Question1: Consider the following cash-flow:
$1000 in t=0,1,2, and 3
$0 in t=4
$1200 in t=5,6,7 and 8
$1500 in t=9,10 and 11
$0 in t=12 and 13
$1700 in t=14.
Find the present value (t=0). Assume the interest rate is 8%
Answer: P= {[1000(F/A,8%,12)–1000(1.087)+200(F/A,8%,7)+300(F/A,8%,3)](1.083)+1700}/1.0814=9165.77
Question2: Consider the following cash-flow:
$1000 in t=1,2, and 3
$0 in t=4
$1200 in t=5,6,7 and 8
$1500 in t=9,10 and 11
$0 in t=12 and 13
$1700 in t=14.
Find the present value (t=0). Assume the interest rate is 8%
Answer:P= {[1000(F/A,8%,11)–1000(1.087)+200(F/A,8%,7)+300(F/A,8%,3)](1.083)+1700}/1.0814=8165.79
Question3: Consider the following cash-flow:
$1000 in t=0,1,2, and 3
$0 in t=4
$1300 in t=5,6,7 and 8
$1500 in t=9,10 and 11
$0 in t=12 and 13
$1700 in t=14.
Find the present value (t=0). Assume the interest rate is 9%
Answer:P= {[1000(F/A,9%,12)–1000(1.097)+300(F/A,9%,7)+200(F/A,9%,3)](1.093)+1700}/1.0914=8929.19
Question 4: You deposit $6000 in a savings account. It will take X years to triple your balance. The
interest rate is 8%. Find X.
Answer: Finding equivalence at VN: 18 = 6 (1.08)N and solve for N: N=14.27
Question 5: You deposit $6000 in a savings account. It will take 11 years to triple your balance. The
interest rate is X%. Find X.
Answer: 18=6(1+i)11 ; i = .1050
Question 6: On the day his baby was born, a father decided to establish a savings account for the child’s
university education. The father will make a series of annual deposits on each of his child’s birthdays,
the deposits growing at 6% annually, from the 1st (at n=1) through to the 18th (n=18) birthday. This will
need to be sufficient so that the child can withdraw $3000 on his 18th birthday, $3300 on his 19th
birthday, $3600 on his 20th birthday, and so on (growing at $300 per year) until (and including) his 30th
birthday.
Note: this is 13 withdrawals overall (including the first one on the 18th birthday).
What is the value of the first deposit (when the baby turns one)? The interest rate is 8%.
Answer: Finding equivalence at V18:
X(P/A1,g=6%,i=8%,18)(1.08)18 = 3k(P/A,8%,13)(1.08)+300(P/G,8%,13)(1.08)
X=670.23
Question: 7: What is the equal payment series with the first payment in t=0 and the final payment in
t=10, with a skipped payment at t=6, that is equivalent to a an increasing payment series where the first
payment is $20,000 at t=2and it is increasing by 8% each year with the final payment at t=10 (none
skipped here)? Interest rate is 8%.
Answer: Equivalence at t=10: X[(F/A,8%,11)-1.084]= 20k(P/A,8%,8%,9)(1.089); X=21,796.92
Question 8: You are about to borrow $8000 from a bank at an interest rate of 8% compounded annually.
You are required to make ten annual repayments but the first year is a grace period, so you don’t make
the first payment until t=2. Note: there are 10 re-payments in all. The interest rate is 8%.
How much will each payment be?
Answer: 8k(1.08)(A/P,8%,10)=1287.36
Question 9: On the day his baby was born, a father decided to establish a savings account for the child’s
university education. Any money that is put into the account will earn an interest rate of 8%. The father
will make a series of annual deposits in equal amounts on each of his child’s birthdays from the day of
birth up to and including the 18th birthday so that the child can make five annual withdrawals from the
account in the amount of $50,000 on each birthday from the 18th up to and including the 22nd
birthday. How much should the annual deposits be?
Answer: Equivalence at t=18: X(F/A,8%,19) = 50k + 50k(P/A,8%,4) ; X=5202.03
Question 10: You are about to borrow $10,000 from a bank at an interest rate of 8%.. You are required
to make 15 equal payments of $____ per year. How much of the 11th payment (t=11) will go
towards principal repayment?
Answer: A= 10k(A/P,8%,15) =10k*.1168=1168; B10=1168(P/A,8%,5)=1168*3.9927=4663.47
PP11=1168 – 4663.47*.08=794.92
Question 11: Find the value of X to make the two cash flows economically equivalent. If a cash-flow is
not represented below then it means the cash-flow is zero. The interest rate is 8%
Cash flow #1
n=0: 100
n=2: 100
n=3: 100
Cash flow #2
n=0: –X
n=1,2 and 3: 2X
Answer: Finding equivalence at t=0: 100+100(1.08)-2 + 100(1.08)-3=X(-1+2(P/A,8%,3); X=63.82
Question 12: Find equivalence between cash flow #1 and cash flow #2 by solving for the unknown value
X, but assume that interest rates are rising: so there is an interest rate of 4% between n=0 and n=1, 6%
between n=1 and n=2 and 8% between n=2 and n=3, and 10% between n=3 and n=4. Assume a missing
time-period means there is no cash-flow at that time. Note: as usual this is easier if you draw the cashflow and then solve.
Cash flow #1
n=0: –100
n=1: 120
n=2: 120
n=3: 120
Cash flow #2
n=0: –100
n=1,2 and 4: X
Answer: Finding equivalence at t=2. 120(1.06) + 120+120/1.08 = X(1.06 + 1 + 1/(1.08*1.10); X=123.48
Question 13: Consider the following two cash flows:
100 at t=0, 200 at t=2, and 300 at t=3
X at t=1, X at t=2, and 2X at t=4
The interest rate rises by 3% every year, so:
between t=0 and t=1: i=0%; between t=1 and t=2: i=3%; between t=2 and t=3: i=6%, etc.
What the value of X that makes the two cash flows equivalent?
Answer: Finding equivalence at t=2: 100(1.03) + 200 + 300/1.06 = X(1.03+1+2/(1.06*1.09)); X=155.81
Question 14: You borrowed $20,000 for buying a new car from a bank at an interest rate of 12%
compounded quarterly. This loan will be repaid in 36 equal monthly instalments over 3 years.
Find the monthly instalments. Immediately after the 20th payment, you desire to pay the
remainder of the loan in a single payment (there are no penalties). Compute this lump sum
amount at this time.
Answer: A=20k(A/P,iM=(1+.12/4)4/12-1 =.009902,36)=20k*.0332=664
B20=664(P/A,.009902,16)=664*14.7297 = 9780.52
Question 15: Rob wants to buy a $700,000 loft. He intends to put a $250,000 down-payment and take a
30-year amortized mortgage on the rest. He will choose a 5-year fixed term with monthly payments. The
current rate is 3% APR quarterly. What will he still owe when the 5-year term is over?
Answer: A=450k(A/P,iM=(1+.03/4)4/12-1=.002494, 12*30=360) = 450k*.004212=1895.40
B5=1895.50( P/A,iM,360-60=300) = 1895.4*211.0431 = 400,011.15
Question 16: You are depositing $1000 a week into a fund that pays 9% compounded monthly. How
much is in the fund at the end of 3 years?
Answer: F=1000(F/A,iw =(1+.09/12)12/52-1=.001726,3*52=156)=178,845
Question 17: Suppose the interest rate on your credit card is 16% compounded quarterly and you
currently owe $2000 on it. If you don’t make any payments for 3 years, how much will you owe in 3
years?
Answer: F=2k(1.04)3*4=12=3202.06
Question 18: Rob buys a 30-year coupon bond with a $250,000 face value at a time when the interest
(coupon) rate is 3%. In just under 3 years (so just before the 3rd coupon payment) he sells the bond to a
neighbour. At that time the interest rate is 5%. How much is the bond’s approximate (within $50)
competitive value at the time of sale?
Answer: P = 7500 + 7500(P/A,5%,27) + 250k/1.0527 = 184,284.58
Question 19: Rob buys a 30-year coupon bond with a $250,000 face value at a time when the interest
(coupon) rate is 5%. In just under 5 years (so just before the 3rd coupon payment) he sells the bond to a
neighbour. At that time the interest rate is 3%. How much is the bond’s approximate (within $50)
competitive value at the time of sale?
Answer: P = 12500 + 12500(P/A,3%,25) + 250k/1.0325 = 349,565.14
Question 20: Rob wishes to sell a bond that has a face value of $1000. The coupon rate on the bond is
4%. Rob bought the bond (from someone else) exactly eight years ago for $920 and wishes to earn a 5%
return (yield) on his investment. At what price would he have to now sell the bond so as to have earned
the 5% return (yield)? (Rob bought the bond immediately after a coupon payment and also sells the
bond immediately after receiving his 8th coupon payment, so he’s held the bond for 8 years).
Answer: Finding equivalence at t=8: 920(1.058)=40(F/A,5%,8) + P; P=977.30
Question 21:
Consider the following debt that is paid off with unequal payments over 4 years. There is a constant rate
of interest and all cash-flows are positive. Find the payment in period 2.
n
An
Bn
0
-1000
1
-1100
2
-1000
3
550
4
Answer: B2(1+i)+A3=B3 so -1000(1+i)+550=-600 therefore i=0.15
B1(1+i)+A2=B2 so -1100(1.15)+A2=-1000 therefore A2 =265
-600
0
Question 22: Consider the following debt that is paid off with unequal payments over 4 years. There is a
constant rate of interest and all cash-flows are positive. Find the payment in period 4.
n
An
Bn
0
-1000
1
-1100
2
-1000
3
520
4
Answer: B2(1+i)+A3=B3 so -1000(1+i)+520=-600 therefore i=0.12
B3(1+i)+A4=B4 so -600(1.12)+A4=0 therefore A4 =672
-600
0