Topic 4 - Valuing Cashflows and Annuities Flashcards
Give the general formula for the PV of a series of discrete cashflows
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
Q) Consider PV of £1,000 payable at end of next 3 yrs, i = 5% pa
PV = 1000v + 1000v^2 + 1000v^3 = £2,723.25
What are the two different types of series of cashflows?
Discrete and Continuous
Give the general formula for the PV of a series of continuous cashflows
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
Q) Calculate the PV of a continuous cashflow of £200 pa if v(t) = 1-0.05t for 0 ≤ t ≤ 3
200x∫0,3(1−0.05t)dt = 200[t –0.025t^2]3,0
£555
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
A(0,5) = 1000 x 1.05^3 x 1.03^2 + 2000 x 1.05 x 1.03^2
£3,456.01
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
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
What is an Annuity?
An Annuity is a regular series of payments e.g. a pension
What is an annuity certain?
Paid for a finite period of time
What is an annuity in arrears?
Regular payments at end of time periods
What is an annuity in advance?
Regular payments at start of time periods
Also known as an annuity-due
What is an Immediate annuity?
Payment made during first time period
What is a Deferred annuity?
Payments made after first time period
Derive the annuity factor using a simple annuity with payments in arrears
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
Calculate 𝑎5¬ at i = 5% pa and verify using tables.
𝑎5¬ = (1-1.05^-5)/0.05 = 4.3295 a5¬ = 4.3295 (page 57 tables)
Derive the annuity factor using a simple annuity with payments in advance
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
State the relationship between the annuity factor for payments in arrears and the annuity factor for payments in advance
ᾃn¬ = (1+i) x 𝒂n¬ ᾃn¬ = (1−v^n)/d = [(1+i)x(1−v^n)]/i = (1+i) 𝑎n¬ = (i/d)𝑎n¬
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
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)
Give the formula for an annuity factor with continuous payments
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
Q)Calculate the present value of an annual pension
of £1,000 payable continuously for 5 years with i =
4% pa using both methods
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)
Give the formula for accumulations in arrears
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
Give the formula for accumulations in advance
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]
Give the formula for accumulations continuously
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
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
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)
Give a summary by stating the formulas for annuities and accumulations payable in arrears, advance, and continuously. State all other relationships if applicable.
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
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
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
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
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
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]
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)
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
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¬
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
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
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
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)
Give a summary by stating the formulas for annuities and accumulations payable pythly in arrears, advance, and continuously. State all other relationships if applicable.
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)
Give the annuity factor annually and pythly for payments in arrears and in advance forever
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)
What are the three different types of annuity?
Level (non increasing)
Deferred
Increasing
What is a Deferred Annuity?
Payments are made after the first time period
State the annuity factor for a deferred annuity with payments in arrears
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)
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
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)
State the annuity factor for a deferred annuity with payments in advance
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
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
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
State the annuity factor for a deferred annuity with payments continuously
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¬
State the annuity factor for a deferred annuity with payments pythly in arrears
m|𝑎n¬(p)
= 𝑎m+n¬(p) - 𝑎m¬(p)
= v^m x 𝑎n¬(p)
State the annuity factor for a deferred annuity with payments pythly in advance
m|ᾃn¬(p)
= ᾃm+n¬(p) - ᾃm¬(p)
= v^m x ᾃn¬(p)
Consider an annuity in which payments are not equal. State the PV of the sum of cashflows
∑n,(j=1) Xiv^ti
If Xi = ti then this is known as an increasing annuity
Using your knowledge of level annuity cashflows derive the formula for an increasing annuity factor in arrears
(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]
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¬
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)
Using your knowledge of level annuity cashflows derive the formula for an increasing annuity factor in advance
(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
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.
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)
What are the two different types of Increasing Annuities payable Continuously
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).
Give the formula for an increasing annuity factor payable continuously with a constant rate of payment r (per unit time)
(𝐼ᾱ)n¬ = ∑n,(r=1) ∫r-1, r (r.v^t) dt (𝐼ᾱ)n¬ = (ᾃn¬ –nv^n)/𝛿 = i/𝛿 x (𝐼a)n¬
Give the formula for an increasing annuity factor payable continuously with a rate of payment t at time t
(Īa)n¬ = ∫0, n (t.v^t) dt (Īa)n¬ = (ᾱn¬ –nv^n)/𝛿
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¬
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
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.
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
Further examples
Increasing annuities can also be used to find the PV of annuities that …
decrease by a fixed amount
Can also look at compound increasing annuities
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.
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)
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.
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
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.
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
What are the three different kinds of increases which occur under an increasing annuity?
Fixed amount
Occurring continuously
Compounding
