2. Next, in a MS Word document, you are to answer the following questions below,

Question

2. Next, in a MS Word document, you are to answer the following questions below, using the online HP10BII financial calculator demo website (weblink also provided below). Be sure to respond to each financial calculation question by using the website to walk you through each calculation. HINT - you will need to know what type of calculation you are conducting in order to get to the correct demonstration and answer! In the end, label each problem with the type of transaction used to solve it online, document your response for each questio

HP10BII calculator website:

http://h22150.www2.hp.com/7B9288F8-563C-4B97-9694-9228808399BF/Model.html

Financial problems to solve (from the website demonstration):

1. If you deposit $100 today, how much is it worth in 20 years, if interest is compounded at a 4% annual rate?

2. If you deposit $50 a month starting one month from today, how much do you have in the account in 10 years, if the account begins with $1,000 in it and earns 5% compounded monthly?

3. Calculate the first year of the annual amortization schedule for a 30 year, $180,000 mortgage at 7.75%, compounded monthly.

4. A company is considering replacing a machine. It will require an initial cash outlay of $20,000 and then is expected to generate cash flows the next 3 years of $10,000, $15,000, and $20,000. If the cost of funds for the company is estimated at 10% what is the Net Present Value? Should the machine be replaced?

5. Jack will make annual deposits of $1,500 into an account paying 4.75%, compounded monthly. How much is in his account in 10 years, assuming he makes the first deposit one year from now?

Explanation / Answer

1. A= p (1+r)^n

       = 100 (1+.04)^20= 100* 2.191http://www.chegg.com/homework-help/expertquestion = $ 219.1

2. Future value= 1000(1+.05/12)10*12 + 50{ (1+.05/12)^10- 1)/.05/12}

                        = 1000 *1.647 + 50 {1.042-1)/ .05/12 = 1647+ 504 = $ 2,151

3. monthly payment, A= {i * p* (1+i)^n}/ (1+i)^n- 1                          i= .775/12= 0.065

     = .065* 180,000 {(1+0.065)^12*30/ (1+.065)^360 - 1}

     = $ 1,289.54

first year amortization schedule

Interest= 180,000 * (0.775/12)%= $ 116.28                     

balance= 180,000- 1173.36= $ 178,826.74

4. computation of present value

Since NPV is positive and favorable therefore the machine should be replaced.

5. A= p {(1+ i/12 )^ 10 - 1 }/ i/12

      = 1500 { ( 1+ 0.475/12)^10 - 1) /0.475/12

     = $ 17,974.40

payment amount principal interest balance 1 1289.54 1173.26 116.28 178,826.74 2 1289.54 1174.02 115.52 177,652.72 3 1289.54 1174.78 114.76 176,477.94 4 1289.54 1175.54 114 175,302.4 5 1289.54 1176.29 113.24 174,126.11 6 1289.54 1177.06 112.48 172,949.05 7 1289.54 1177.82 111.72 171,771.23 8 1289.54 1178.58 110.96 170,592.65 9 1289.54 1179.34 110.20 169,413.31 10 1289.54 1180.10 109.44 168,233.21

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