# 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 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.

• 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.

• 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%.

• 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