Question

This month, June, your parents wish to save $10,000 for your wedding that will be on 16th December next year. They start deposit by the end of June and earn j₁₂ = 6% p.a.
a) What is the regular deposit required?

b) Immediately after your five deposits interest rates fall to j₁₂ = 5%. What new deposit size is required to meet your original target?

Answer 1

a) Regular deposit required:

p=regular deposit, n=This year end of june to next year end of november = 18 months, Maturity Value (M)=$10,000, r=6%pa = 0.5%per month

M=(p*n)+{p*[n*(n+1)/2]*[r/100]}

10000 = (p*18)+p*[18*(18-1)/2]*[0.5/100]

10000 = 18p+p*0.005*[18*17/2]

10000 = 18p+0.005p*9*17

10000 = 18p+0.765p

10000 = 18.765p

p = 10000/18.765

p = $532.91

Note: Assumed it will be withdrawn at the end of november month & no interest for december month.

b) New deposit size required:

Total amount deposited for first five deposits along with interest = (p*n)+{p*[n*(n+1)/2]*[r/100]}

where, p=$532.91, n=5, r=0.5

= (532.91*5)+{532.91*[5*(5-1)/2]*[0.5/100]}

= 2664.55+{532.91*0.005*[5*4/2]}

= 2664.55+26.6455

= 2691.1955 = $2,691.20

Remaining amount required after interest rate change = $10,000-$2,691.20 = $7,308.80

New deposit size required after interest rate changes,

M=(p*n)+{p*[n*(n+1)/2]*[r/100]} where, m=$7,308.80, r=5%pa = 5%/12 per month, n=18-5=13

7308.80 = (p*13)+{p*[13*(13-1)/2]*[5/(12*100)]}

7308.80 = 13p+{p*[13*12/2]*[0.05/12]}

7308.80 = 13p+{p*[13*6*0.05/12]}

7308.80 = 13p+0.325p

7308.80 = 13.325p

p = 7308.80/13.325

p = $548.50