Topic 4 - Valuing Cashflows and Annuities

Question 1

Give the general formula for the PV of a series of discrete cashflows

Answer 1

If cashflows are ct1,..,ctn due at time t1,…,tn
PV = ∑n,(j=1) ctjv^tj
PV of series is sum of individual present values
Subscript notation available page 4 Week 3 notes

Question 2

Q) Consider PV of £1,000 payable at end of next 3 yrs, i = 5% pa

Answer 2

PV = 1000v + 1000v^2 + 1000v^3 = £2,723.25

Question 3

What are the two different types of series of cashflows?

Answer 3

Discrete and Continuous

Question 4

Give the general formula for the PV of a series of continuous cashflows

Answer 4

If cashflows are payable at a rate ρ(t) per unit time
PV of each cashflow is v(t)ρ(t)dt
PV of series of cashflows ∫0,T v(t)ρ(t) dt

Question 5

Q) Calculate the PV of a continuous cashflow of £200 pa if v(t) = 1-0.05t for 0 ≤ t ≤ 3

Answer 5

200x∫0,3(1−0.05t)dt = 200[t –0.025t^2]3,0

£555

Question 6

Accumulating cashflows at different interest rates
Q) Calculate A(0,5) for following payments
£1,000 at t = 0
£2,000 at t = 2
i = 5% 0≤t≤3
i = 3% t>3

Answer 6

A(0,5) = 1000 x 1.05^3 x 1.03^2 + 2000 x 1.05 x 1.03^2

£3,456.01

Question 7

Q) Calculate total PV for following payments
£1,000 at t = 2
£2,000 at t = 4
£3,000 at t = 5
i = 4% 0≤t≤3
i = 2% t>3

Answer 7

1000v^2@4% + 2000v^3@4%v@2% + 3000v^3@4%v^2@2%
1000 x 1.04^-2 + 2000 x 1.04^-3 x 1.02^-1 + 3000 x 1.04^-3 x 1.02^-2
£5,231.11
Full timeline available on page 8 Week 3 lecture notes

Question 8

What is an Annuity?

Answer 8

An Annuity is a regular series of payments e.g. a pension

Question 9

What is an annuity certain?

Answer 9

Paid for a finite period of time

Question 10

What is an annuity in arrears?

Answer 10

Regular payments at end of time periods

Question 11

What is an annuity in advance?

Answer 11

Regular payments at start of time periods

Also known as an annuity-due

Question 12

What is an Immediate annuity?

Answer 12

Payment made during first time period

Question 13

What is a Deferred annuity?

Answer 13

Payments made after first time period

Question 14

Derive the annuity factor using a simple annuity with payments in arrears

Answer 14

𝑎n¬ = v + v^2 + v^3 + … + v^n
𝑎n¬ = v(1-v^n)/(1-v)
𝑎n¬ = (1-v^n)/i
Timeline showing payments in arrears available page 10 Week 3 lecture notes

Question 15

Calculate 𝑎5¬ at i = 5% pa and verify using tables.

Answer 15

𝑎5¬ = (1-1.05^-5)/0.05 = 4.3295
a5¬  = 4.3295 (page 57 tables)

Question 16

Derive the annuity factor using a simple annuity with payments in advance

Answer 16

Annuities with payments in advance (annuity-due)
Timeline showing payments in advance available page 12 Week 3 lecture notes
ᾃn¬ = 1 + v + v^2 + … + v^n-1
ᾃn¬ = (1−v^n)/1−v
ᾃn¬ = (1−v^n)/d

Question 17

State the relationship between the annuity factor for payments in arrears and the annuity factor for payments in advance

Answer 17

ᾃn¬ = (1+i) x 𝒂n¬
ᾃn¬ = (1−v^n)/d = [(1+i)x(1−v^n)]/i = (1+i) 𝑎n¬ =  (i/d)𝑎n¬

Question 18

Q) Calculate the present value of an annual pension of
£1,000 payable in arrears for 5 years with i = 4% pa,
using formulae and using tables
Q) Calculate the present value of an annual pension of
£1,000 payable in advance for 5 years with i = 6% pa,
using formulae and using the relationship between the annuity factor for payments in arrears and the annuity factor for payments in advance

Answer 18

1000𝑎5¬ @ i=4%
1000 x (1-1.04^-5)/0.04
£4,451.82
Using tables
1000 x 4.4518 = £4,451.80 (rounding)

Solution –annuity in advance
Method 1 - calculation
1000 ᾃ5¬ @i = 6%, d = 0.06/1.06 = 0.056604
1000 x (1-1.06^-5)/0.056604
£4,465.09
Method 2 –using relationship
1000 x 1.06 x a5¬ @ i=6%
1000 x 1.06 x 4.2124 = £4,465.14 (rounding)

Question 19

Give the formula for an annuity factor with continuous payments

Answer 19

ᾱn¬ = ∫0,n(1.v^t)dt= ∫0,n[e^(−𝛿t)]dt= [-1/𝛿 x e^−𝛿t]0,n
= (1−e^[−𝛿n])/𝛿
= (1−v^n)/𝛿 = i/𝛿 𝑎n¬ for δ≠ 0

Question 20

Q)Calculate the present value of an annual pension
of £1,000 payable continuously for 5 years with i =
4% pa using both methods

Answer 20

Solution - continuously
Method 1 - calculation
1000ᾱ5¬ @i = 4% δ= ln(1.04) = 0.039221
1000 x (1-1.04^-5)/0.039221
£4,540.24
Method 2 –using relationship
1000 x 0.04/0.039221 x 𝑎5¬ @ i=4%
1000 x 0.04/0.039221 x 4.4518 = £4,540.22 (rounding)

Question 21

Give the formula for accumulations in arrears

Answer 21

𝑠n¬ = (1+i)^(n-1) + (1+i)^(n-2) + … + 1
sn¬ = (1+i)^n x 𝑎n¬
sn¬ = = [(1+i)^(n)−1]/i
Timeline available week 3 lecture notes page 20

Question 22

Give the formula for accumulations in advance

Answer 22

ṡn¬ = (1+i)^n + (1+i)^(n-1) + … + (1+i)
ṡn¬ = (1+i)^n x ᾃn¬
ṡn¬ = [(1+i)^(n)−1]/d = (1+i) 𝑠n¬
Timeline available week 3 lecture notes page 21
[Note that the singular dot over the ‘s’ should in fact be two dots]

Question 23

Give the formula for accumulations continuously

Answer 23

ӯn¬ = ∫0,n e^(δ(n−t)) dt
ӯn¬ = (1+i)^n x ᾱn¬ 
ӯn¬ = [(1+i)^(n)-1]/δ =  (i/𝛿) x 𝑠n¬ for δ≠ 0
[ӯ = s bar]

Question 24

Question
Calculate the accumulated value of an annual pension
of £1,000 payable for 5 years with i = 4% pa, payable:
- In arrears (example)
- In advance
- Continuously
Using the formula i.e. [(1+i)^(n)−1]/X & the relationships

Answer 24

Example –In arrears
Method 1 - Calculation
1000𝑠5¬ @ i=4%
1000 x (1.04^5 -1)/0.04
£5,416.32
Method 2 - Using relationship/tables
1000 x 5.4163 = £5,416.30 (rounding)

Solution –In advance 
Method 1 - calculation
1000 ṡ5¬ @i = 4%, d = 0.04/1.04 = 0.038462
1000 x (1.04^5  -1)/0.038462
£5,632.91
Method 2 –using relationship
1000 x 1.04 x s5¬ @ i=4%
1000 x 1.04 x 5.4163 = £5,632.95 (rounding)

Solution - continuously
Method 1 - calculation
1000ӯ5¬ @i = 4%, δ= ln(1.04) = 0.039221
1000 x (1.04^5  -1)/0.039221
£5,523.90
Method 2 –using relationship
1000 x 0.04/0.039221 x 𝑠5¬ @ i=4%
1000 x 0.04/0.039221 x 5.4163 = £5,523.88 (rounding)

Question 25

Give a summary by stating the formulas for annuities and accumulations payable in arrears, advance, and continuously. State all other relationships if applicable.

Answer 25

𝑎n¬ = (1−v^n)/i
ᾃn¬ = (1−v^n)/d
ᾱn¬ = (1−v^n)/δ
𝑠n¬ = [(1+i)^(n)−1]/i
ṡn¬ = [(1+i)^(n)−1]/d
ӯn¬ = [(1+i)^(n−1)]/𝛿 

ᾃn¬ = 1 + 𝑎n-1¬ for n ≥ 2
ṡn¬ = 𝑠n+1¬ -1

Question 26

If p and n are positive integers, then we can use them to
denote level annuity payments payable at time 1/p, 2/p,…, n giving us the formula for the annuity factor for pthly payments in arrears and in advance

Answer 26

𝑎n¬(p) = (1−v^n)/i(p) = [i/i(p)] x 𝑎n¬
ᾃn¬(p) = (1−v^n)/d(p) = [i/d(p)]𝑎n¬ = (1+i)^(1/p) 𝑎n¬(p)

Annuties payable pthly are not tabulated but:
i/i(p)
i/d(p)
Are tabulated for a range of p

Question 27

Example
Calculate the PV of a series of payments of £100
payable monthly in arrears
For a period of 5 years with i = 5% pa, using the
formulae and relationships

Answer 27

Example - monthly in arrears
Method 1 - Calculation
1200𝑎5¬(12) @i=5%, i(12) = 12 x (1.05^(1/12) –1) = 0.048889
1200 x (1-1.05^-5)/0.048889
£5,313.44
Method 2 - Using relationship/tables
1200 x i/i(12) x 𝑎5¬ @ i=5% 
1200 x 0.05/0.048889 x 4.3295= £5,313.47 (rounding)

Question 28

Question
Calculate the PV of a series of payments of £100
payable quarterly in advance
For a period of 5 years with i = 5% pa, using the
formulae and relationships
Use d(p) = p[1-(1+i)^-1/p]

Answer 28

Solution –quarterly in advance 
Method 1 - calculation
400 ᾃ5¬(4) @i = 5%, d(4) = 4 x (1-1.05^(-1/4)) = 0.048494
400 x (1-1.05^-5)/0.048494
£1,785.57
Method 2 –using relationship
400 x i/d(4) x 𝑎5¬ @ i=5%
400 x 0.05/0.048494 x 4.3295 = £1,785.58 (rounding)

Question 29

If p and n are positive integers, then we can use them to denote annuity payments payable at time 1/p, 2/p,…, n giving us the formula for the accumulations of pthly payments in arrears and in advance

Answer 29

𝑠n¬(p) = (1+i)^n x 𝑎n¬(p) = [(1+i)^(n)−1]/i(p) = i/i(p) x 𝑠n¬
ṡn¬(p) = (1+i)^n x ᾃn¬(p) = [(1+i)^(n)−1]/d(p) = i/d(p) x 𝑠n¬

Question 30

Example
Calculate the accumulated value of a series of payments of £300 payable quarterly in arrears
For a period of 4 years with i = 4% pa, using the
formulae and relationships

Answer 30

Example –quarterly in arrears
Method 1 - Calculation
1200𝑠4¬(4) @i=4%, i(4) = 4 x (1.04^(1/4) –1) = 0.039414
1200 x (1.04^(4)-1)/0.039414
£5,171.52
Method 2 - Using relationship/tables
1200 x i/i(4) x 𝑠4¬ @ i=4% 
1200 x 0.04/0.039414 x 4.2465= £5,171.56 (rounding)

Question 31

Question
Calculate the accumulated value of a series of payments of £300 payable monthly in advance
For a period of 4 years with i = 4% pa, using the
formulae and relationships

Answer 31

Solution –monthly in advance
Method 1 - calculation
3600 ṡ4¬(12) @i = 4%, d(12) = 12 x (1-1.04^(-1/12)) = 0.039157
3600 x (1.04^(4)-1)/0.039157
£15,616.39
Method 2 –using relationship
3600 x i/d(12) x 𝑠4¬ @ i=4%
3600 x 0.04/0.039157 x 4.2465 = £15,616.52 (rounding)

Question 32

Give a summary by stating the formulas for annuities and accumulations payable pythly in arrears, advance, and continuously. State all other relationships if applicable.

Answer 32

𝑎n¬(p) = (1−v^n)/i(p) 
ᾃn¬(p) = (1−v^n)/d(p)
𝑠n¬(p) = [(1+i)^(n)−1]/i(p)
ṡn¬(p) = [(1+i)^(n)−1]/d(p)

Question 33

Give the annuity factor annually and pythly for payments in arrears and in advance forever

Answer 33

Payable in perpetuity
Annuities payable forever i.e. n-->∞:
a∞¬ = 1/i
ᾃ∞¬ = 1/d
a∞¬(p) = 1/i(p)
ᾃ∞¬ (p) = 1/d(p)

Question 34

What are the three different types of annuity?

Answer 34

Level (non increasing)
Deferred
Increasing

Question 35

What is a Deferred Annuity?

Answer 35

Payments are made after the first time period

Question 36

State the annuity factor for a deferred annuity with payments in arrears

Answer 36

m| 𝑎n¬ = v^(m+1) + v^(m+2) + v^(m+3) + … + v^(m+n)
= (v + v^2 + v^3 + … + v^m+n) - (v + v^2 + v^3 + … + v^m)
= 𝑎m+n¬ - 𝑎m¬ OR
= v^m x (v + v^2 + v^3 + … + v^n)
= v^m x 𝑎n¬
Timeline on page 8 Week 4 lecture notes (pay attention to m| notation and placement)

Question 37

Example
Calculate the PV of an annual pension of £2,000 payable (in arrears) for 6 years, with the first payment deferred for 3 years with i = 4% pa, using both methods
Hint –determine what m & n are first

Answer 37

Solution
Method –1 
2000 x 3|𝑎6¬  @ 4% = 2000 x (𝑎9¬ - 𝑎3¬ )
2000 x (7.4353 –2.7751)
£9,320.40
Method –2
2000 x 3|𝑎6¬  @ 4% = 2000 x v^3 𝑎6¬ 
2000 x 1.04^-3 x 5.2421 
£9,320.42 (rounding)

Question 38

State the annuity factor for a deferred annuity with payments in advance

Answer 38

m|ᾃn¬ = v^(m) + v^(m+1) + v^(m+2) + … + v^(m+n-1)
= ( 1 + v + v^2 + … + v^(m+n-1) ) - (1 + v + v^2 + … + v^(m-1))
= ᾃm+n¬ - ᾃm¬ OR
= v^m x ( 1 + v + v^2 + … + v^(n-1) )
= v^m x ᾃn¬
Timeline on page 11 Week 4 lecture notes (pay attention to m| notation and placement)

Question 39

Question
A person currently aged 45 will receive an annual
pension of £5,000 (payable in advance), which begins
on their 60th birthday and will cease on their 69th
birthday inclusive. Calculate the PV of this pension using i = 5% pa.
Determine what m & n are first

Answer 39

Solution
m = 60-45 = 15, n = 10
5000 x15|ᾃ10¬ = 5000 x v^15 x ᾃ10
5000 x 1.05^-15 x 1.05 x 7.7217
£19,499.92

Question 40

State the annuity factor for a deferred annuity with payments continuously

Answer 40

m|ᾱn¬ = ∫m,m+n (e^−𝛿t) dt
= ∫0,m+n (e^−𝛿t) dt - ∫0,m (e^−𝛿t) dt
= ᾱm+n¬ - ᾱm¬ OR
= e^(−𝛿m) x ∫0,n e^(−𝛿s) ds
=  v^m x ᾱn¬

Question 41

State the annuity factor for a deferred annuity with payments pythly in arrears

Answer 41

m|𝑎n¬(p)
= 𝑎m+n¬(p) - 𝑎m¬(p)
= v^m x 𝑎n¬(p)

Question 42

State the annuity factor for a deferred annuity with payments pythly in advance

Answer 42

m|ᾃn¬(p)
= ᾃm+n¬(p) - ᾃm¬(p)
= v^m x ᾃn¬(p)

Question 43

Consider an annuity in which payments are not equal. State the PV of the sum of cashflows

Answer 43

∑n,(j=1) Xiv^ti

If Xi = ti then this is known as an increasing annuity

Question 44

Using your knowledge of level annuity cashflows derive the formula for an increasing annuity factor in arrears

Answer 44

(Ia)n¬ = v + 2v^2 +3v^3+..+nv^n (1)
(1+i)(Ia)n¬ = 1 + 2v + 3v^2+..+nv^(n-1) (2)
(2) - (1) = i(Ia)n¬ = 1+v + v^2 +..+v^(n-1) –nv^n = ᾃn¬ –nv^n
(Ia)n¬ = (ᾃn¬ –nv^n)/i [Tabulated]

Question 45

Typical question
Calculate the PV at t=0 of a series of 8 annual payments starting at £1,000 at t=1 and increasing by £200 pa, with i = 5% pa.
We can think of the cashflows as follows
Level annuity of £800 and 
Increasing annuity of £200
So PV = 800𝑎8¬ + 200(Ia)8¬

Answer 45

So PV = 800𝑎8¬ + 200(Ia)8¬
800𝑎8¬ +200[(ᾃ8¬ –8v^8)/i]
800 x 6.4632 + 200[(1.05x6.4632−8x1.05^−8)/0.05]
£10,657.14 OR
800 x 6.4632 + 200 x 27.4332 = £10,657.20 (using tables)

Question 46

Using your knowledge of level annuity cashflows derive the formula for an increasing annuity factor in advance

Answer 46

(Iᾃ)n¬ = 1 + 2v + 3v^2 +..+ nv^(n-1)
However, we already know that (1+i)(Ia)n¬ = 1 + 2v + 3v^2+..+nv^(n-1)
Therefore, (Iᾃ)n¬ = (1+i)(Ia)n¬
(Iᾃ)n¬ = (1+i) [(ᾃn¬  –nv^n)/i]
(Iᾃ)n¬ = (ᾃn¬  –nv^n)/d

Question 47

Question
Calculate the PV at t=0 of a series of 6 annual payments starting with £750 at t=0 and increasing by £250 pa, with i = 4% pa.

Answer 47

Solution
PV = 500ᾃ6¬ + 250(Iᾃ)6¬
= 500ᾃ6¬ + 250[ᾃ6¬ –6v^6)/d]
500x1.04x5.2421+250x1.04x[(1.04x5.2421−6x1.04^−6)/0.04]
= £7,340.23 OR
500x1.04x5.2421 + 250x1.04x17.7484 = £7,340.48 (tables)

Question 48

What are the two different types of Increasing Annuities payable Continuously

Answer 48

Need to distinguish between annuities which have:
A constant rate of payment r (per unit time) –(𝐼ᾱ)n¬
A rate of payment t at time t - (Īa)n¬ [bar over I and a]

A bar over the a indicates that the payments are made
continuously
A bar over the I indicates that the increases occur
continuously (rather than at the end of the year).

Question 49

Give the formula for an increasing annuity factor payable continuously with a constant rate of payment r (per unit time)

Answer 49

(𝐼ᾱ)n¬ = ∑n,(r=1) ∫r-1, r (r.v^t) dt
(𝐼ᾱ)n¬ = (ᾃn¬  –nv^n)/𝛿 = i/𝛿 x (𝐼a)n¬

Question 50

Give the formula for an increasing annuity factor payable continuously with a rate of payment t at time t

Answer 50

(Īa)n¬ = ∫0, n (t.v^t) dt
(Īa)n¬ = (ᾱn¬  –nv^n)/𝛿

Question 51

Example
Calculate the PV at t=0 of a series of continuous
payments made at a rate of £200t payable for 5 years,
with i = 4% pa
Hint –200 x (Ī𝑎)n¬

Answer 51

Example –continuous rate of payment
PV = 200 x (Ī𝑎)n¬
PV = 200(ᾱ5¬ –5v^5)/𝛿
PV = 200x [(4.4518x0.04/ln(1.04)  −5x1.04^−5)/ln(1.04)]
£2,195.87

Question 52

Question
Calculate the PV at t=0 of a series of continuous
payments payable for 5 years, starting at £5,000 and
increasing by £500 pa, with i = 6% pa.

Answer 52

Solution - constant rate of payment
PV = 4500 ᾱ5¬ + 500(Iᾱ)5¬
PV = 4500 ᾱ5¬ + 500[(ᾃ5−5v^5)/𝛿]
PV = 4500x [0.06/ln(1.06)]x4.2124 +500x[(1.06x4.2124−5x1.06^−5)/ln(1.06)]
£25,773.17 OR
4500x[0.06/ln(1.06)]x4.2124 +500x[0.06/ln(1.06)]x12.1469 = £25,772.84

Question 53

Further examples

Increasing annuities can also be used to find the PV of annuities that …

Answer 53

decrease by a fixed amount

Can also look at compound increasing annuities

Question 54

Calculate the PV at t=0 of a series of 6 annual payments starting at £1,000 at t=1 and decreasing by £100 pa, with i = 5% pa.

Answer 54

We can think of the cashflows as follows
Level annuity of £1,100 LESS
Increasing annuity of £100
So PV = 1100𝑎6¬ - 100(Ia)6¬ 
PV = 1100𝑎6¬ - 100[(ᾃ6¬−6v^6)/i]
PV = 1100 x 5.0757 - 100[(1.05x5.0757−6x1.05^−6)/0.05]
£3,878.88 OR
1100 x 5.0757 - 100 x 17.0437 = £3,878.90 (using tables)

Question 55

Calculate the PV at t=0 of a series of 15 annual
payments with £500 at t=1 and increasing by 5% pa
compound each subsequent year, with i = 10% pa.

Answer 55

Compound increasing annuity
Look at cashflows …
PV = 500v + 500x1.05v^2 + 500x1.05^2v^3+..+ 500x1.05^14v^15
PV = 500/1.05(1.05v + 1.05^2v^2 + 1.05^3v^3 + .. + 1.05^15v^15)
PV = 500/1.05(v’ + v’^2 + 1.05^3v^3 + .. + 1.05^15v^15)
where v’= 1.05v
𝑎n¬ = v + v^2 + v^3 +…+ v^n @ v= 1/1+i
𝑎n¬’ = v’ + v’^2 + v’^3 +…+ v’^n @ v’= 1.05v = 1/1+i’
PV = (500/1.05)𝑎15¬’ i’= (1.1/1.05)-1 = 4.762%
PV = (500/1.05)(1-1.04762-15)/0.04762 = £5,023.18

Alternatively,
PV = 500v + 500x1.05v^2 + 500x1.05^2v^3+..+ 500x1.05^14v^15
= 500v(1 + 1.05v + 1.05^2v^2 + 1.05^3v^3 + .. + 1.05^14v^14)
= 500v(1 + v’ + v’^2 + v’^3 +…+ v’^14) v’ = 1.05v
ᾃn¬ = 1 + v + v^2 + v^3 + … + v^(n-1) @ v = 1/1+i
ᾃn¬’ = 1 + v’ + v’^2 + v’^3 + … + v’^(n-1) @ v’ = 1/1+i’ = 1.05v
= 500vᾃ15¬’ i’ = (1.1/1.05) -1 = 4.762%
= (500/1.1) x 1.04762(1–1.04762^-15)/0.04762 = £5,023.18

Question 56

Question
Calculate the PV at t=0 of a series of 5 annual payments with £100 at t=0 and increasing by 4% pa compound each subsequent year, with i = 8% pa.

Answer 56

PV = 100 + 100x1.04v + 100x1.04^2v^2+..+ 100x1.04^4v^4
= 100(1 + 1.04v + 1.04^2v^2 + 1.04^3v^3 + 1.04^4v^4)
= 100(1+ v’ + v’^2 + v’^3 + v’^4) v’= 1.04v
ᾃn¬ = 1 + v + v^2 + v^3 + … + v^(n-1) @ v = 1/1+i
ᾃn¬’ = 1+ v’+ v’2 + v’3 + … + v’^(n-1) @ v’= 1/1+i’ = 1.04v
= 100 ᾃ5¬’ i’= (1.08/1.04)-1 = 3.846%
 =100x1.03846x(1-1.03846^-5)/0.03846 = £464.31

Question 57

What are the three different kinds of increases which occur under an increasing annuity?

Answer 57

Fixed amount
Occurring continuously
Compounding