I m having difficulties solving this problem with "Linear" Model. can you guys help?

The Aberdeen Development Corporation (ADC) is trying to complete its investment plans for the next five years.

Currently ADC has $6.1 million available for investment. At 1 year intervals, ADC expects the following

income streams from other business ventures over the next 4 years: $2.8 million (one year from now), $2.4

million (two years from now), $2.3 million (three years from now), and $2.7 million (four years from now).

This income stream is also available for investment.

There are three development projects that ADC is considering. The first is the Aberdeen Resort

Hotel, a four-star deluxe hotel on the picturesque banks of Grays Harbor. If ADC participates fully in this

project, it would have the following projected cash flow streams at one year intervals over the next four years

(negative numbers represent investments, positive numbers represent income): –$6 million (now); –$2.1

million (one year from now); $1.5 million (two years from now); $1.9 million (three years from now); and $2.4

million (four years from now). The estimated value of the project five years from now would be $8 million.

A second project, the Aberdeen South Shore Mall, would be a shopping complex adjacent to the

Aberdeen Resort Hotel. The cash flow stream for this project, if ADC participates fully, would be (at 1 year

intervals): –$5 million (now); –$3.1 million (one year from now); $2.1 million (two years from now); $2.4

million (three years from now); $2.6 million (four years from now). The estimated value of the project five

years from now would be $7 million.

The third project, the Bank of Aberdeen building, is a 30-story office building to be built in

downtown Aberdeen starting in two years, after the demolition of several smaller buildings. This project

would have the following cash flow stream (at 1 year intervals) if ADC participates fully: $0 (now); –$0.5

million (one year from now); –$3.5 million (two years from now); –$2.1 million (three years from now); and

–$1.3 million (four years from now). The estimated value of the project five years from now would be $10

million.

a. Assume that ADC may participate either fully, fractionally (with a partner), or not at all in any or all of the

three projects. If ADC participates in a project at less than 100%, all the cash flows, and the final value of

that project for ADC are reduced proportionally. For example, if ADC participates at 50% in the Aberdeen

Resort Hotel, the cash flows would be –$3 million (now); –$1.05 million (one year from now); $0.75 million

(two years from now); $0.95 million (three years from now); and $1.2 million (four years from now). The

estimated value to ADC five years from now would be $4 million. ADC can borrow money at 6% interest per

year. At most $5 million can be borrowed in any one year. The loan must be paid back the following year with

interest (e.g., if $1 million is borrowed, $1.06 million is due the following year). However, a new loan can be

taken out the following year. ADC can invest surplus funds, and earn a 4% return per year (e.g., if $1 million

in surplus funds are invested in year 3, then $1.04 million are available in year 4). The goal is to maximize

their net worth in five years (the value of their developments plus any surplus funds and interest minus any

borrowed principal and interest due). You may assume that all income from other business ventures, loan

proceeds, savings interest, or current projects are received at the start of each year, followed immediately by

any required cash outflows. Set up and solve a linear programming spreadsheet model for this problem.

b. Now suppose that there are no other potential investors in the Aberdeen Resort Hotel project. Thus, ADC

must participate either fully or not at all in this particular project. The other two projects have other

developers that are interested, so these projects can be undertaken by ADC either in full, fractionally, or not

at all. Suppose further that ADC would only consider investing in the Mall (at any level) if they are also

developing the Aberdeen Resort Hotel. Finally, suppose that ADC does not want to participate in all three

projects. They would consider pursuing at most two of them (at any
level). revise the model to incorporate these changes. The resulting
model must be linear.

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Posted on Jan 02, 2017

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SOURCE: solving word problems involving linear equation

A positive number is 5 times another number.If 21 is added to both the numbers,then one of the new numbers becomes twice the other number.What are the numbers

Posted on Jun 30, 2008

SOURCE: online project on java

please send the project for online shopping mall. Send Immediately. Language is JAVA

Posted on Sep 18, 2008

SOURCE: Office 2007 wont upgrade from project 2007

Sorry to tell you this but Project 2007 isn't listed as an option to upgrade to the Office Professional. If you had Project through an Office Suite previously it would have been. Go to the URL pasted below and it will show products elegible for the upgrade. Good luck.

http://office.microsoft.com/en-us/products/FX101754511033.aspx

Posted on May 08, 2009

SOURCE: Calculate Quarters between any two dates

not sure if this is what you are after entirely but should at least help you out some what

=DATEDIF(A3,C3,"m")/3

As you said you cant just divide by 3 but I have divided the formula by 3 to see how many 3 months will go into the output you could maybe have another formula to divide it by 2 to see if it will divide by 2 to see if you get a whole number or not, if you do then you can just return the first part before the dot ( using split function ) and if the value is less then 1 then obviously the 3 months has not passed between the 2 dates

Posted on May 09, 2009

No, manufacturers do not distribute parts of keypads. The keypad will have to be replaced entirely.

Mar 28, 2014 | Linear Corporation ACP00878 Linear MDKP...

In equation mode, you have system of linear equations (3 unknown) you have polynomial (quadratic and cubic), and solver. Use the solver foe any type of equation (nonlinear, polynomial of order higher than 4, trigonometric, exponential, logs).

May 21, 2012 | Casio Scientific Calculator Fx-570 Fx570...

Thhe Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Mar 17, 2012 | Casio FX-9860G Graphic Calculator

The Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Mar 17, 2012 | Casio Office Equipment & Supplies

If you call one n (as you did), the other is 2n-3000. The sum of the two is 91500. Thus

2n+n-3000=91500. Combine the terms with n to get 3n, Add 3000 to both sides (members) of the equation. Finally

3n=91500+3000=94500 . Divide both members by 3 to get

**n=94500/3=31500**

The other is 2n-3000=2(31500)-3000=60000.

Some calculators have a solver that you can use to solve linear and nonlinear equations, However, since you no not disclose the brand or model number, nobody will be able to tell you how to set it up. Sorry.

2n+n-3000=91500. Combine the terms with n to get 3n, Add 3000 to both sides (members) of the equation. Finally

3n=91500+3000=94500 . Divide both members by 3 to get

The other is 2n-3000=2(31500)-3000=60000.

Some calculators have a solver that you can use to solve linear and nonlinear equations, However, since you no not disclose the brand or model number, nobody will be able to tell you how to set it up. Sorry.

Jan 12, 2012 | Office Equipment & Supplies

Calculators come in multiple shapes and makes. It would certainly be helpful to know a calculator's manufacturer and Model number.

The Simul button allows you to use a solver of simultaneous linear equations.

You must have enter one of the simutaneous equation solver.

Enter the value of a_1, press =. For each symbol with a ? enter a number and press =

It will give you the solutions.

You must get out of this mode to do other types of calculations. Read your user's guide to know how.

The Simul button allows you to use a solver of simultaneous linear equations.

You must have enter one of the simutaneous equation solver.

Enter the value of a_1, press =. For each symbol with a ? enter a number and press =

It will give you the solutions.

You must get out of this mode to do other types of calculations. Read your user's guide to know how.

Sep 12, 2011 | Office Equipment & Supplies

This calculator doesn't really support a solver like that. You can use Woflram Alpha to solve equations like that really quickly. Example

There are more expensive calculators that will solve these types of equations, but I would recommend just using Wolfram Alpha whenever you need a solution to a math problem.

There are more expensive calculators that will solve these types of equations, but I would recommend just using Wolfram Alpha whenever you need a solution to a math problem.

Mar 02, 2011 | Casio FX-115ES Scientific Calculator

There are a few routines to solve simultaneous LINEAR equations that are embedded in the Operating system of the calculator.

If you need download and install instructions, you will find them here.

- solve(
- Simult(

If you need download and install instructions, you will find them here.

Aug 01, 2010 | Texas Instruments TI-89 Calculator

You must have pressed the MODE key followed by SOLV , QUAD or SIMUL-2 or SIMUL-3 . If the calculator has these a programs they solve a Quadratic equation, a simultaneous system of two linear equations (SIMUL-2) . a system of 3 linear equations (SIMUL-3).

To use these solvers (IF YOU ARE INTERESTED), you select the solver you want and you will be prompted for a first coefficient (a, or a1). You enter the value and press the DATA key and you will be prompted for the next coefficient.

To exit the Solver mode press the MODE key followed by the key next to which there is DEC or >DEC. It might also be marked COMP.

Sorry, I do not have that very calculator but an other AUREX Scientific calculator (DS-752C).

To use these solvers (IF YOU ARE INTERESTED), you select the solver you want and you will be prompted for a first coefficient (a, or a1). You enter the value and press the DATA key and you will be prompted for the next coefficient.

To exit the Solver mode press the MODE key followed by the key next to which there is DEC or >DEC. It might also be marked COMP.

Sorry, I do not have that very calculator but an other AUREX Scientific calculator (DS-752C).

May 31, 2010 | Casio fx-300ES Calculator

Hi,

You have two ways to solve an equation (not just linear) in the TI83Plus

**The solve( command**

**The interactive solver**

Press [MATH][0:Solver]. You see what is on the first picture (I entered K-0.5*V^2). Press [ENTER]. You see what is on the right picture.

In the interactive solver you enter a value for all the parameters, constants,**for all, save the entity you are solving for, here V.** So **you must clear the 0 to the right of the = sign after the V**. when finished you press [ALPHA][ENTER] (SOLVE).

Here is what I get for the small exemple I chose.

Left -rt=0 shows that the root was found to within the calculator's tolerance.

Hope it helps.

Thank you for using FixYa.

You have two ways to solve an equation (not just linear) in the TI83Plus

- The solve( command the syntax of which is solve( expr, var,guess, {lower_limit, upper_limit}). Here expr is the expression (without =0), var is the variable you solve for, guess is you estimate of the solution, lower_limit and upper_limit are the bounds of the domain where you are looking for a solution. Of course your guess must be between theses limits.
- The Interactive Solver.

- The solve command is accessed through the catalog of commands [2nd][0] (CATALOG).
- Press [LN] (S) to let the cursor jump to the first command that starts with s.
- Scroll down to reach the solve( command.
- Select it and press [ENTER]. The command echoes on main screen.
- Enter the different arguments: expr, var, guess, etc.
- Close the right parenthesis and press [ENTER].
- Wait for the solution to be displayed.

Press [MATH][0:Solver]. You see what is on the first picture (I entered K-0.5*V^2). Press [ENTER]. You see what is on the right picture.

In the interactive solver you enter a value for all the parameters, constants,

Here is what I get for the small exemple I chose.

Left -rt=0 shows that the root was found to within the calculator's tolerance.

Hope it helps.

Thank you for using FixYa.

Nov 19, 2009 | Texas Instruments TI-30XA Calculator

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