# Fundamentals of Numeric Computing

## CPAN 112 || Lab 03

**Please read the following instruction very carefully before answering any questions:**

- Please read all the questions very carefully.
- Please provide your answers in the boxes below each question, and do not change the text colour.
- Your answer MUST show the solution procedure.
- There is no credit if you only state the final answer.
- Please underline your final answer to each question.
- Please keep the naming conventions requested in this lab and each question.
- Once you complete your lab, rename your word document file to the (CPAN112_LabXX_FirstName_LastName). Replace XX with the lab number (e.g. 01). Replace FirstName and LastName with your first name and last name, respectively.

*It will be a 10% mark deduction if you do not follow the guidelines mentioned above*

**What rate of interest did you receive over a period of 1yr and 60 days if your principal was $1500 and it has a maturity value of $2500?**- Principal amt = $1500
- Maturity Value = $2500
- Time = 1 yr + 60/365
- = 1.16438 yr
- Interest = 2500-1500 = 1000
- I = P*T*R
- R = I/P*T
- R = 1000 /(1500*1.16438)
- R = 0.57255
- R=57.26%

**What sum of money invested at 11% p.a., compounded quarterly, will grow to $7500.00 in 5 years and 9 months?**- A = 7500
- t = 5 years and 9 months (69/12 = 23/4)
- r = 11%
- A = P*(1 + r/n)n*t
- 7500 = P*(1 + 0.11/4)
- 4*23/4
- 7500 = P*(1.0275)
- 23
- 7500 = P*1.8663
- P = 7500/1.8663
- P = 4018.64
- Therefore the, principal of $4018.64 will grow to 7500 in 5 years and 9 months.

**A debt of $9200 due today is to be settled by two equal payments due three months from now, and 9 months from now respectively. What is the size of the equal payments at 7% compounded quarterly?**- PV = 9200
- r = 7%
- PV = A/(1 + r/n)t
- 9200 = A/(1 + 0.07/4)1 + A/(1 + 0.07/4)3
- 9200 = A/(1.0175) + A/(1.0534)
- 9200 = {(A*1.0534) + (A*1.0175)}/{1.0534 * 1.0175}
- 9200 = 2.0709A / 1.0719
- 9200*1.0719 = 2.0709A
- 9,861.48 = 2.0709A
- A = 9,861.48 / 2.0709
- A = 4,761.93
- Therefore, the size of the equal payments will be $4,761.93.

**John started a registered retirement savings plan on January 1, 2008, with a deposit of $5000. He added $3500 on January 1, 2009, and $7500 on January 1,2010. What is the accumulated value of his RRSP account on July 1, 2010, if interest is 9% compounded quarterly?**- Principal amt = $5000
- rate = 9% (0.09)
- n = 4 (quarterly payments)
- t = 1 year
- A = P*(1 + r/n)n*t
- A = 5000*(1 + 0.09/4)4*1
- A = 5000*(1.0225)4
- A = 5000*1.0931
- A = 5465.5
- After a year a deposit of $3500 was made
- So, Principal = $3500 + $5465.5 = $8965.5
- Accumulated value for January 1st, 2009:
- A = P*(1 + r/n) 4*1
- A = 8965.5*(1 + 0.09/4)1*4
- A = 9800.04
- Another deposit of $7500 was made 1 year later
- So, Principal = $7500 + $9800.04 = $17,300.04
- Accumulated value for July 1st, 2010:
- t = 0.5 (Since the given time is of 6 months)
- A = P*(1 + r/n)n*t
- A = 17,300.04*(1 + 0.09/4) 4*0.5
- A = 17,300.04*(1.0225)2
- A = 17,300.04*1.0455

A = 18,087.3 - Therefore, accumulated value of his RRSP account on July 1, 2010 is $18,087.3

**In ten months, you want to buy a car. You can invest $24 000 at 3.3% now. The car you want to purchase has a price of $22 225 plus $750 for freight and $1200 for air conditioning. GST of 6% is charged on all items.****How much money will you have and is it enough?****At the current rate of simple interest, how much longer will you have to wait?****While you are saving for the car a new model comes out and it has 1.2% price increase. The freight and the air conditioning do not have a price increase. You will be able to afford this car if he invests money with a private lender. What rate of interest the was offered by the private lender, Use the length of time from part (b).?**

- Total car price = (22,225 + 750 + 1200)* (1+0.06)

Total car price = 24,175 * 1.06

Total car price = 25,625.5

a) Investment Price = $24000

r = 3.3%

t = 10 months (10/12)

A = P*(1 + r*t)

A = 24,000*(1 + 0.033*10/12)

A = 24,000*1.0275

A = 24,660

Total money required to pay the car is $25,625.5 but we will only have

$24,660 by 10 months.

So, there will not be enough money to pay the car.

- Total car price = (22,225 + 750 + 1200)* (1+0.06)

- A = P*(1 +r*t)

25,625.5 = 24,000*(1 + 0.033 * t/12)

25,625.5/24,000 = 1 + 0.033t/12

1.0677 = 1 + 0.033t/12

1.0677 – 1 = 0.033t/12

0.0677*12 = 0.033t

t = 0.0677*12 / 0.033

t = 24.6182 months

So, they will have to wait another 24.62 – 10 = 14.62 months.

- A = P*(1 +r*t)

- New interest rate for the car =1.2%

t = 14.63 months

Car price = ((22,225*(1 + 0.012) + 750 + 1200) * (1 + 0.06))

Car price = $25,908.20

Private lender interest rate.

25,908.2 = 24,000*(1 + r/100 * 24.62/12)

25,908.2/24,000 = 1 + 2.0517r/100

1.0795 =1 + 2.0517r/100

0.0795 = 2.0517r/100

7.95 = 2.0517r

r = 3.88%

Therefore, the required interest rate is 3.88%.

- New interest rate for the car =1.2%

**A debt payment of $14000.00 due today, $5100.00 due in twenty one months, and $19000.00 due in 4.5 years are to be combined into a single payment due three years from now. What is the size of the single payment if interest is 6.50% p.a. compounded quarterly?**- r = 6.5%
- t = 36 months
- PV = FV/(1 + r/n)n*t
- 1st due: 14,000
- 2nd due:
- PV = 5100/(1 + 0.065/4)21/3
- PV = 5000/1.1194
- PV = $4555.8169

- 3rd due:
- FV = 19000/(1 + 0.065/4)54/3
- FV = 19000/(1.3366)
- FV = 14214.8914

- Single payment:
- 14000 + 4555.8169 + 14214.8914= x/(1 + 0.065/4)36/3
- 32770.7083= x/1.2134
- x = 39764.23

- Therefore, the size of the single payment would be $39,764.23