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Sta 2191 ;Financial Mathematics Question Paper
Sta 2191 ;Financial Mathematics
Course:Financial Engineering
Institution: Jomo Kenyatta University Of Agriculture And Technology question papers
Exam Year:1
STA 2191 Jane Akinyi Aduda
STA 2191: FINANCIAL MATHEMATICS I Problem set
1. Calculate the time in days it takes for 3600 to accumulate to 4000 at:
(a) A simple rate of interest of 6% p.a. [2 marks]
(b) A compound rate of interest of 6% p.a. convertible quarterly. [2 marks]
(c) A compound rate of interest of 6% p.a. convertible monthly. [1 mark]
2. At a certain simple rate of interest, 1,000 will accumulate to 1,110 after a certain period of time. Find
the accumulated value of 500 at a simple rate of interest three fourths as great over twice as long a
period of time. [2 marks]
3. Express d(4) as a function of i(3) [3 marks]
4. You deposit 1,000 today and another 2,000 in ve years into a fund that pays simple discount at 5%
per year. Your friend makes the same deposits into another fund but at times n and 2n, respectively.
the fund credits interest at an annual e ective rate of interest of 10%. At the end of 10 years, the
accumulated value of your deposits is exactly the same as the accumulated value of your friends
deposits. Calculate n [3 marks]
5. The force of interest (t) at time t is at+bt2 where a and b are constants. An amount of 1000 invested
at time t = 0 accumulates to 1500 at time t = 5 and 2300 at time t = 10. Determine the value of a
and b. [5 marks]
6. The force of interest,(t); is a function of time and at any time t, measured in years, is given by the
formula:
(t) =
8<
:
0:06 0 t 5
0:01(t2 ?? t) 5 < t
(a) Calculate the present value of a unit sum of money due at time t = 10. [4 marks]
(b) Calculate the e ective rate of interest over the period t = 9 to t = 10. [3 marks]
(c) In terms of t, determine an expression for v(t), the present value of a unit sum of money due
during the period 0 < t 5. [1 mark]
(d) Calculate the present value of a payment stream paid continuously for the period 0 < t 5,
where the rate of payment, (t), at time t, is e0:04t. [4 marks]
7. You are given that the nominal rate of discount per annum convertible every 2 months is 15%. Calculate
the product of the equivalent nominal rate of interest per annum convertible every three months i(4)
and the force of interest . [6 marks]
8. A fund is earning 7% simple interest. Find the year when this will be equivalent to an e ective rate
of 4:7%. [3 marks]
9. June borrows KSh.50,000 for 10 years at an annual e ective interest rate of 10%. She can repay this
loan using the amortization method with payments of KSh.8,137.25 at the end of each year. Instead,
June repays the KSh.50,000 using a sinking fund that pays an annual e ective interest rate of 14%.
1
STA 2191 Jane Akinyi Aduda
The deposits to the sinking fund are equal to KSh.8,137.25 minus the interest on the loan and are
made at the end of each year for 10 years. Determine the balance in the sinking fund immediately
after repayment of the loan. [5 marks]
10. 1000 is deposited into Fund X, which earns an annual e ective rate of 6%. At the end of each year,
the interest earned plus an additional 100 is withdrawn from the fund. At the end of the tenth year,
the fund is depleted. The annual withdrawals of interest and principal are deposited into Fund Y ,
which earns an annual e ective rate of 9%. Determine the accumulated value of Fund Y at the end of
year 10. [6 marks]
11. Olga buys a 5-year increasing annuity for X. Olga will receive 2 at the end of the rst month, 4 at
the end of the second month, and for each month thereafter the payment increases by 2. The nominal
interest rate is 9% convertible quarterly. Calculate X. [6 marks]
12. Sally borrows X for four years at an annual e ective interest rate of 8%, to be repaid with equal
payments at the end of each year. The outstanding loan balance at the end of the third year is 559.12.
Calculate the principal repaid in the rst payment. [4 marks]
13. An annuity is paid half-yearly in arrears at a rate of KSh.18,000 per annum, for 20 years. The rate
of interest is 6% per annum e ective in the rst 13 years and 7% per annum convertible quarterly for
the remaining 7 years. Calculate the accumulation of the annuity at the end of 20 years. [4 marks]
14. A payment of KSh.24,000 is due in 8 years time in return for KSh.4,000 now, KSh.8,000 in 5 years
and KSh.X in 10 years. If i = 8%, nd KSh.X, such that the value of both options is equal. [4 marks]
15. The force of interest (t) at time t is at+bt2 where a and b are constants. An amount of 1000 invested
at time t = 0 accumulates to 1500 at time t = 5 and 2300 at time t = 10. Determine the value of a
and b. [6 marks]
16. The force of interest (t) is a function of time and at any time t (measured in years) is given by the
formula
(t) =
8>>><
>>>:
0:03 + 0:001t2 0 t < 7
0:01t 7 t < 10
0:1 10 t
(a) An investment of 1 is made at time t = 4. Find the value to which it will have accumulated by
time t = 6:5.
(b) Find the present value (at time t = 0) of an investment of 15 due at time t = 20.
(c) Find the constant force of interest which would lead to the same present value in (ii) being
obtained.
(d) What is the e ective rate of interest from time t = 8 to time t = 9. [14 marks]
17. An annuity is paid half-yearly in arrears at a rate of KSh.18,000 per annum, for 20 years. The rate
of interest is 6% per annum e ective in the rst 13 years and 7% per annum convertible quarterly for
the remaining 7 years. Calculate the accumulation of the annuity at the end of 20 years. [4 marks]
2
STA 2191 Jane Akinyi Aduda
18. A payment of KSh.24,000 is due in 8 years time in return for KSh.4,000 now, KSh.8,000 in 5 years
and KSh.X in 10 years. If i = 8%, nd KSh.X, such that the value of both options is equal.[4 marks]
19. David can receive one of the following two payment streams:
(a) 150 at time 0, 300 at time n, and 450 at time 2n
(b) 900 at time 10
At an annual e ective interest rate of i, the present values of the two streams are equal. Given
vn = 0:76, determine i. [4 marks]
20. Jane takes a home improvement loan of $11; 000 over 5 years. She makes monthly repayments in
arrears and the bank charges an e ective rate of interest of 6% p.a.
(a) what is the monthly repayment. [4 marks]
(b) How much interest does she pay in the 3rd year. [4 marks]
21. Show that
(a) an =
1 ?? vn
i
[2 marks]
(b) an =
1 ?? vn
[3 marks]
(c) (Ia)n =
an ?? nvn
[5 marks]
22. Assuming an interest rate of 12% pa convertible monthly,
(a) Calculate the combined present value of an immediate annuity payable monthly in arrears such
that payments are KSh.100,000 pa for the rst 6 years and KSh.40,000 pa for the next 4 years,
together with a lump sum of KSh.200,000 at the end of the 10 years. [3 marks]
(b) Calculate the amount of the level annuity payable continuously for 10 years having the same
present value as the payments in (i). [3 marks]
(c) Calculate the accumulated values of the rst 7 years'' payments at the end of the 7th year for the
payments in (i) and (ii). [4 marks]
23. The force of interest (t) is a function of time and at any time t (measured in years) is given by the
formula
(t) =
8>>><
>>>:
0:03 + 0:001t2 0 t < 7
0:01t 7 t < 10
0:1 10 t
(a) An investment of 1 is made at time t = 4. Find the value to which it will have accumulated by
time t = 6:5. [5 marks]
(b) Find the present value (at time t = 0) of an investment of 15 due at time t = 20. [4 marks]
(c) Find the constant force of interest which would lead to the same present value in (ii) being
obtained. [3 marks]
(d) What is the e ective rate of interest from time t = 8 to time t = 9. [2 marks]
3
STA 2191 Jane Akinyi Aduda
24. You are given that the nominal rate of discount per annum convertible every 2 months is 15%. Calculate
the nominal rate of interest per annum convertible every three months. [3 marks]
25. A fund is earning 7% simple interest. Calculate the e ective interest rate in the 8th year. [2 marks]
26. At a certain simple rate of interest, 1,000 will accumulate to 1,110 after a certain period of time. Find
the accumulated value of 500 at a simple rate of interest three fourths as great over twice as long a
period of time. [2 marks]
27. De ne the accumulation factor A(t; t + h) and give the formula for the force of interest (t) per unit
time in terms of the accumulation factor. [3 marks]
28. Given a nominal interest rate of 7:5% convertible semi-annually, determine the sum of the force of
interest, ; and nominal discount rate compounded monthly, d(12). [5 marks]
29. Given that an = 7:029584 and a2n = 10:934563, nd the rate of interest i and n. [8 marks]
30. Express d(4) as a function of i(3) [2 marks]
31. David can receive one of the following two payment streams:
(i) 100 at time 0, 200 at time n, and 300 at time 2n
(ii) 600 at time 10
At an annual e ective interest rate of i, the present values of the two streams are equal. Given
vn = 0:76, determine i. [2 marks]
32. Let A(t) represent the value of a fund at time t and in represent the nth year e ective rate of interest.
If A(4) = 2000 and in = 0:01n, nd A(8) [4 marks]
33. Janny and Robbie each open up new bank accounts at time 0. Janny deposits 100 into his bank
account, and Robbie deposits 50 into his. Each account earns the same annual e ective interest rate.
The amount of interest earned in Bruce''s account during the 11th year is equal to X. The amount of
interest earned in Robbie''s account during the 17th year is also equal to X. Calculate X. [5 marks]
34. At time 0, K is deposited into Fund X, which accumulates at a force of interest (t) = 0:006t2 . At
time m, 2K is deposited into Fund Y , which accumulates at an annual e ective interest rate of 10%.
At time n, where n > m, the accumulated value of each fund is 4K. Determine m. [6 marks]
35. Sam, Janice, and Laura each borrow KSh.500,000 for ve years at a nominal interest rate of 12%,
compounded semi-annually. Sam has interest accumulated over the ve years and pays all the interest
and principal in a lump sum at the end of ve years. Janice pays interest at the end of every six-month
period as it accrues and the principal at the end of ve years. Laura repays her loan with 10 level
payments at the end of every six-month period. Calculate the total amount of interest paid on all
three loans. [6 marks]
36. Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of j, compounded
semi-annually. Mike deposits 2X into a di erent savings account at time 0, which pays simple interest
at an annual rate of j. Eric and Mike earn the same amount of interest during the last 6 months of
the 8th year. Calculate j. [3 marks]
4
STA 2191 Jane Akinyi Aduda
37. At time t = 0, 1 is deposited into each of Fund X and Fund Y . Fund X accumulates at a force of
interest (t) = t2
k . Fund Y accumulates at a nominal rate of discount of 8% per annum convertible
semi-annually. At time t = 5, the accumulated value of Fund X equals the accumulated value of Fund
Y . Determine k. [4 marks]
38. The force of interest (t) at time t is at+bt2 where a and b are constants. An amount of 1000 invested
at time t = 0 accumulates to 1500 at time t = 5 and 2300 at time t = 10. Determine the value of a
and b. [5 marks]
39. The force of interest,(t); is a function of time and at any time t, measured in years, is given by the
formula:
(t) =
8<
:
0:06 0 t 5
0:01(t2 ?? t) 5 < t
(i) Calculate the present value of a unit sum of money due at time t = 10.
(ii) Calculate the e ective rate of interest over the period t = 9 to t = 10.
(iii) In terms of t, determine an expression for v(t), the present value of a unit sum of money due
during the period 0 < t 5.
(iv) Calculate the present value of a payment stream paid continuously for the period 0 < t 5,
where the rate of payment, (t), at time t, is e0:04t. [12 marks]
40. Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest
of 4% convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter''s
account is credited interest at a force of interest of . After 7.25 years, the value of each account is
the same. Calculate . [4 marks]
41. An amount X is deposited in an account that grows interest at an annual e ective rate of interest 6%.
Another amount X
2 is deposited in another account that earns interest at an annual e ective rate of
discount d. After 10 years, the total interest earned by both accounts is equal. Find d [5 marks]
42. A business venture is negotiated in January 2010, that has the following cash-
ows.
Date (1stJanuary) Cash
ow (shs)
2011 120,000
2012 150,000
2013 160,000
2014 180,000
2015 255,000
Find the value of this business venture as at 1st of January 2013, if you are given that
v(t) = 0:94 ?? 0:06e(??0:1t) for all t > 0
[5 marks]
43. For a rate of interest of 7% per annum, convertible monthly, calculate the equivalent rate of interest
per annum convertible half yearly, and the equivalent rate of discount per annum convertible monthly
5
STA 2191 Jane Akinyi Aduda
[4 marks]
44. Payments are made at a continuous rate of (8k+tk), where 0 t 10 for 10 years. Interest is credited
at a force of interest (t) = 1
8+t . The present value of the fund is shs20,000. Calculate k. [6 marks]
45. Braun invested a certain sum of money at 8% p.a. simple interest for n years. At the end of n years,
Braun got back 4 times his original investment. What is the value of n ? [4 marks]
46. Shawn invested one half of his savings in an account that paid simple interest for 2 years and received
$550 as interest. He invested the remaining in an account that paid compound interest, interest being
compounded annually, for the same 2 years at the same rate of interest and received $605 as interest.
What was the value of his total savings before investing in these two accounts ? [5 marks]
47. If 3a(2)
n = 2a(2)
2n
= 45s(2)
1
, nd i [5 marks]
48. Project P requires an investment of KSh.400,000 at time 0. The investment pays KSh.200,000 at time
1 and KSh.400,000 at time 2. Project Q requires an investment of X at time 2. The investment pays
KSh.200,000 at time 0 and KSh.400,000 at time 1. The net present values of the two projects are
equal at an interest rate of 10%. Calculate X. [3 marks]
THE END
6
STA 2191 Jane Akinyi Aduda
STA 2191: Financial Mathematics I solutions
1a.
3600
1 +
di
365
= 4000
1 +
0:06d
365
=
4000
3600
0:06d =
4000
3600
?? 1
365 = 40:555
and d = 675:926 days
1b. If i(4) = 0:06 then the quarterly e ective rate is i(4)
4 = 0:06
4 = 0:015. Therefore
3600 (1:015)n = 4000 ) (1:015)n =
4000
3600
n ln (1:015) = ln (1:11111)
and n = 7:0759 quarters ) 645:675 days
1c. If i(12) = 0:06 then the monthly e ective rate is i(12)
12 = 0:06
12 = 0:005. Therefore
3600 (1:005)n = 4000 ) (1:005)n =
4000
3600
n ln (1:005) = ln (1:11111)
and n = 21:1245 months ) 642:5368 days
2.
1000(1 + in) = 1100 ) (1 + in) = 1:11
in = 0:11
we require
3
4
i 2n =
3
2
in )
3
2
in = 0:165
therefore 500(1 + 0:165) = 582:5
3.
1 ??
d(4)
4
!??4
=
1 +
i(3)
3
!3
1 ??
d(4)
4
=
1 +
i(3)
3
!??3
4
d(4)
4
= 1 ??
1 +
i(3)
3
!??3
4
therefore d(4) = 4
2
41 ??
1 +
i(3)
3
!??3
4
3
5
7
STA 2191 Jane Akinyi Aduda
4.
1000(1 ?? 0:05(10))??1 + 2000(1 ?? 0:05(5))??1 = 1000(1:10)10??n + 2000(1:10)10??2n
1000
0:5
+
2000
0:75
= 1000(1:10)10vn + 2000(1:10)10v2n
4666:67 = 2593:742vn + 5187:485v2n
solving this quadratically we obtain
vn =
??2593:742
p
2593:7422 ?? 4(5187:485)(??4666:67)
2(5187:485)
= 0:7306
so (1:10)n = 0:7306??1
and n = 3:3
5.
(t) = at + bt2
A(0; t) = exp
Z t
0
(t)dt
= exp
Z t
0
(at + bt2)dt
= exp
at2
2
+
bt3
3
t
0
= exp
at2
2
+
bt3
3
A(0; 5) = exp
25a
2
+
125b
3
and A(0; 5) = exp
100a
2
+
1000b
3
1500 = 1000 exp(12:5a + 41:667b)
2300 = 1000 exp(50a + 333:333b)
dividing both expressions by 1000 and taking logs we obtain
ln(1:5) = 12:5a + 41:667b
ln(2:3) = 50a + 333:333b
multiply the rst expression by 4 and subtract fron the second one
ln(2:3) ?? 4 ln(1:5) = 166:667b ) b = ??0:0047337
thus a =
ln(2:3) ?? 333:333 ??0:0047337
50
a = 0:0482162
6a.
p:v: = A(10; 0) = A(10; 5)A(5; 0)
A(5; 0) = exp
Z 5
0
??0:06dt = exp
??j0:06tj5
0
= exp(??0:3)
= 0:740818
A(10; 5) = exp
Z 10
5
??(0:01t2 ?? 0:01t)dt = exp
"
??
0:01t3
3
??
0:01t2
2
10
5
#
= exp [(3:333 ?? 0:5) ?? (0:4167 ?? 0:125)] = exp(??2:4387)
= 0:087274
so p:v = 0:064654
8
STA 2191 Jane Akinyi Aduda
6b.
A(9; 10) = exp
Z 10
9
(0:01t2 ?? 0:01t)dt
= exp [(3:333 ?? 0:5) ?? (2:43 ?? 0:405)] = 2:24342
therefore (1 + i) = 2:24342 ) i = 1:24342 = 124:342%
6c.
v(t) = e??
R 5
0 0:06dt = e??0:06t
6d.
Z 5
0
(t)v(t)dt =
Z 5
0
e0:04te??0:06tdt
=
Z 5
0
e??0:02tdt = ??
e??0:02t
0:02
5
0
= ??
e??0:1
0:02
??
1
0:02
= ??(45:2418709 ?? 50) = 4:7581291
7.
1 +
i(4)
4
!4
=
1 ??
d(5)
5
!??5
1 +
i(4)
4
!
=
1 ??
d(5)
5
!??5=4
i(4) = 4
2
4
1 ??
d(5)
5
!??5=4
?? 1
3
5
i(4) = 0:1552324462
= ln
1 ??
d(5)
5
!??5
= ln(1 ?? 0:03)??5
= 0:1522960374
so i(4) = 0:02364128644
8. We require n when the e ective interest rate in the nth year is in which is given by
in =
i
1 + i(n ?? 1)
0:047 =
0:07
1 + 0:07(n ?? 1)
1 + 0:07(n ?? 1) =
0:07
0:047
= 1:489361702
0:07(n ?? 1) = 0:489361702
0:07n = 0:489361705 + 0:07 = 0:559361702
n = 7:99 8 years
9
STA 2191 Jane Akinyi Aduda
9. The payment using the amortization method is KSh.8,137.25.
The periodic interest is 0:1(50000) = 5; 000. Thus, deposits into the sinking fund are 8; 137:25??5000 =
3; 137:25.
Then, the amount in sinking fund at end of 10 years is 3; 137:25s10 @0:14
Amount in sinking fund = 3; 137:25
1:1410 ?? 1
0:14
= 60; 665:92906
Balance in sinking fund = 60; 665:92906 ?? 50; 000
= 10; 665:92906
10. The amount of interest paid into fund X reduces by 6 every year from 60; 54; 48; 42; : : : ; 0 as the capital
is reduced by 100 each year until everything is nished. So the accumulated value in fund Y is obtained by
A(0; 10) = 6(Ds)10 @0:09 + 100s10 @0:09
(Ds)10 =
10(1:09)10 ?? s10
0:09
s10 =
1:0910 ?? 1
0:09
= 15:19292972
A(0; 10) = 6
10 1:0910 ?? 15:19292972
0:09
+ 100 15:19292972
= 6 94:23007807 + 1519:292972
= 565:3804684 + 1519:292972
= 2; 084:67344
11. Let j be the monthly e ective rate of interest, so that
(1 + j)3 =
1 +
i(4)
4
!
where
i(4)
4
is the quarterly e ective rate
= (1:0225)
j = 1:02251=3 ?? 1 = 0:007444
Therefore X = 2(Ia)60 @0:007444
(Ia)60 =
a60 ?? 60(1:007444)??60
0:007444
a60 =
1 ?? 1:007444??60
0:007444
1:007444 = 48:608118
(Ia)60 =
48:608118 ?? 38:45000216
0:007444
= 1; 364:604492
so X = 2; 729:208984
12. Given the term of the loan is 4 years, and the outstanding balance at end of third year = 559.12,
the amount of principal repaid in the 4th payment is 559.12. But given level payments, the principal
repaid forms a geometric progression and thus the principal repaid in the rst year is v3 times the principal
10
STA 2191 Jane Akinyi Aduda
repaid in the fourth year = v3 559:12. Interest on the loan is 8%, thus principal repaid in rst year is
1
(1:08)3 559:12 = 443:85.
Alternatively, let each instalment be X, so
Xv = 559:12 ) X = 603:9144 and capital repaid in year 1 = 1:08??4 603:9144 = 443:8951125
13. The accumulated value is given by
18; 000
s(2)
13
@6% (1:07)7 + s(2)
7
@7%
s(2)
13
@6% = (1:06)13??1
2
h
1:06
12
??1
i = 19:16124412 and s(2)
7
@7% = (1:07)17??1
2
h
1:07
12
??1
i = 8:802905054. Therefore the accumulated
value is
18; 000 (19:16124412 1:605781476 + 8:802905054) = 712; 290:1665
14. The equation of value is given by
24; 000 (1:08)??8 = 4; 000 + 8; 000 (1:08)??5 + X (1:08)??10
12; 966:45323 = 4000 + 5; 444:665576 + 0:463193488X
3; 521:787654 = 0:463193488X
X = 7; 603:275403
15.
(s) = as + bs2
A(0; t) = exp
Z t
0
(s)ds
= exp
Z t
0
(as + bs2)ds
= exp
as2
2
+
bs3
3
t
0
= exp
at2
2
+
bt3
3
A(0; 5) = exp
25a
2
+
125b
3
and A(0; 5) = exp
100a
2
+
1000b
3
1500 = 1000 exp(12:5a + 41:667b)
2300 = 1000 exp(50a + 333:333b)
dividing both expressions by 1000 and taking logs we obtain
ln(1:5) = 12:5a + 41:667b
ln(2:3) = 50a + 333:333b
multiply the rst expression by 4 and subtract from the second one
ln(2:3) ?? 4 ln(1:5) = 166:667b
) b = ??0:0047337
thus a =
ln(2:3) ?? 333:333 ??0:0047337
50
a = 0:0482162
11
STA 2191 Jane Akinyi Aduda
16a. We are interested in A(4; 6:5) = A(0;6:5)
A(0;4) for 0 t < 7
A(0; t) = exp
Z t
0
0:03 + 0:001s2dt
= exp
0:03s +
0:001s3
3
t
0
= exp
0:03t +
0:001t3
3
A(0; 6:5) = exp (0:195 + 0:09154) = exp (0:28654)
A(0; 4) = exp (0:12 + 0:02133) = exp (0:14133)
Therefore A(4; 6:5) = exp(0:14521) = 1:15628
16b. We are interested in p:v: = 15v(t)
v(t) = exp??
Z 7
0
0:03 + 0:001tdt
exp??
Z 10
7
0:01tdt
exp??
Z 20
10
0:1dt
= exp??
0:03t +
0:001t3
3
7
0
exp??
0:005t210
7 exp??[0:1t]20
10
= exp(??0:32433) exp(??0:255) exp(??1)
= exp(??1:57933) = 0:206113
therefore p:v: = 15 0:206113 = 3:091695
= 3:10
16c. For the constant force of interest
e??(t) = 0:206113 where t = 20
therefore ?? 20 = ln 0:206113 = ??1:579330717 and
=
??1:579330717
??20
= 0:078966535
= 7:896%
16d.
A(8; 9) = exp
Z 9
8
0:01tdt = exp
0:005t29
8 = exp(0:085) = 1:0887
so 1 + i = 1:0887 ) i = 0:0887
= 8:87%
17. The accumulated value is given by
18; 000
s(2)
13
@6% (1:07)7 + s(2)
7
@7%
s(2)
13
@6% = (1:06)13??1
2
h
1:06
12
??1
i = 19:16124412 and s(2)
7
@7% = (1:07)17??1
2
h
1:07
12
??1
i = 8:802905054. Therefore the accumulated
value is
18; 000 (19:16124412 1:605781476 + 8:802905054) = 712; 290:1665
12
STA 2191 Jane Akinyi Aduda
18. The equation of value is given by
24; 000 (1:08)??8 = 4; 000 + 8; 000 (1:08)??5 + X (1:08)??10
12; 966:45323 = 4000 + 5; 444:665576 + 0:463193488X
3; 521:787654 = 0:463193488X
X = 7; 603:275403
19.
900(1 + i)??10 = 150 + 300vn + 450v2n
900v10 = 150 + 300vn + 450v2n
= 150 + 300(0:76) + 450(0:762)
= 637:92
v10 = (1 + i)??10 =
637:92
900
= 0:7088
therefore i = 0:035017 = 3:5017%
20a. Let the monthly instalment be X. So
12Xa(12)
5
@6% = 11; 000
12X =
11; 000
a(12)
5
=
11; 000
4:326985135
= 2; 542:185761 and
X = 211:8488135
20b. Loan outstanding after the second year is L2 = 12 211:8488135 a(12)
3
= 12 211:8488135
2:745960213 = 6; 980:740957
Loan outstanding after the third year is L3 = 12211:8488135a(12)
2
= 12211:84881351:883280581 =
4; 787:649078
The capital repaid during the third year is 6; 980:740957 ?? 4; 787:649078 = 2; 193:091879.
Therefore total interest repaid is total instalment minus total capital =(12211:8488135)??2; 193:091879 =
349:093883
21a.
an = v + v2 + : : : + vn
Multiply both sides by (1 + i) and then subtract an
(1 + i)an = 1 + v + v2 + : : : + vn??1
(1 + i)an ?? an = 1 ?? vn
ian = 1 ?? vn
an =
1 ?? vn
i
13
STA 2191 Jane Akinyi Aduda
21b.
an =
Z n
0
vtdt
=
vt
ln v
n
0
=
vn ?? 1
ln v
=
1 ?? vn
since ln v = ??
21c.
(Ia)n =
Z n
0
tvtdt
Using integration by parts where
R
udv = uv ??
R
vdu
let u = t s.t. du = dt and let dv = vtdt s.t. v = vt
ln v
therefore (Ia)n =
t:vt
ln v
n
0
??
Z n
0
vt
ln v
dt
=
nvn
ln v
n
0
??
1
ln v
Z n
0
vtdt
(Ia)n = ??
nvn
+
1
vn ?? 1
??
= ??
nvn
+
an
=
an ?? nvn
22a. Working in monthly time periods, using a rate of interest of 1% per month, the required value is:
P.V. =
1
12
100; 000 a72 +
1
12
40; 000v72a48 + 200; 000v120
=
1
12
100; 000 51:1504 +
1
12
40; 000 0:48850 37:974 + 200; 000 0:30299
= 426; 253:3333 + 61834:33 + 60; 598
= KSh.548; 685:6633
22b. Let the level payment be X so that
548; 685:6633 =
X
12
a120 =
X
12
1 ?? 1:01??120
ln(1:01)
=
X
12
70:0484
so X =
548; 685:6633 12
70:0484
= KSh.93; 995:34566
22c. For (i) we have
A.V = (1:01)84
1
12
100; 000 a72 +
1
12
40; 000v72a12
= 2:306722744
1
12
100; 000 51:1504 +
1
12
40; 000 0:48850 11:2550775
= 2:306722744(426; 253:3333 + 18; 327:01786)
= KSh.1; 025; 523:608
14
STA 2191 Jane Akinyi Aduda
For (ii) we have
A.V =
93; 995:34566
12
s84 =
93; 995:34566
12
1:0184 ?? 1
ln(1:01)
=
93; 995:34566
12
131:3245523
= KSh.1; 028; 658:057
23a. We are interested in A(4; 6:5) = A(0;6:5)
A(0;4) for 0 t < 7
A(0; t) = exp
Z t
0
0:03 + 0:001s2dt
= exp
0:03s +
0:001s3
3
t
0
= exp
0:03t +
0:001t3
3
A(0; 6:5) = exp (0:195 + 0:09154) = exp (0:28654)
A(0; 4) = exp (0:12 + 0:02133) = exp (0:14133)
Therefore A(4; 6:5) = exp(0:14521) = 1:15628
23b. We are interested in p:v: = 15v(t)
v(t) = exp??
Z 7
0
0:03 + 0:001tdt
exp??
Z 10
7
0:01tdt
exp??
Z 20
10
0:1dt
= exp??
0:03t +
0:001t3
3
7
0
exp??
0:005t210
7 exp??[0:1t]20
10
= exp(??0:32433) exp(??0:255) exp(??1)
= exp(??1:57933) = 0:206113
therefore p:v: = 15 0:206113 = 3:091695
= 3:10
23c. For the constant force of interest
e??(t) = 0:206113 where t = 20
therefore ?? 20 = ln 0:206113 = ??1:579330717 and
=
??1:579330717
??20
= 0:078966535
= 7:896%
23d.
A(8; 9) = exp
Z 9
8
0:01tdt = exp
0:005t29
8 = exp(0:085) = 1:0887
so 1 + i = 1:0887 ) i = 0:0887
= 8:87%
15
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