we have to do is find and test

each corner point by plugging them into the objective function! Usually, you will have to

use algebra to find the intersection of the constraints (so you can test those vertices). In

this question, they have been given to you.

(0,10)

(10,0)

(3.33,3.33)

Feasible

Region

(5,0) (0,0)

20

(0,5)

10

5 10

MAX z = 2x1 + 7x2

(0,0) z = 2(0) + 7(0) = 0

(0,5) z = 2(0) + 7(5) = 35 ✔

(3.33,3.33) z = 2(3.33) + 7(3.33) = 30

(5,0) z = 2(5) + 7(0) = 10

Therefore the optimal value of 35 occurs

at (0,5)

Note: Make sure to test ALL vertices of

the FR. Don’t just assume the optimal

solution is the intersection of the two

“main” constraints

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MAX z = 2x1 + 7x2

Method One: Test each vertex

Question: What does it mean if two of my vertices have the same optimal solution? (i.e. I

plug each corner point into the optimal solution and two of them get the same response)

Answer: It means the objective function is parallel to one of the constraints…stay tuned

for more info!

(0,10)

(10,0)

(3.33,3.33)

Feasible

Region

(5,0) (0,0)

20

(0,5)

10

5 10

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MAX z = 2x1 + 7x2

Method Two: Graph the objective function

An alternate way to find the optimal solution is to graph the objective function. However,

we have a problem: we don’t have enough information to graph the OF since we don’t

know what z is yet.

(0,10)

(10,0)

(3.33,3.33)

Feasible

Region

(5,0) (0,0)

20

(0,5)

10

5 10

Graph z = 2x1 + 7x2

z = 2x1 + 7x2

-7x2 = 2x1 – z

x2 = -(2/7)x1 + (z/7)

We know the slope of the OF: -2/7

We don’t know the y-intercept of the OF,

since we don’t know what z is.

However, we know that we want to maximize z.

Therefore, if we want to maximize z, in the

process we are also maximizing z/7 (the y-

intercept). Therefore, if we can maximize the y-

intercept, we can maximize the OF along the

way!

→ y = mx+b

Slope

Y-Intercept

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MAX z = 2x1 + 7x2


We know our slope…we just don’t know our y-intercept. All we know is that we want the

biggest y-intercept we can manage while still being in the feasible region.

(3.33,3.33)

20

(0,5)

10

5 10

Slope is -(2/7)

(this means we go down 2, over 7)

(0,7)

↓2

→7

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MAX z = 2x1 + 7x2


We know our slope…we just don’t know our y-intercept. All we know is that we want the

biggest y-intercept we can manage while still being in the feasible region.

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(0,5)

10

5 10 (0,7)

We keep “sliding” the OF up

until we can’t slide it up

anymore (because we wouldn’t

be in the FR). The point(s)

which the objective function

touch are the optimal solution

(in this case it’s 0,5).

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MAX z = 2x1 + 7x2

Redundant Constraints

Let’s say we added in the orange constraint:

20

10

5 10

Adding this constraint has not changed the

feasible region; therefore it is redundant.

Redundant constraints are any constraints which

do not affect the feasible region (in other words, if I

were to take it away, the feasible region would

remain unchanged). Note that redundant

constraints have nothing to do with the optimal

solution.

Note: In almost all situations, redundant

constraints can be identified because they never

touch the feasible region

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MAX z = 2x1 + 7x2

Special Situation: Unbounded Problems

What would happen if the feasible region was on the other side?

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(0,5)

10

5 10 (0,7)

By maximizing the y-intercept,

we keep on going…and

going…and going…and going…

Therefore this problem

is unbounded. There is

no real solution (the

solution is infinity).

Note: Any open-ended feasible region runs the risk of being unbounded.

Careful though…it depends on if it’s a MAX or MIN problem!

• If it’s open to the right or top (or both) and it’s a MAX, it’s unbounded

• If it’s open to the left or bottom (or both) and it’s a MIN, it’s unbounded

In fact, the solution to this

problem is infinity, because we

can go infinitely high.

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MAX z = x1 + 2x2

Special Situation: Infinite number of solutions

What would happen if we changed the objective function to MAX z = x1 + 2x2?

(3.33,3.33)

20

(0,5)

10

5 10 (0,7)

↓1

→2

Graph z = x1 + 2x2

z = x1 + 2x2

-2x2 = x1 – z

x2 = -(1/2)x1 + (z/2)

→ y = mx+b

Slope

Y-Intercept

We know the slope of the OF: -1/2

We don’t know the y-intercept of the OF,

since we don’t know what z is.

Again, we know that we want to maximize z.

Therefore, if we want to maximize z, in the

process we are also maximizing z/2 (the y-

intercept). Therefore, if we can maximize the y-

intercept, we can maximize the OF!

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MAX z = x1 + 2x2


By following the same procedure and “sliding” the objective function upwards, we

maximize the y-intercept (and therefore the objective function) at (0,5) and (3.33,3.33),

as well as all points inbetween

(3.33,3.33)

20

(0,5)

10

5 10 (0,7)

Since there are many points on

that last objective function

(coloured purple) which lead to

a maximized y-intercept (and

therefore objective function),

there are many (an infinite

number) of solutions to this

answer.

All points on this line segment

are optimal. However, there is

only one optimal value: 10 (if

you plug in any point along that

line, you’ll get a value of 10).

If you have more than one optimal solution, it means your objective function

and one of your constraints have the same slope (i.e. they are parallel)

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If the ratio between x1 and x2 in the objective function is the same as the ratio between

x1 and x2 in one of your constraints, it is highly possible that there are infinite solutions.

(0,14) Feasible

Region

(0,0)

20

20 40

10

(12,0)

(0,15)

(0,20)

(15,0)

(18,0)

(6.75, 8.75)

We are trying to minimize our cost

MIN z = 20x1 + 12x2

s.t.

5x1 + 3x2 ≥ 60

2x1 + 2x2 ≥ 30 x1, x2 ≥ 0

7x1 + 9x2 ≥ 126

Right away we can see that 20/12 = 1.67 = 5/3

Therefore this constraint and the OF have the

same slope and are therefore parallel. This

means there is a good chance we have multiple

optimal solutions (you still need to solve

though) Note: only the portion of the line within the FR represent the

optimal solution (i.e. it’s not the whole line!)

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Here’s a summary

• You can graph an intercept by finding its x and y intercepts and then connecting them with a

ruler

• To determine what side of the line the feasible region is on, you can use the (0,0) trick

• You should always use algebra (substitution or elimination) to find the exact point that two

lines intersect at

• To graph the objective function, find the slope with a ruler and “slide” it upwards (if MAX) or

downwards (if MIN) until you reach a point where if you were to slide it anymore it would leave

the feasible region

• A constraint which does not change the feasible region is known as redundant. Usually you

can tell a constraint is redundant if it does not touch the feasible region

• If a feasible region is “open” (not enclosed) to the right or top and the problem is a MAX

problem, the problem is unbounded and there is no optimal solution (same thing if it’s open to

the bottom or left and it’s a MIN problem)

• When a binding constraint and an objective function have the same slope (they’re parallel),

there will be infinite optimal solutions (but only one optimal value). The optimal solutions occur

on the line segment defined by the constraint (but only the part within the feasible region!)

That’s too Confusing!

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Practice Problem #1 1. Consider the following linear programming problem:

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

a) Graph the constraint lines and mark them clearly with numbers (1), (2), and (3). Darken the feasible region.

b) Without graphing, find the optimal solution. Make sure to solve for the exact values of co-ordinates.

c) If the co-efficient of x2 in the objective function was “a”, what would be the value of “a” so that the iso-profit [objective function] line would be parallel to constraint (1) and there would be multiple optimal solutions? What would be the value of the objective function?

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1. Consider the following linear programming problem:

a) Graph the constraint lines and mark them clearly with

numbers (1), (2), and (3). Darken the feasible region.

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

To graph each line, let’s find the intercepts:

x1 + 2x2 = 13

x1 + 2(0) = 13

x1 = 13


x1 + 2x2 = 13

1(0) + 2x2 = 13

2x2 = 13

x2 = 6.5

The x2 intercept is (0,6.5)

-2x1 + x2 = 4

-2x1 + 2(0) = 4

-2x1 = 4

x1 = -2

The x1 intercept is (-2,0)

-2x1 + x2 = 4

-2(0) + x2 = 4

x2 = 4


2x1 – 3x2 = 12

2x1 – 3(0) = 12

2x1 = 12

x1 = 6


2x1 – 3x2 = 12

2(0) – 3x2 = 12

-3x2 = 12

x2 = -4

The x2 intercept is (0,-4)

To determine which side of the line the FR is on, let’s test out the point (0,0) in each constraint:

x1 + 2x2 ≤ 13

(0) + 2(0) ≤ 13

0 ≤ 13 ✔

-2x1 + x2 ≤ 4

-2(0) + 2(0) ≤ 4

0 ≤ 4 ✔

2x1 – 3x2 ≤ 12

2(0) – 3(0) ≤ 12

0 ≤ 12 ✔

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a) Graph the constraint lines and mark them clearly with

numbers (1), (2), and (3). Darken the feasible region.

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

The intercepts of constraint 1 are

(13,0) and (0,6.5); the intercepts

of constraint 2 are (-2,0) and

(0,4); the intercepts of constraint

3 are (6,0) and (0,-4)

We know the Feasible Region is

on the side of the line with (0,0)

(this is true for all constraints in

this problem)

We know these vertices:

(0,0) (0,4) (6,0)

However, to solve this problem,

we also need to find the

intersection of (1) and (2), as

well as (1) and (3)

5

10

20 10 -10 -20 (6,0)

(0,6.5)

(-2,0) (13,0)

(0,4)

(0,-4)

(1)

(2)

(3)

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b) Without graphing, find the optimal solution. Make sure to

solve for the exact values of co-ordinates.

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

(1) and (2)

2x(1) 2x1 + 4x2 = 26

+ (2) -2x1 + x2 = 4

5x2 = 30

x2 = 6

-2x1 + x2 = 4

-2x1+ (6) = 4

-2x1 = -2

-2x1 = 1

Therefore they

intersect at (1, 6)

(1) and (3)

2x(1) 2x1 + 4x2 = 26

- (2) 2x1 - 3x2 = 12

7x2 = 14

x2 = 2

2x1 - 3x2 = 12

2x1 - 3(2) = 12

2x1 = 18

x1 = 9

Therefore they

intersect at (9, 2)

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b) Without graphing, find the optimal solution. Make sure to

solve for the exact values of co-ordinates.

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

We now know all the vertices of

the feasible region, allowing us

to test each corner point.

5

10

20 10 -10 (6,0)

(0,6.5)

(-2,0) (13,0)

(0,4)

(0,-4)

(1)

(2)

(3) (1,6)

(9,2)

MAX z = 3x1 + 4x2

(0,0) z = 3(0) + 4(0) = 0

(0,4) z = 3(0) + 4(4) = 16

(1,6) z = 3(1) + 4(6) = 27

(9,2) z = 3(9) + 4(2) = 35 ✔

(6,0) z = 3(6) + 4(00) = 18

Therefore the objective function is

maximized at 35 at the point (9,2)

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c) If the co-efficient of x2 in the objective function was “a”,

what would be the value of “a” so that the iso-profit

[objective function] line would be parallel to constraint (1)

and there would be multiple optimal solutions? What would

be the value of the objective function?

MAX z = 3x1 + 4x2

Subject to

(1) x1 + 2x2 ≤ 13

(2) -2x1 + x2 ≤ 4

(3) 2x1 - 3x2 ≤ 12 x1, x2 ≥ 0

We want the line z = 3x1 + ax2 to be parallel (same slope) as the line x1 + 2x2 = 13.

Let’s put both equations into y = mx + b form to easily see what m, the slope, is:

z = 3x1 + ax2

-ax2 = 3x1 – z

x2 = -(3/a)x1 + (z/a)

13 = x1 + 2x2

-2x2 = x1 – 13

x2 = -(1/2)x1 + (13/2)

We must make the slopes of each line equal for these two lines to be parallel:

-(3/a) = -(1/2)

a = 6

If a = 6 and the objective function is MAX z = 3x1 + 6x2, we would have infinite solutions.

The optimal value can be found by plugging in any point on constraint (1), such as (1,6)

or (9,2).

z = 3(1) + 6(6) = 3(9) + 6(2) = 39

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There are a number of problems you may encounter on your midterm. It’s

important you have a general sense of how to solve each one:

• Product-Mix problem

• Blending Problems

• Transportation Problems

Problems you Might Encounter

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Product-Mix Problems

Product-Mix problems are the most basic type of problem you can expect to encounter.

They involve choosing a combination of products to produce that maximize profit while

minimizing cost.

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Practice Problem #2

2. The Primo Insurance Company is introducing new product lines: special risk insurance

and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per

unit on mortgages. Management wishes to establish sales quotas for the new product

lines to maximize total expected profit. The work requirements are shown below.

Formulate this model algebraically.

35

Work-Hours per Unit

Department Special

Risk

Mortgage Work-Hours

Available

Underwriting 3 2 2400

Administration 0 1 800

Claims 2 0 1200

2.12 The Primo Insurance Company is introducing new

product lines: special risk insurance and mortgages. The

expected profit is $5 per unit on special risk insurance and

$2 per unit on mortgages. Management wishes to establish

sales quotas for the new product lines to maximize total

expected profit. The work requirements are shown here:

d) Formulate this same model algebraically.

Work-Hours per Unit

Department Special

Risk

Mortgage Work-Hours

Available

Underwriting 3 2 2400

Administration 0 1 800

Claims 2 0 1200

Let x1 = the amount of special risk units we should seek to sell

Let x2 = the amount of mortgages units we should seek to sell

MAX z = 5x1 + 2x2

3x1 + 2x2 ≤ 2400

(1)x2 ≤ 800

2x1 ≤ 1200

x1,x2 ≥ 0

Step 1: Define decision variables

Step 2: Create objective function

Step 3: State constraints

Blending Problem

Blending problems can be a bit confusing because they typically require the use of

percentages – and a percentage of the final result must meet a certain requirement.

For these problems, it might be a good idea to review decimal rules, and perhaps try

some practice problems which use decimals.

A key characteristic of this type of problem is that there will be a “total amount” on the

right hand side multiplied by a percentage (because a max/min percentage of the total

must meet a requirement); note that this “total amount” might be a number or a variable.

For example: if you wanted to know how much lemonade to produce, but you knew that

80% of total lemonade produced must be lemon juice, then the constraint would be

Lemon Juice = (0.8)(Lemonade Produced), where Lemonade Produced is a variable.

Alternatively, if you had a set amount of lemonade that needed to be produced, then

Lemonade Produced would be a set number (like 5 liters).

37

Practice Problem #3

3. Agri-Pro is a company that sells agricultural products to farmers in a number of states.

One service it provides to customer is custom feed mixing, whereby a farmer can order a

specific amount of livestock feed and specify the amount of corn, grain, and minerals the

feed should contain. This is an important service because the proper feed for various

farm animals changes regularly depending on the weather, pasture conditions, and so

on. Agri-Pro has just received an order from a local chicken farmer for 8,000 pounds of

feed. The farmer wants this feed to contain at least 20% corn, 15% grain, and 15%

minerals. Minimize cost using the following information:

Feed 1 Feed 2 Feed 3 Feed 4 Minimum

Requirement

Corn 30% 5% 20% 10% 20%

Grain 10% 30% 15% 10% 15%

Minerals 20% 20% 20% 30% 15%

Cost per

1000

pounds

$250 $300 $320 $150

38

3 Agri-Pro is a company that sells agricultural products to

farmers in a number of states. One service it provides to

customer is custom feed mixing, whereby a farmer can

order a specific amount of livestock feed and specify the

amount of corn, grain, and minerals the feed should

contain. This is an important service because the proper

feed for various farm animals changes regularly depending

on the weather, pasture conditions, and so on. Agri-Pro has

just received an order from a local chicken farmer for 8,000

pounds of feed. The farmer wants this feed to contain at

least 20% corn, 15% grain, and 15% minerals. Minimize

cost using the following information:


We are trying to decide how much feed to use

Let x1 = the amount of Feed 1 in thousands of pounds to use in the mix




Step 2: Create objective function

We are trying to minimize cost

MIN z = 250x1 + 300x2 + 320x3 + 150x4


Requirement

Corn 30% 5% 20% 10% 20%

Grain 10% 30% 15% 10% 15%

Minerals 20% 20% 20% 30% 15%

Cost per 1000 pounds

$250 $300 $320 $150

39

3 Agri-Pro is a company that sells agricultural products to

farmers in a number of states. One service it provides to

customer is custom feed mixing, whereby a farmer can

order a specific amount of livestock feed and specify the

amount of corn, grain, and minerals the feed should

contain. This is an important service because the proper

feed for various farm animals changes regularly depending

on the weather, pasture conditions, and so on. Agri-Pro has

just received an order from a local chicken farmer for 8,000

pounds of feed. The farmer wants this feed to contain at

least 20% corn, 15% grain, and 15% minerals. Minimize

cost using the following information:


The farmer has certain requirements for his feed mix:

x1 + x2 + x3 + x4 = 8 (farmer wants 8000 pounds)

0.3x1 + 0.05x2 + 0.2x3 + 0.1x4 ≥ 8(0.2) (20% must be corn)

0.1x1 + 0.3x2 + 0.15x3 + 0.1x4 ≥ 8(0.15) (15% must be grain)

0.2x1 + 0.2x2 + 0.2x3 + 0.3x4 ≥ 8(0.15) (15% must be minerals)

xi ≥ 0


Requirement

Corn 30% 5% 20% 10% 20%

Grain 10% 30% 15% 10% 15%

Minerals 20% 20% 20% 30% 15%

Cost per 1000 pounds

$250 $300 $320 $150

The “total amount” (8) on the

RHS is multiplied by a

percentage. In other blending

problems, this “total” could be a

number or a variable.

40

Transportation Problem

A transportation problem involves moving resources between two or more areas;

typically the goal is to minimize cost.

The key to a transportation problem is that the amount pumped into the network must

equal the amount pumped out of the network. In other words, the total resource flow into

a node equals the total resource flow out of a node.

There may be additional constraints as well, such as a limit on what can travel certain

routes.

41

Practice Problem #4

4. The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron

ore is shipped to either of two storage facilities. When needed, it is then shipped on to

the company’s steel plant. The diagram below depicts the distribution network, where M1

and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel

plant. The diagram also shows the monthly amounts produced at the mines and needed

at the plant, as well as the shipping cost and the maximum amount that can be shipped

per month through each shipping lane. Management now wants to determine the most

economic plan for shipping the iron ore from the mines through the distribution network

to the steel plant. Express this model in algebraic form.

M1

M2

S1

S2

P

40 Tons

Produced

60 Tons

Produced

$2,000/ton

$1,100/ton

30 tons max

50 tons max

42

3.18 The Fagersta Steelworks currently is working two

mines to obtain its iron ore. This iron ore is shipped to

either of two storage facilities. When needed, it is then

shipped on to the company’s steel plant. The diagram

below depicts the distribution network, where M1 and

M2 are the two mines, S1 and S2 are the two storage

facilities, and P is the steel plant. The diagram also

shows the monthly amounts produced at the mines and

needed at the plant, as well as the shipping cost and the

maximum amount that can be shipped per month

through each shipping lane. Management now wants to

determine the most economic plan for shipping the iron

ore from the mines through the distribution network to

the steel plant. Summarize the model in algebraic form.

43

M1

M2

S1

S2

P

40 Tons

Produced

60 Tons

Produced

$2,000/ton

$1,100/ton

30 tons max

50 tons max


We are trying to decide how much to ship on each route

Let XM1S1 = the number of tons heading from M1 to S1




Let XS1P = the number of tons heading from S1 to P

Let XS2P = the number of tons heading from S2 to P















44

M1

M2

S1

S2

P

40 Tons

Produced

60 Tons

Produced

$2,000/ton

$1,100/ton

30 tons max

50 tons max

Step 2: Define objective function

We want to minimize cost

MIN z = 2000XM1S1 + 1700XM1S2 + 1600XM2S1 + 1100XM2S2 + 400XS1P + 800XS2P















45

M1

M2

S1

S2

P

40 Tons

Produced

60 Tons

Produced

$2,000/ton

$1,100/ton

30 tons max

50 tons max


Everything that goes in must come out

XM1S1 + XM1S2 = 40

XM2S1 + XM2S2 = 60

XM1S1 + XM2S1 = XS1P

XM1S2 + XM2S2 = XS2P















46

M1

M2

S1

S2

P

40 Tons

Produced

60 Tons

Produced

$2,000/ton

$1,100/ton

30 tons max

50 tons max


Each route has a capacity limit

XM1S1 ≤ 30

XM1S2 ≤ 30

XM2S1 ≤ 50

XM2S2 ≤ 50 Xij ≥ 0

XS1P ≤ 70

XS2P ≤ 70

It’s important that you understand how spreadsheets are constructed, as well

as how certain formulas are built.

• Changing Cells – These are your “decision variables”. Solver will change

these cells. These cells will not have formulas, but will be used within other

formulas.

• Target Cell – This is your “objective function.” This is the cell you want

solver to maximize or minimize. Generally, the target cell will be a profit cell

(if maximizing profit) or a cost cell (if minimizing cost).

• Note: Only one cell can be a target cell.

• Coefficient Cells – These form the basis of your constraints and your

objective function. They are numbers which are multiplied by variables (i.e.

your changing cells) to reach the right-hand side of a constraint or the

objective function.

• These cells are also known as parameters because they never

change

Spreadsheets and Sensitivity Analysis

47

It’s important that you understand how spreadsheets are constructed, as well

as how certain formulas are built.

• Blue Cells – Coefficient cells (parameters)

• Orange Cells – Target cell (total profit/cost cell)

• Yellow Cells – Changing cells (decision variables)


48

It’s also important that you know how solver works, as well as certain formulas:

• SUM(A1:B5) – Will sum all cells in the rectangle formed by A1 and B5.

• SUMPRODUCT(A1:A3,B1:B3) – Multiplies the first cell in the first range

with the first cell in the second range (A1 times B1), the second cell in the

first range with the second cell in the second range (A2 times B2), etc. and

then sums them all together (don’t get confused about the syntax!)

• IF(Condition, Result if True, Result if False) – Not as commonly used

• Solver – Self-explanatory (see picture)

If this spreadsheet had blanks, could you fill them in?


49

Sensitivity Analysis

Allows us to examine how changes in variables/constraints will affect the final solution

50


Final Value of Target Cell – Gives

us the optimal value (usually profit or

cost). In this case, optimal profit is

$208,200

Final Value of Adjustable Cells –

(“Adjustable” is the same as

“changing”) Gives us the optimal

solution. In this case, the optimal

solution is 0 digital boxes, 380 PVRs,

0 TVs, and 1060 recorders.

Status – Binding means that the

optimal solution touches that

constraint (i.e. there’s no extra

“slack” or “surplus”). Non-binding

means that the constraint was not

met (i.e. there was a little wiggle

room). In this case, we’ve used all

our electronic modules and

assembly hours, but still have 560 non-electronic models left for use.

51


Objective Coefficient – The

coefficient that the variable in that

row is multiplied by in the objective

function. In this case, each digital box

produces $87 of profit

Allowable Increase/Decrease – The

amount the objective coefficient could be

increased or decreased without changing

the optimal solution (note: the optimal

value, i.e. profit or cost, will be changed).

In this case, increasing the profit per PVR

by up to 120 or decreasing it by 5 will not

change the optimal solution. The optimal

value (profit) will change by the coefficient

change multiplied by the final value. If

outside this range, we cannot know

exactly how the optimal solution/value will

change without rerunning the model.

Reduced Cost – The amount an

objective coefficient would have to

“improve” to make the variable

non-zero. In this case, profit per

HD TV would have to increase by

24 (alternatively, reduce by -24)

per unit before we would consider

producing TVs. *reminder: optimal solution refers to the decision variables; optimal value refers to the objective

function. Our optimal solution, plugged into the objective function, gives us our optimal value.

Reduced Cost – Alternate

interpretation: the amount that an

optimal value would increase by if we

were to force a one unit increase in

the optimal solution. In this case, if

we were to force the company to

produce 1 HD TV, profit would

increase by -24 (i.e. decrease by 24)

52


Shadow Price – The amount that the

optimal value (the objective function)

would increase (decrease) if we

added (subtracted) 1 extra unit to the

RH side. In this case, an extra

electronic module would lead to $6

extra in profit. This only holds within

the allowable increase/decrease.

Constraint RHS – Simple: it’s the

right hand side of the constraint. In

this case, we only have 4500 non-

electronic modules to use.

Allowable Increase/Decrease – Tells

us the range that the shadow price is

valid in. In this case, we know that an

increase of up to 466.67 hours will

increase profit by $72/hour. We also

know a decrease of up to 1325 hours

will decrease profit by $72/hour.

Outside this range, though, we cannot

make any solid conclusions.

53

Note: Shadow Price is easy to

calculate! If given two scenarios with

different RHS constraints, you can

easily calculate the Shadow Price by

dividing the difference in the optimal

values with the difference in the RHS

constraints (think of two points on a

straight line; the SP is the slope).

*reminder: don’t confuse “Allowable Increase/Decrease” in this section with the one in the Adjustable Cells section. Here, it refers

to the range the shadow price is valid in (this has nothing to do with the optimal solution!). In the previous section, it referred to the

range which the objective coefficient could be changed without impacting the optimal solution.

100% Rule

If two or more elements are changing at the same time, you can use the 100% rule to

determine how things will change.

• If two or more objective coefficients are changing, calculate each change as a

percentage of the allowable increase (if increasing) or decrease (if decreasing). If

the TOTAL sum of the percentage changes is greater than 100%, then we deem this

to be outside the feasible range (and therefore we cannot determine exactly how the

optimal solution and value will change). If it is less than 100%, then this is inside the

feasible range and the optimal solution will remain the same.

• Similarly, if two or more RHS constraints are changing, calculate each change in

the same way (as a percentage of the allowable increase/decrease). If the TOTAL

sum of the percentage changes is greater than 100%, then the shadow prices are

no longer valid and we can’t determine how the optimal value will change. If it is less

than 100%, then the shadow prices are still valid and we can therefore determine

how the optimal value will change.

• See Practice Question 5, part B for an example.

54

Practice Problem #5 5. HiTech Electronics produces four consumer electronic products. They are Digital Boxes, PVRs, HD TVs, and HD Recorders. Each product uses a certain number of electronic modules, non-electronic modules, and assembly hours. Available are 4700 electronic modules, 4500 non-electronic modules, and 2500 assembly hours.

The linear programming problem is the following:

x1 to 4 = units produced (in the same order as above)

MAX z = 87x1 + 96x2 + 216 x3 + 162x4

s.t.

3x1 + 4x2 + 4 x3 + 3x4 ≤ 4700 (electronic)

2x1 + 2x2 + 4 x3 + 3x4 ≤ 4500 (non-electronic)

x1 + 1x2 + 3 x3 + 2x4 ≤ 2500 (assembly hours)

xi ≥ 0

55

5. a) Determine the optimal

production schedule of HTE for the

four products. What is the optimal

profit of such optimal strategy?

Answer: According to this report, the

optimal solution is to produce no digital

boxes or HD TVs, 380 PVRs, and 1060 HD

Recorders. This will lead to an optimal

value (profit) of $208,200

56

5. b) What would be the impact on the

production schedule and the optimal

profit if the local supplier decreased

the purchasing cost of electronic

modules by $0.75? Justify,

Answer: If we look at the original problem two slides

ago, we see that digital boxes use 3 electronic

modules, PVRs use 4 electronic modules, HD TVs use

4 electronic modules, and recorders use 3 electronic

modules (see previous slides).

This corresponds to an INCREASE in the profit

coefficients (decreased cost = higher profit) of 3(0.75) =

$2.25 per unit for digital boxes and HD recorders and

4(0.75) = $3 per unit for PVRs and HD TVs. Because

we are changing multiple coefficients, we must use the

100% rule. Since 2.25/3 + 3/120 + 3/24 + 2.25/30 =

0.975 (less than 100%), the optimal solution will not

change (we will still produce the same amount).

The optimal value WILL change. We now make $2.25

more profit per unit for digital boxes and HD recorders,

and $3 more profit per unit for PVRs and HD TVs.

Z = $208,200 + 2.25(0) + 3(380) + 3(0) + 2.25(1060)

Z = $211,725

Therefore profit will increase by $3,525 to $211,725

Recall: The allowable increase of

the objective coefficient tells us

how much the objective coefficient

(i.e. profit per unit) can increase

without causing a change in the

optimal solution (i.e. we would still

produce the same amount)

57

5. c) In order to start producing HD

TVs, what would need to happen to

the unit profit of HD TVs?

Answer: Recall that the reduced cost

tells us how much a coefficient needs

to “improve” before the variable would

become non-zero

In this case, the objective coefficient

(profit per unit) of HD TVs would need

to improve by $24 before we would

consider producing HD TVs.

Therefore, unit profit would need to

increase $24.

58

5. d) What is the shadow price for

assembly time? If the assembly time

available were to be 2600 hours

instead of the current value, what

would be the new profit and what

would happen to the product mix?

Answer: Recall that the shadow price tells us how

much the optimal solution would

increase/decrease if we were to add one unit to the

right-hand side (RHS) or a constraint (so long as

it’s within the allowable increase/decrease)

In this case, an increase of 100 assembly hours

would yield additional profit of $72 x 100 = $7,200.

Note that the increase of 100 is within the

allowable increase of 466.67, so the shadow price

is valid.

The product mix will almost certainly change;

however, the sensitivity report does not give us

information as to how a change in constraints will

change the optimal solution.

59

5. e) If the availability of non-

electronic modules increases by 500

units, what effect will it have on the

product mix? Why is its shadow price

zero?

Answer: Recall that non-electronic modules

is a non-binding constraint, which means

we already aren’t even using all the non-

electronic modules we have available.

Therefore, an increase in non-electronic

modules is useless to us (we already have

some we’re not using, we don’t need

more!). The optimal value and solution (i.e.

the product mix) will be unchanged.

The shadow price tells us how much an

increase/decrease in a constraint will

increase/decrease the objective function

(profit). Clearly, since extra non-electronic

modules are useless to us, an extra non-

electronic module is of no use and will not

improve profit in any way (i.e. it will bring $0

of extra value).

60

5. f) If the available electronic

modules were only 4000, how would it

impact total profit? What can you say

about the product mix?

Answer: Electronic modules is a binding

constraint, meaning we are using all that

we have available.

Recall that the shadow price tells us how

much the optimal value (profit) will

increase/decrease given an

increase/decrease in the right hand side of

a constraint. A 700 unit decrease in

electronic modules would lead to a $6 x

700 = $4200 decrease in profit (note that

700 is within the allowable decrease of 960,

so the shadow price is valid).

Again, the product mix will likely change,

but we cannot determine how (since the

sensitivity report gives us no information

about how the change in a constraint will

change the optimal solution).

61

Tells us about the optimal value and optimal

solution

Gives us information on how changes to

the objective coefficient will change the

optimal solution (i.e. product mix), so long

as the changes are with the allowable

increase or decrease (we can use that

information to determine the effect on the

optimal value, i.e. profit/cost).

Gives us information on how changes to a

constraint will affect the optimal value (i.e.

profit). Gives no information on how

changes to constraints will affect the

optimal solution (i.e. product mix). We

cannot make any conclusions beyond the

allowable range.

62

63

Part of

Report

If Within the Allowable

Range

If Outside the Allowable

Range

Adjustable

Cells

• Optimal Solution will not

change

• Optimal Value will change

by the coefficient change

multiplied by the final value

• Optimal Solution and

Value will most likely

change (we can’t know by

how much without re-

running the model)

Constraints • Shadow Price remains

valid; therefore, Optimal

Value will change by

change times shadow price

• Optimal Solution will

probably change; we can’t

know by how much without

re-running the model

• Shadow Price is no longer

valid; therefore, Optimal

Value and Solution will

most likely change, but we

can’t know by how much

without re-running the

model

Question: If you were given a

sensitivity report, would you be able

to use things like shadow price,

reduced cost, etc. to work backwards

and calculate what the original values

were?

Question: Given certain information,

could you calculate things like

shadow price, reduced cost, etc.

yourself?

Question: Do you know the difference

between “Allowable

Increase/Decrease” in the Adjustable

Cells part of the report versus the

Constraints part of the report?

Answer: You should!

64

Practice Problem #6 6. Which of the following is not true:

• A feasible solution satisfies all constraints

• An optimal solution satisfies all constraints

• An infeasible solution violates all constraints

• A feasible solution points does not have to lie on the boundary of the feasible region

The amount that the objective function coefficient of a decision variable would have to

improve before that variable would have a positive value in the solution is the

• Shadow price

• Allowable increase

• Reduced cost

• Slack variable

65

Practice Problem #6 6. Which of the following is not true:

• A feasible solution satisfies all constraints

• An optimal solution satisfies all constraints

• An infeasible solution violates all constraints

• A feasible solution points does not have to lie on the boundary of the feasible region

The amount that the objective function coefficient of a decision variable would have to

improve before that variable would have a positive value in the solution is the

• Shadow price

• Allowable increase

• Reduced cost

• Slack variable

66

Practice Problem #6 When formulating transportation problem on a spreadsheet (e.g. Excel), which of the

following are necessary?

a) parameter table.

b) network representation.

c) solution table.

d) a and b only.

e) a and c only.

f) all of the above.

Let Xij = the production of product i in period j. To specify that production of product 1 in

period 3 and in period 4 differs by no more that 100 units.

• X13 – X14 <100; X14 – X13 <100.

• X13 – X14 < 100; X13 – X14 > 100.

• X13 – X14 < 100; X14 – X13 > 100.

• X13 – X14 > 100; X14 – X13 > 100.

67

Practice Problem #6 When formulating transportation problem on a spreadsheet (e.g. Excel), which of the

following are necessary?

a) parameter table.

b) network representation.

c) solution table.

d) a and b only.

e) a and c only.

f) all of the above.

Let Xij = the production of product i in period j. To specify that production of product 1 in

period 3 and in period 4 differs by no more that 100 units.

• X13 – X14 <100; X14 – X13 <100.

• X13 – X14 < 100; X13 – X14 > 100.

• X13 – X14 < 100; X14 – X13 > 100.

• X13 – X14 > 100; X14 – X13 > 100.

68

Practice Problem #6 There is shadow price for every decision variable in a model

a) True

b) False

Because the shadow price represents the improvement in the value of the optimal

solution per unit increase in right-hand side, a shadow price cannot be negative

a) True

b) False

Demand constraints frequently take the form: Beginning inventory + production - ending

inventory = demand

a) True

b) False

69

Practice Problem #6 There is shadow price for every decision variable in a model

a) True

b) False

Because the shadow price represents the improvement in the value of the optimal

solution per unit increase in right-hand side, a shadow price cannot be negative

a) True

b) False

Demand constraints frequently take the form: Beginning inventory + production - ending

inventory = demand

a) True

b) False

70

Any questions?

71

Good luck!

72

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