MoreAnnuitiesProblems:Varying- Interestrate.Payment

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ChangingInterestRates

• Intheprevioussection,therateofcompoundinterestremainunchangedthroughouttheperiod.

• Inpractice,interestrateofinterestvarywithconsiderablefrequency.

Forinstance:Banksvarytheirdepositrateswithchangesinmarketconditions.

(compoundinterest)

Example(changeofinterestrate- compoundinterest)

Find thepresentvalueofRM10000due in10years’timeif𝑗" = 8% forthefirst3years,𝑗& = 10% forthenext5yearsand 𝑗) = 9% forthelast2years,

RM10,000

Time 0 3 8 10𝑗" = 8%𝑚 = 4, 𝑡 = 3n=12

è𝑖 = 34" = 0.02

𝑗& = 10%𝑚 = 2, 𝑡 = 5n=10

è𝑖 = 37& = 0.05

𝑗) = 9%𝑚 = 1, 𝑡 = 2n=2

è𝑖 = 38) = 0.09

ie thepresentvalueofRM10K:

ValueofRM10Kin3years’time= 8416.80 1+ 0.05 ;)<

= 𝑅𝑀5167.19

ValueofRM10KNow= 5167.19 1+ 0.02 ;)&

= 𝑅𝑀4074.29

ValueofRM10Kin8years’time= 10000 1+ 0.09 ;&

= 𝑅𝑀8416.80

= 1000 1 + 0.12 ;)& 1+ 0.05 ;)< 1+ 0.09 ;& = 𝑅𝑀4074.29

Previously,inevaluatingthePV(orFV)whentheinterestratevariesè thegeneralstrategyisto evaluatewheneverthereisachangeintherateofinterest

MoreAnnuitiesProblems• Findingtheperiodicpayment• Findingtheinterestrate• Findingthetermofannuity• Changesintheinterestrate• Annuitieswherepaymentsvary

Similartotheevaluationofvalueswheninterestratesvaries.

è Itisnecessarytovalueannuitieswheninterestrateschangeduringtheannuitiesterm

è Similartothepreviousmethod,• Eachsuccessiveintermediatevalueshouldhaveafocaldatethatiscloserto

thefinalfocaldate.• Theaimistomovetowardsthefinalfocaldate,gatheringuptheannuity

paymentsaswego.• Eachstepcomprisesthevaluationoftheannuitypaymentswithinthe

appropriateperiodtogetherwiththevalueofthepreviousintermediatevalueatthenewrateofinterest

Findingperiodicpayment

Example1Findthevalueat1Jan2000ofanannuityofRM1000papayablefrom1Jan1991to1Jan2000inclusive,iftheinterestrateis9%papriorto1Jan1997and8%pathereafter

R=1000

FV?

0 1

RM1K

2

RM1K

3

RM1K

1Jan1991

1Jan1992

1Jan1993

4

RM1K

5

RM1K

6

RM1K

1Jan1994

1Jan1995

1Jan1996

7

RM1K

1Jan1997

8

RM1K

9

RM1K

10

RM1K

1Jan1998

1Jan1999

1Jan2000

𝑖 = 0.09n=7

𝑖 = 0.08n=3

𝐹𝑉 == 𝑅1 + 𝑖 𝑛− 1

𝑖𝑛𝑖𝑠𝑅

At1Jan1997

= 10001+ 0.09 7 − 1

0.09

70.09𝑠= 1000

= 𝑅𝑀9200.43

𝐹𝑉 = 𝑛𝑖𝑠+𝑅

At1Jan2000

= 11.589.89 + 10001 + 0.08 3 − 1

0.08

30.08𝑠+1000

= 𝑅𝑀14836.29

BringforwardRM9200.43

= 𝑅𝑀9200.43(1 + 0.08)J

= 11.589.89 + 3246.40

Example2(a)CalculatethePVofa10-yearannuitywithhalf-yearlypaymentsofRM50each,assumingj2=8%foryears1,2,and3;j2=7%foryears4and5;j2=6%foryears6,7,8,9,and10.

Time 0 6𝑗& = 8%𝑚 = 2,𝑡 = 3n=6

è𝑖 = 37&= 0.04

𝑗& = 7%𝑚 = 2,𝑡 = 2n=4

è𝑖 = 37&= 0.035

RM50 RM50 RM50RM50 RM50

10

RM50RM50

RM50 RM50 RM50

20𝑗& = 6%𝑚 = 2,𝑡 = 5n=10

è𝑖 = 37&= 0.03

𝑛𝑖𝑃𝑉 = 𝑅𝑎 = 𝑅

1− 1+ 𝑖 ;M

𝑖AtYear 5

= 50𝑎100.03

= 𝑅𝑀426.51

𝑃𝑉 =𝑛𝑖𝑎+𝑅

AtYear 3

= 𝑅𝑀555.33

Bringbward RM426.51

= 𝑅𝑀426.51(1+ 0.035);"

= 371.68+ 183.6540.035𝑎+50

𝑃𝑉 =𝑛𝑖𝑎+𝑅

Now

= 𝑹𝑴𝟕𝟎𝟎. 𝟗𝟗

Bringbward RM555.33

= 𝑅𝑀555.33(1+ 0.04);S

= 438.89+ 262.1060.04𝑎+50

PV?

RM50 RM50RM50

RM50RM50

RM50 RM50 RM50 RM50 RM50

Example2(cont’d)(b)From(a),whatunchangedrateofinterestthroughoutthe10yearswouldprovidethesamePV?

PV=RM700.99;R=50;m=2;t=10è n=20

𝑛𝑖𝑃𝑉 = 𝑅𝑎 = 𝑅

1− 1+ 𝑖 ;M

𝑖

700.99 = 50𝑎20𝑖

= 𝑅𝑀14.019810𝑖𝑎

200.03𝑎 = 14.877

Fromtable

200.04𝑎 = 13.590

20𝑖𝑎 = 14.0198

Thus,linearinterpolationformulais

𝑖 − 0.030.04− 0.03 =

14.0198− 14.87713.590− 14.877

𝑖 − 0.030.01 =

−0.8572−1.287

𝑖 = 0.0367ie.3.67%

𝑗2 = 𝑖𝑚 = 7.34%

Exercise1. Aworkeraged25investsRM150everyyearfor40years.Howmuchwillshe

retirewithiftheinvestmentearns:• 10%painyears1-5• 8%painyears6-15• 9%painyears16-30• 3%painyears31-40Whatlevelofinterestratethroughout thetermwouldprovide thesameretirementbenefits?

(𝑅𝑀27953.68)

MoreAnnuitiesProblems• Findingtheperiodicpayment• Findingtheinterestrate• Findingthetermofannuity• Changesintheinterestrate• Annuitieswherepaymentsvary

Annuitieswherepaymentsvary

• Inevaluatingannuities,sizeofpaymentswithinthetermofannuityneedtobeconsidered

• Previously,weonlyconsidereduniformsizeofpayments.

• Therearetwomethodsusedwhenthepaymentsvaryintermsofaconstantratio.

Example3Mr AdamswantstobuyanannuityofRM1000ayearfor10yearsthatisprotectedagainstinflation.TheXYZtrustCompanyofferstosellhimanannuitywherepaymentsincreaseeachyearbyexactly10%.Inparticular, thepaymentswillbeRM1100attheendofyear1,RM1210attheendofyear2,RM1331attheendofyear3,andsoonfor10years.Find thecostofthisannuityiftherateofinterestis13%pa.

0 101

1000(1+0.1)1=1100

2 3 9PVie cost

1000(1+0.1)2=1210

1000(1+0.1)3=1331

1000(1+0.1)9 1000(1+0.1)10

Note:Thisdiagramissimilar tothecompound interesttimeline.WeevaluatePVbasedonthecompound interestevaluationie.ateachperiodtothefocaltimeandaccumulateitsamount.

𝑃𝑉 = 𝑆(1 + 𝑖);M

𝑃𝑉 = 𝑆(1 + 𝑖);) + 𝑆(1 + 𝑖);& + 𝑆(1 + 𝑖);J +…+𝑆(1 + 𝑖);(M;)) + +𝑆(1+ 𝑖);M

= [1000 1.1 ](1+ 0.13);)+[1000(1.1)2](1+ 0.13);& +⋯+ [ 1000 1.1 Z] 1+ 0.13 ;Z

+ [ 1000 1.1 )<] 1+ 0.13 ;)<

Example3- Method1𝑃𝑉 = 𝑆(1 + 𝑖);) + 𝑆(1 + 𝑖);& + 𝑆(1 + 𝑖);J +…+𝑆(1 + 𝑖);(M;)) + +𝑆(1 + 𝑖);M

Note:RHsideoftheequationisthesumofgeometricprogression

𝑎1 − 𝑟M

1 − 𝑟Thus,

𝑃𝑉 = 𝑆(1 + 𝑖);) + 𝑆(1 + 𝑖);& + 𝑆(1 + 𝑖);J +…+𝑆(1 + 𝑖);(M;)) + +𝑆(1 + 𝑖);M

= [1000 1.1 ](1+ 0.13);)+[1000(1.1)2](1 + 0.13);& +⋯+ [ 1000 1.1 )<] 1+ 0.13 ;)<

𝑎 = 100 1.1 (1 + 0.13);)

𝑟 = 1.1 (1 + 0.13);)𝑛 = 10

𝑃𝑉 = 100 1.1 (1 + 0.13);)1− 1.1 (1 + 0.13);) )<

1− 1.1 (1 + 0.13);)= 𝑅𝑀8650.17

Example3- Method2

Noticewecansimplify theaboveas:

Revisit

𝑃𝑉 = 𝑆(1 + 𝑖);) + 𝑆(1 + 𝑖);& + 𝑆(1 + 𝑖);J +…+𝑆(1 + 𝑖);(M;)) + +𝑆(1 + 𝑖);M

= [1000 1.1 ](1+ 0.13);)+[1000(1.1)2](1 + 0.13);& +⋯+ [ 1000 1.1 )<] 1+ 0.13 ;)<

𝑃𝑉 = 10001.11.13

+1.11.13

&+

1.11.13

J+ ⋯+

1.11.13

)<

Rewritingwehave

𝑃𝑉 = 10001.11.13

+1.11.13

&+

1.11.13

J+ ⋯+

1.11.13

)<

𝑖\Letbethenewrateofinterest

1 + 𝑖\ =1.11.13

;)

=1.131.1

𝑖\=0.027

= 1000 1+ 𝑖 \ ;) + 1+ 𝑖 \ ;& + 1+ 𝑖 \ ;J + ⋯+ 1+ 𝑖 \ ;)<

Solveusingordinaryannuity:

𝑖

𝑃𝑉 = 𝑅𝑛𝑖 ’𝑎 𝑃𝑉 = 1000 = 𝑹𝑴𝟖𝟔𝟓𝟎.𝟏𝟕

10𝑖 ’𝑎ie

Example4Find thePVofpaymentsmadeattheendofeachyearfor10yearsusing12%paifthefirstpaymentisRM100,thesecondisRM200,thethirdRM300andsoon

0 101

100

2 3 9PV

200 300 900 1000

Note:Thisdiagramissimilar tothecompound interesttimeline.WeevaluatePVbasedonthecompound interestevaluationie.ateachperiodtothefocaltimeandaccumulateitsamount.

𝑃𝑉 = 𝑆(1 + 𝑖);M

𝑃𝑉 = 𝑆(1 + 𝑖);)+ 𝑆(1 + 𝑖);&+…. +𝑆(1 + 𝑖);(M;))+ +𝑆(1+ 𝑖);M

= [100(1+ 0.12);)+200(1+ 0.12);& + ⋯+ 900 1+ 0.12 ;Z + 1000 1+ 0.12 ;)<

Multiply bothsides by(1+i)i.e 1.12

1.12𝑃𝑉 = 100 +200(1+ 0.12);) +⋯+900 1+ 0.12 ;a + 1000 1+ 0.12 ;)<

(1)

(2)

Multiply bothsides by(1+i)i.e 1.12

Example4(cont’d)𝑃𝑉 = 𝑆(1 + 𝑖);)+ 𝑆(1 + 𝑖);&+…. +𝑆(1 + 𝑖);(M;))+ +𝑆(1+ 𝑖);M

𝑃𝑉 = [100(1+ 0.12);)+200(1+ 0.12);& + ⋯+ 900 1+ 0.12 ;Z + 1000 1+ 0.12 ;)<

Multiply bothsides by(1+i)i.e 1.12

1.12𝑃𝑉 = 100 +200(1+ 0.12);) +⋯+900 1+ 0.12 ;a + 1000 1+ 0.12 ;)< (2)

Subract (2)from(1)

0.12𝑃𝑉 = 100 +100(1+ 0.12);) +⋯+100 1+ 0.12 ;Z−1000 1+ 0.12 ;)<

(1)

NowontheRhside wehavetheformofa10yearannuitydue, ie.

0.12𝑃𝑉 = 100 +100(1+ 0.12);) +⋯+100 1+ 0.12 ;Z+ 100 1+ 0.12 ;)<

𝑃𝑉 = 𝑅𝑛𝑖𝑎

0.12𝑃𝑉 = 100100.12𝑎

0.12𝑃𝑉 = 100(1+ 0.12)<+ 100(1+ 0.12);) + ⋯+ 100 1+ 0.12 ;Z + 100 1+ 0.12 ;)<

−1000 1+ 0.12 ;)<

𝑷𝑽 = 𝑹𝑴𝟐𝟓𝟗𝟎.𝟒𝟑

ExerciseAcourtwastryingtodetermine thePVofthefuture incomeofamanparalysed inacaraccident.Atthetimeoftheaccident,themanwasearningRM25000ayearandanticipatedreceivingan8%riseeachyear.Heis30yearsawayfromretirement.Ifmoney isworth10%pa,whatisthePVofhisfuture income.

(𝑅𝑀571486.98)