Quadratic and logarithmic models (new post)

In summary, the conversation discussed a problem regarding internet provider's broadband usage packs. They are considering discontinuing the Quest pack and replacing it with two new options. The first option is a Quadratic Model with a monthly access fee of $20, included gigabytes of 20, and a cost per additional gigabyte that increases quadratically after 20 gigabytes. The second option is a Logarithmic Model with a monthly access fee of $30, included gigabytes of 25, and a cost per additional gigabyte that increases logarithmically after 25 gigabytes. The conversation also mentioned the possibility of removing the "included gigabyte usage" from each pack, and tasks were assigned to graph the current and replacement pack options and develop
  • #1
needalgebra
45
0
Here is my previous post! please look at pages 3 - 4 to see what I am talking about.

http://mathhelpboards.com/pre-algebra-algebra-2/mathematical-modeling-6006-4.html

I figured out how to input the quadratic and logarithmic equations into wolfram, this is what i got. Let me know if its correct please!

f(x)=1/80x2+15, f(x)=30 1+ln(x/25)ln(12/5) where x= 0 to 50

f(x)=1/80x2+15, f(x)=30 1+ln(x/25)ln(12/5) where x= 0 to 50 - Wolfram|Alpha

Now for the no included gigabytes graph, i first have to get the equations. I was wondering if these are set up correctly before solving them.

Quadratic model:

f(0) = a (0)^2 + b = b = 20

f(60) = a (60)^2 + b = a + b = 60

Logarithmic model:

f(0) = a + b Ln (0 + 1) = a = 30

f(60) = a + b Ln(60 + 1) = a + b Ln = 60

Thanks
 
Last edited:
Mathematics news on Phys.org
  • #2
ok.. so I've got new answers. is this correct?

Quadratic formula

cost = a*(gig)² + b
20 = a*(0)² + b ----> b=20.
60 = a*(60)² + b.
b=20,
a = 40/(60²) = 1/90.
cost = (gig²)/90 + 20.

Logarithmic formula:

30 = a + b*LN(0 + 1),
a = 30.
60 = a + b*LN(60 + 1)
a=30
b= 30/LN(61) = approx 7.298
 
  • #3
I am going to go through this problem from beginning to end, mostly because it has been a while since I have looked at it, and this will get me back up to speed with it. I will only address the purely mathematical aspects, as the conclusions drawn should be left up to you to write in your own words. Much of what I will post is also posted elsewhere, but I think it will be nice to finally have everything in one place.

Here is the problem statement:

An internet provider is reviewing its monthly broadband usage packs. Currently customers can choose one of the three following options:
  • Appetizer: monthly access fee \$7, included gigabytes 10, cost per additional gigabyte \$2.
  • Quest: monthly access fee \$40, included gigabytes 40, cost per additional gigabyte \$1.
  • Voyager: monthly access fee \$60, included gigabytes 60, cost per additional gigabyte \$0.50.

These packs comprise a monthly access fee which includes usage up to a certain number of gigabytes plus an additional cost per gigabyte over and above that.

The internet provider intends to discontinue the quest pack. They are considering two new pack options to replace it.

Replacement pack options:

Option 1: Quadratic Model - monthly access fee \$20, included gigabytes 20, cost per additional gigabyte will increase quadratically after 20 gigabytes with customers paying the same amount as Voyager customers at 60 gigabytes.

Option 2: Logarithmic Model - monthly access fee \$30, included gigabytes 25, cost per additional gigabyte will increase logarithmically in the form $f(x)=a+b\ln(x)$ after 25 gigabytes with customers also paying the same amount as Voyager customers at 60 gigabytes.

In addition, they are also looking into the possibility of removing the 'included gigabyte usage' from each pack. This would mean that the customer would pay the monthly access fee plus the cost of each gigabyte used.

This activity requires you to write a report to the internet provider including:

(i) A recommendation of an appropriate new pricing pack for medium users to replace the quest pack.

(ii) An analysis of the effect of the cost to the customer of removing the "included gigabytes usage" from your recommended replacement of the quest pack option and the two remaining packs, appetizer and voyager.

Tasks:
  • Graph the current pack options - Appetizer, Quest and Voyager.
  • Graph the two replacement pack options - option 1 - quadratic model, option 2 - logarithmic model.
  • Develop equations that model the three current and the two replacement pack options, including the domain for each equation - full working must be shown.
  • Find the intersection points of at least 3 different models.
  • Model the effects on the cost to the customer of removing the "included gigabytes usage."

When analyzing the results of your modeling, you will need to:
  • Compare and contrast the different plans, describing the positive and negative aspects of each plan for potential customers.
  • Recommend which of the replacement pack options should replace the quest pack option, with justification.
  • State any assumptions, generalizations and reflections you make during your modeling.
  • Discuss the effect on the cost to the customer of removing the "included gigabyte usage."
  • Provide clear recommendations to the internet provider based on your findings.

Task 1: Graph the current pack options -Appetizer, Quest and Voyager.

The functions that model these packages will need to be piecewise defined, since the cost remains constant from 0 gigabytes used up to and including a given "included gigabytes usage" number, and then for additional usages, the coast increases linearly at a given rate. We will let the independent variable $x$ be the usage in gigabytes, and so the dependent variables will be the cost associated with each package.

Appetizer: monthly access fee \$7, included gigabytes 10, cost per additional gigabyte \$2.

For this package, we will let $A(x)$ represent the cost. We see that we need $A(x)=7$ for $0\le x\le10$. When $10<x$, we know the slope of the line is 2, and it begins at the point (10,7), and so using the point-slope formula, this portion of the function is:

\(\displaystyle A(x)-7=2(x-10)\)

\(\displaystyle A(x)=2x-13\)

And so we may then write the piecewise definition of the cost function as:

\(\displaystyle A(x)=\begin{cases}7 & 0\le x\le10\\ 2x-13 & 10<x \\ \end{cases}\)

Quest: monthly access fee \$40, included gigabytes 40, cost per additional gigabyte \$1.

For this package, we will let $Q(x)$ represent the cost. We see that we need $Q(x)=40$ for $0\le x\le40$. When $40<x$, we know the slope of the line is 1, and it begins at the point (40,40), and so using the point-slope formula, this portion of the function is:

\(\displaystyle Q(x)-40=x-40\)

\(\displaystyle Q(x)=x\)

And so we may then write the piecewise definition of the cost function as:

\(\displaystyle Q(x)=\begin{cases}40 & 0\le x\le40\\ x & 40<x \\ \end{cases}\)

Voyager: monthly access fee \$60, included gigabytes 60, cost per additional gigabyte \$0.50.

For this package, we will let $V(x)$ represent the cost. We see that we need $V(x)=60$ for $0\le x\le60$. When $60<x$, we know the slope of the line is $\dfrac{1}{2}$, and it begins at the point (60,60), and so using the point-slope formula, this portion of the function is:

\(\displaystyle V(x)-60=\frac{1}{2}(x-60)\)

\(\displaystyle V(x)=\frac{1}{2}x+30\)

And so we may then write the piecewise definition of the cost function as:

\(\displaystyle V(x)=\begin{cases}60 & 0\le x\le60\\ \frac{1}{2}x+30 & 60<x \\ \end{cases}\)

To get a plot of the three cost functions, enter the following command at W|A:

piecewise[{{7,0<=x<=10},{2x-13,10<x}}],piecewise[{{40,0<=x<=40},{x,40<x}}],piecewise[{{60,0<=x<=60},{x/2+30,60<x}}] where x=0 to 100

Task 2: Graph the two replacement pack options - option 1 - quadratic model, option 2 - logarithmic model.

We are implicitly given two points through which both curves must pass.

For the quadratic model, we know the quadratic portion of the function begins at (20,20), but must also pass through the point (60,60), since this is the cost for Voyager customers at 60 gigabytes.

So, using:

\(\displaystyle f(x)=ax^2+b\)

we obtain the 2X2 linear system:

\(\displaystyle f(20)=a(20)^2+b=400a+b=20\)

\(\displaystyle f(60)=a(60)^2+b=3600a+b=60\)

Now we may determine the values of the parameters $a$ and $b$. Subtracting the first equation from the second, we eliminate $b$ and get:

\(\displaystyle 3200a=40\,\therefore\,a=\frac{1}{80}\)

Now, substituting for $a$ into the first equation, we find:

\(\displaystyle 400\cdot\frac{1}{80}+b=20\,\therefore\,b=15\)

If we let the quadratic model be $q(x)$, then we may state:

\(\displaystyle q(x)=\begin{cases}20 & 0\le x\le20\\ \frac{1}{80}x^2+15 & 20<x \\ \end{cases}\)

For the logarithmic model, we are told it must pass through the points (25,30) and (60,60).

Hence, using:

\(\displaystyle f(x)=a+b\ln(x)\)

we obtain the system:

\(\displaystyle f(25)=a+b\ln(25)=30\)

\(\displaystyle f(60)=a+b\ln(60)=60\)

Subtracting the first equation from the second, we eliminate $a$ to obtain:

\(\displaystyle b\ln(60)-b\ln(25)=30\)

\(\displaystyle b\ln\left(\frac{60}{25} \right)=b\ln\left(\frac{12}{5} \right)=30\)

\(\displaystyle b=\frac{30}{\ln\left(\frac{12}{5} \right)}\)

Substituting for $b$ into the first equation, we find:

\(\displaystyle a+\frac{30\ln(25)}{\ln\left(\frac{12}{5} \right)}=30\)

\(\displaystyle a=30-\frac{30\ln(25)}{\ln\left(\frac{12}{5} \right)}=30\left(1-\frac{\ln(25)}{\ln\left(\frac{12}{5} \right)} \right)\)

Thus, the logarithmic portion of the function is:

\(\displaystyle f(x)=30\left(1-\frac{\ln(25)}{\ln\left(\frac{12}{5} \right)} \right)+\frac{30}{\ln\left(\frac{12}{5} \right)}\ln(x)=30\left(1+\frac{\ln\left(\frac{x}{25} \right)}{\ln\left(\frac{12}{5} \right)} \right)\)

If we let the logarithmic model be $\ell(x)$, then we may state:

\(\displaystyle \ell(x)=\begin{cases}30 & 0\le x\le25\\ 30\left(1+\frac{\ln\left(\frac{x}{25} \right)}{\ln\left(\frac{12}{5} \right)} \right) & 25<x \\ \end{cases}\)

To obtain a plot of the two models at W|A, enter the command:

piecewise[{{20,0<=x<=20},{x^2/80+15,20<x}}],piecewise[{{30,0<=x<=25},{30(1+ln(x/25)/ln(12/5)),25<x}}] where x=0 to 75

Task 3: Develop equations that model the three current and the two replacement pack options, including the domain for each equation - full working must be shown.

We have already done this, as it was necessary to develop the functions first so we could plot them. As a recap, we have:

Current packages:

Appetizer: \(\displaystyle A(x)=\begin{cases}7 & 0\le x\le10\\ 2x-13 & 10<x \\ \end{cases}\)

Quest: \(\displaystyle Q(x)=\begin{cases}40 & 0\le x\le40\\ x & 40<x \\ \end{cases}\)

Voyager: \(\displaystyle V(x)=\begin{cases}60 & 0\le x\le60\\ \frac{1}{2}x+30 & 60<x \\ \end{cases}\)

Replacement package options:

Quadratic model: \(\displaystyle q(x)=\begin{cases}20 & 0\le x\le20\\ \frac{1}{80}x^2+15 & 20<x \\ \end{cases}\)

Logarithmic model: \(\displaystyle \ell(x)=\begin{cases}30 & 0\le x\le25\\ 30\left(1+\frac{\ln\left(\frac{x}{25} \right)}{\ln\left(\frac{12}{5} \right)} \right) & 25<x \\ \end{cases}\)
 
  • #4
I had to begin a new post as the character limit was exceeded...

Task 4: Find the intersection points of at least 3 different models.

We were required to let the quadratic and logarithmic models both have the same cost as the Voyager package at 60 gigabytes of usage, so these three models all pass through the point (60,60).

Using the graph as a guide for the current models, we may find the intersection points as follows:

Appetizer and Quest:

\(\displaystyle 2x-13=40\)

\(\displaystyle x=\frac{53}{2}\)

Intersection point: \(\displaystyle \left(\frac{53}{2},40 \right)\)

Appetizer and Voyager:

\(\displaystyle 2x-13=60\)

\(\displaystyle x=\frac{73}{2}\)

Intersection point: \(\displaystyle \left(\frac{73}{2},60 \right)\)

Voyager and Quest:

\(\displaystyle x=60\)

Intersection point: \(\displaystyle \left(60,60 \right)\)

Task 5: Model the effects on the cost to the customer of removing the "included gigabytes usage."

We are told that "the customer would pay the monthly access fee plus the cost of each gigabyte used."

So, for the current linear models, the $y$-intercept for each function is the monthly access fee, and the slope is the cost per additional gigabytes used. Hence, we have:

\(\displaystyle A(x)=2x+7\)

\(\displaystyle Q(x)=x+40\)

\(\displaystyle V(x)=\frac{1}{2}x+60\)

To see a plot of these functions, at W|A, enter the command:

y=2x+7,y=x+40,y=(1/2)x+60 where x=0 to 50

For the quadratic and logarithmic replacement models, we may use the fact that at $x=0$, the cost function is equal to the monthly access fee, and at $x=60$, we want $f(60)=V(60)=90$.

Quadratic model:

\(\displaystyle f(x)=ax^2+b\)

\(\displaystyle f(0)=b=20\)

\(\displaystyle f(60)=a(60)^2+b=3600a+20=90\,\therefore\,a=\frac{7}{360}\)

Thus, we find:

\(\displaystyle q(x)=\frac{7}{360}x^2+20\)

Logarithmic model:

Because $\ln(0)$ is undefined, let's use:

\(\displaystyle f(x)=a+b\ln(x+1)\)

\(\displaystyle f(0)=a=30\)

\(\displaystyle f(60)=a+b\ln(61)=30+b\ln(61)=90\,\therefore\,b= \frac{60}{\ln(61)}\)

And so we have:

\(\displaystyle \ell(x)=30+\frac{60\ln(x+1)}{\ln(61)}= 30\left(1+\frac{2\ln(x+1)}{\ln(61)} \right)\)

To see a plot of these functions, at W|A, enter the command:

y=(7/360)x^2+20,y=30(1+(2ln(x+1))/ln(61)) where x=0 to 75

Now, I will leave the analysis of the results of the models to you, but feel free to post your thoughts for review.
 
  • #5
Thanks mark!

Analysis:

This activity requires you to write a report to the internet provider including:

(i) a recommendation of an appropriate new pricing pack for medium users to replace the quest pack.

(ii) an analysis of the effort on the cost to the customer of removing the "included gigabyte usage' from your recommended replacement of the quest pack option and the two remaining packs, appetizer and voyager.

-----------------------------------------------------------------------------------------------------

(i) For replacement of the quest pack, i would recommend switching it to the included GB logarithmic model. I would choose using the Logarithmic model because its a little over 15 dollars cheaper per every 100 gigabytes used.

(ii) If we were to remove the included gigabytes from the current pack options, the prices would change depending on the customers usage. For the appetizer pack, the change in price would be about 25 dollars more per 100 gigabytes. For the quest pack about 40 dollars more per 100 gigabytes used and for the voyager pack about 30 dollars more per 100 gigabytes used.
If we compare the ' no included gigabyte' quest model to the logarithmic with INCLUDED gigabytes model, there is a little more than 60 dollars in savings per 100 gigabytes used!

---------------------------------------------------------------------------------------------------

***When analyzing the results of your modeling, you will need to:*Compare and contrast the different plans, describing the positive and negative aspects of *each plan for potential customers.
*Recommend which of the replacement pack options should replace the quest pack option, with justification.
*State any assumptions, generalizations and reflections you make during your modeling.
*Discuss the effect on the cost to the customer of removing the "included gigabyte usage."
*Provide clear recommendations to the internet provider based on your findings.----------------------------------------------------------------------------------------------------

If we compare both current pack options, we see that the monthly access fee stays the same, but the price steadily climbs for the " no include " gigabyte packs. If we compare both replacement pack options we have the same result.

Now, if we compare both 'No included gigabyte' packs we see that the quadratic model is more expensive than the appetizer, quest and voyager packs by about 3 dollars, 50 dollars and 90 dollars respectively per 100 gigabytes used and the logarithmic model is cheaper than all three by 100 dollars, 40 dollars and 10 dollars respectively. The positive aspect of this is that the customer could save more if he/she decides to pick a with more gigabytes (most expensive packs) but that's only if they use most of the gigabytes. The customer would lose money if he/she doesn't use many gigabytes.

I would recommend the company to use the Logarithmic model in replace of the quest pack, because medium users are just that, medium users. The pack consists of 25 gigabytes for 30 dollars, but an average of 15 dollars less per 100 gigabytes than the quest pack.

The effect customers would have if we remove the "included gigabyte" options are that they wouldn't have any pre available gigabytes, costing them a little more depending on the plan they choose.

I would recommend the internet provider to make two new packs and to keep the quest pack. The quadratic & logarithmic models could be used as packs for people who haven't decided what kind of pack they want. alternatively they could be used for people who just need internet but have no idea how many gigabytes they are going to use in a months time.

At the end of the activity page it says: You should show your calculations to 2 decimal places, where appropriate. eehh i don't know where i would have to change the answers to 2 decimal places..
 
  • #6
Compare and contrast the different plans, describing the positive and negative aspects of each plan for potential customers.

To do this, let's look at the graphs of the current packages and the replacement options, with the included usage:

View attachment 1273

As we can see, for the current packages, the Appetizer package is the best deal for those whose usage ranges from 0 to 26.5 GB, the Quest package is the best deal from 26.5 to 60 GB and then for any usage above 60 GB, the Voyager package is the best deal.

Appetizer package: Low initial cost, but for usage above the included amount, additional usage is the most costly of the the three current packages.

Quest package: Moderate initial cost, and moderate additional usage cost.

Voyager package: High initial cost, but for additional usage, this package is the least costly. This is the best deal for high usage customers.

Recommend which of the replacement pack options should replace the quest pack option, with justification.

As for the replacement options, I think that while the quadratic model is unrealistic for usage above 60 GB, as customers should expect the cost for additional usage to decrease rather than increase, it is however the best deal for the package it is intended to replace, as users who intend to go beyond 60 GB of usage will opt for the Voyager package instead.

If the two replacements were intended to replace the Voyager package, the I would choose the logarithmic model, since its rate of change decreases as the usage increases, and customers would find the quadratic model prohibitive as its rate of change actually increases as usage increases. Customers would find this cost prohibitive.

Discuss the effect on the cost to the customer of removing the "included gigabyte usage."

View attachment 1274

What I notice here is that the range of usage values for which the Quest package is the best deal is made quite a bit smaller. Also for each option, a user will be much less sure what his/her billing amount is going to be for each month as it is completely dependent on the usage, which could vary considerably from month to month. You have already addressed this point.

In addition to this, the logarithmic model becomes the best deal at a much lower usage amount than with the included usage.

If I were going to make a recommendation to the internet provider, I would advise them to not remove the included usage, and to replace the Quest package with the quadratic model, as it is the better deal over the usage range it is intended to serve.

I would say if you are going to recommend two new models, you would have to state what they are and why they would benefit the customer.

Now, as far as showing your calculations to two decimal places, this is most likely intended for when you are computing costs at particular usage levels, especially for the logarithmic replacement model. It is possible that you should also state the replacement models using decimal approximations.
 

Attachments

  • needalgebra01.jpg
    needalgebra01.jpg
    19.6 KB · Views: 55
  • needalgebra02.jpg
    needalgebra02.jpg
    19.2 KB · Views: 55

Related to Quadratic and logarithmic models (new post)

1. What are quadratic and logarithmic models?

Quadratic and logarithmic models are mathematical functions used to represent relationships between two variables. Quadratic models are expressed as y = ax^2 + bx + c, while logarithmic models are expressed as y = a + b ln(x). These models are commonly used in various fields such as physics, economics, and biology.

2. How are quadratic and logarithmic models different?

The main difference between quadratic and logarithmic models is the shape of their graphs. Quadratic models produce a parabolic curve, while logarithmic models produce a curve that increases slowly at first and then becomes steeper. Additionally, quadratic models are used to model relationships where one variable is squared, while logarithmic models are used for relationships where one variable is the logarithm of the other.

3. When should I use a quadratic model?

A quadratic model should be used when you want to represent a relationship between two variables where one variable is squared. This can be seen in relationships such as the distance covered by an object under constant acceleration, or the area of a square with a certain side length.

4. When should I use a logarithmic model?

A logarithmic model should be used when you want to represent a relationship between two variables where one variable is the logarithm of the other. This can be seen in relationships such as population growth, where the number of individuals increases exponentially over time.

5. How do I determine which model to use?

The best way to determine which model to use is to analyze the data and look for patterns. If the data shows a quadratic relationship, a quadratic model would be appropriate. If the data increases exponentially, a logarithmic model would be a better fit. It is also important to consider the context of the data and the variables involved in the relationship.

Similar threads

Replies
37
Views
5K
Replies
3
Views
1K
Replies
1
Views
747
Replies
3
Views
835
  • General Math
Replies
2
Views
798
  • General Math
Replies
6
Views
1K
Replies
9
Views
1K
Replies
0
Views
9K
Replies
38
Views
6K
Back
Top