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.

 

• • • 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