DISCOUNTING AND ACCUMULATING Week 5 WHAT WE SHALL DO TODAY ο§Progress Where are we on course material What are the plans as we proceed ο§Three sessions Discounting and accumulating a series of cashflows Session One Discrete cashflows Session Two Continuously payable cashflows Session Three Valuing cashflows at t not equal to 0 PV of payment stream using force of interest WHERE ARE WE SO FAR 1 Introduction and time value of money 2 The theory of interest rates 3 Discounting and accumulating We are here! 4 Level annuities 5 Deferred and increasing annuities 6 Loan schedules 7 Project appraisal 8 Immunisation Should be here/ need 2 make up lectures to catch up/ next 2 weeks? WHERE ARE WE SO FAR 1 Introduction and time value of money 2 The theory of interest rates 3 Discounting and accumulating We are here! CAT 1 / 1hr written next week 4 Level annuities 5 Deferred and increasing annuities 6 Loan schedules CAT 2 7 Project appraisal 8 Immunisation CAT 3 Should be here/ need 2 make up lectures to catch up/ next 2 weeks? DISCRETE CASHFLOWS So far we have looked at the present value of a single payment C due at time t The present value is Cvt where v = 1/(1+i) is the discounting factor i is the effective rate of interest per unit time period (typically one yr) DISCRETE CASHFLOWS So far we have looked at the present value of a single payment C due at time t The present value is Cvt where v = 1/(1+i) is the discounting factor i is the effective rate of interest per unit time period (typically one yr) Suppose now we had two payments C1 due at time t1 and C2 due a time t2 In other words What would the present value be In other words What would the present value be πΆ1 π£ π‘1 + πΆ2 π£ π‘2 In other words What would the present value be πΆ1 π£ π‘1 + πΆ2 π£ π‘2 The present value is the summation of the individual present values More generally for a series of payments π1 , π2 , . . . , ππ More generally for a series of payments π1 , π2 , . . . , ππ due at times π‘1 , π‘2 , . . . , π‘π More generally for a series of payments π1 , π2 , . . . , ππ due at times π‘1 , π‘2 , . . . , π‘π The present value is EXAMPLE EXAMPLE EXAMPLE EXAMPLE EXAMPLE EXAMPLE EXAMPLE Do you get £465.56? Do you get £465.56? Do you get £465.56? Do you get £465.56? If the effective rate of interest is not constant but varies over time, then the present value becomes If the effective rate of interest is not constant but varies over time, then the present value becomes The series of payments can also go on indefinitely, in which case the upper limit of the summation is ∞ . If the effective rate of interest is not constant but varies over time, then the present value becomes The series of payments can also go on indefinitely, in which case the upper limit of the summation is ∞ . We shall see this for an annuity that is payable in perpetuity. Solution ππ‘1 π£ π‘1 + ππ‘2 π£ π‘2 +. . . +ππ‘π π£ π‘π Solution ππ‘1 π£ π‘1 + ππ‘2 π£ π‘2 +. . . +ππ‘π π£ π‘π = ππ‘1 π − π‘1 0 πΏ π ππ Solution ππ‘1 π£ π‘1 + ππ‘2 π£ π‘2 +. . . +ππ‘π π£ π‘π = ππ‘1 π − π‘1 0 πΏ π ππ + ππ‘2 π − π‘2 0 πΏ π ππ Solution ππ‘1 π£ π‘1 + ππ‘2 π£ π‘2 +. . . +ππ‘π π£ π‘π = ππ‘1 π − π‘1 0 πΏ π ππ + ππ‘2 π − π‘2 0 πΏ π ππ +. . . +ππ‘π π − π‘π 0 πΏ π ππ Solution ππ‘1 π£ π‘1 + ππ‘2 π£ π‘2 +. . . +ππ‘π π£ π‘π = ππ‘1 π − π‘1 0 πΏ π ππ + ππ‘2 π − π π‘π ππ‘π π − 0 πΏ π ππ = π=1 π‘2 0 πΏ π ππ +. . . +ππ‘π π − π‘π 0 πΏ π ππ CONTINUOUSLY PAYABLE CASHFLOWS Suppose that T>0 CONTINUOUSLY PAYABLE CASHFLOWS Suppose that T>0 Between times 0 and T, an investor will be paid money continuously, and CONTINUOUSLY PAYABLE CASHFLOWS Suppose that T>0 Between times 0 and T, an investor will be paid money continuously, and The rate of payment at time t is ρ(t) per unit time CONTINUOUSLY PAYABLE CASHFLOWS Suppose that T>0 Between times 0 and T, an investor will be paid money continuously, and The rate of payment at time t is ρ(t) per unit time What is the present value of this cashflow? First we need to be clear what is meant by this rate of payment Rate of payment per unit time ρ(t) Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt -Consider this example If the rate of payment is 24pa then in any one year the total amount paid is 24, but this payment is spread through the year Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt -Consider this example If the rate of payment is 24pa then in any one year the total amount paid is 24, but this payment is spread through the year In half a year the total paid is 24 * ½ = 12 Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt -Consider this example If the rate of payment is 24pa then in any one year the total amount paid is 24, but this payment is spread through the year In half a year the total paid is 24 * ½ = 12 In one month the total paid is 24 * 1/12 = 2 Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt -Consider this example If the rate of payment is 24pa then in any one year the total amount paid is 24, but this payment is spread through the year In half a year the total paid is 24 * ½ = 12 In one month the total paid is 24 * 1/12 = 2 In a very small period dt the total paid is 24dt Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt -Consider this example If the rate of payment is 24pa then in any one year the total amount paid is 24, but this payment is spread through the year In half a year the total paid is 24 * ½ = 12 In one month the total paid is 24 * 1/12 = 2 In a very small period dt the total paid is 24dt And therefore summation through the yr is π‘ 24ππ‘ = 24π‘|10 = 24 0 Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt If the total payment made between times 0 and t is M(t) Then π‘ π π‘ = π π ππ 0 Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt If the total payment made between times 0 and t is M(t) Then π‘ π π‘ = π π ππ 0 And π½ π π½ −π πΌ = π π‘ ππ‘ πΌ Rate of payment per unit time ρ(t) Rate of payment at a moment ρ(t)dt π‘ Total payment from 0 to t π π ππ 0 The integral sums up lots of small payments each of amount ρ(t)dt If the total payment made between times 0 and t is M(t) Then π‘ π π‘ = π π ππ 0 And Why? π½ π π½ −π πΌ = π π‘ ππ‘ πΌ EXAMPLE A 2nd year BBS student, Maestro, develops an investment product and sells it to his peers across Kenya. The young investors pay weekly contributions of KES 500. Assuming that the Maestro sells to 1,000 new clients over each year and that no client stops paying contributions, what will be the rate of contribution income for the student during the first few years? EXAMPLE A 2nd year BBS student, Maestro, develops an investment product and sells it to his peers across Kenya. The young investors pay weekly contributions of KES 500. Assuming that the Maestro sells to 1,000 new clients over each year and that no client stops paying contributions, what will be the rate of contribution income for the student during the first few years? After t years the Maestro will have sold to 1000t new clients EXAMPLE A 2nd year BBS student, Maestro, develops an investment product and sells it to his peers across Kenya. The young investors pay weekly contributions of KES 500. Assuming that the Maestro sells to 1,000 new clients over each year and that no client stops paying contributions, what will be the rate of contribution income for the student during the first few years? After t years the Maestro will have sold to 1000t new clients The weekly contribution income will be 1000t * 500 = 500,000t EXAMPLE A 2nd year BBS student, Maestro, develops an investment product and sells it to his peers across Kenya. The young investors pay weekly contributions of KES 500. Assuming that the Maestro sells to 1,000 new clients over each year and that no client stops paying contributions, what will be the rate of contribution income for the student during the first few years? After t years the Maestro will have sold to 1000t new clients The weekly contribution income will be 1000t * 500 = 500,000t There are 52.18 weeks in a year, Annual rate of contribution income is 52.18 * 500,000t = 26,090,000t EXAMPLE A 2nd year BBS student, Maestro, develops an investment product and sells it to his peers across Kenya. The young investors pay weekly contributions of KES 500. Assuming that the Maestro sells to 1,000 new clients over each year and that no client stops paying contributions, what will be the rate of contribution income for the student during the first few years? After t years the Maestro will have sold to 1000t new clients The weekly contribution income will be 1000t * 500 = 500,000t There are 52.18 weeks in a year, Annual rate of contribution income is 52.18 * 500,000t = 26,090,000t = ρ(t) Calculate the total contribution income that Maestro will have received in the first 3 years. 3 π π‘ ππ‘ 0 Calculate the total contribution income that Maestro will have received in the first 3 years. 3 3 π π‘ ππ‘ 0 = 26,090,000π‘ππ‘ 0 Calculate the total contribution income that Maestro will have received in the first 3 years. 3 3 π π‘ ππ‘ 0 = 26,090,000π‘ππ‘ 0 π‘2 3 = 26,090,000 |0 2 Calculate the total contribution income that Maestro will have received in the first 3 years. 3 3 π π‘ ππ‘ 0 = 26,090,000π‘ππ‘ 0 π‘2 3 = 26,090,000 |0 2 32 = 26,090,000 2 Calculate the total contribution income that Maestro will have received in the first 3 years. 3 3 π π‘ ππ‘ 0 = 26,090,000π‘ππ‘ 0 π‘2 3 = 26,090,000 |0 2 32 = 26,090,000 2 = 117,405,000 NOW, HOW DO WE GET THE PRESENT VALUE The actual amount of payment at time t is ρ(t)dt We get the present value of this amount as v(t)ρ(t)dt Multiply by the discounting factor NOW, HOW DO WE GET THE PRESENT VALUE The actual amount of payment at time t is ρ(t)dt We get the present value of this amount as v(t)ρ(t)dt Multiply by the discounting factor And yet there is a continuous stream of payments between time 0 and t, NOW, HOW DO WE GET THE PRESENT VALUE The actual amount of payment at time t is ρ(t)dt We get the present value of this amount as v(t)ρ(t)dt Multiply by the discounting factor And yet there is a continuous stream of payments between time 0 and t, Therefore we get sum of present values of this stream of payments as π‘ π£ π π π ππ 0 GENERALISATION We can combine the results for discrete and continuous cashflows to have ∞ ππ‘ π£(π‘) + π£ π‘ π π‘ ππ‘ 0 GENERALISATION We can combine the results for discrete and continuous cashflows to have ∞ ππ‘ π£(π‘) + π£ π‘ π π‘ ππ‘ 0 If the interest rate is constant we have ππ‘ π£π‘ ∞ π£ π‘ π π‘ ππ‘ + 0 Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest π −πΏπ‘ = π −0.08π‘ Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest Therefore π −πΏπ‘ = π −0.08π‘ Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest Therefore π −πΏπ‘ = π −0.08π‘ π −0.08 × 0.8 π‘ 5 = 100 | ln π −0.08 × 0.8 0 Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest Therefore π −πΏπ‘ = π −0.08π‘ π −0.08 × 0.8 π‘ 5 = 100 | ln π −0.08 × 0.8 0 100 = π −0.08 × 0.8 −0.08 + ln0.8 π‘ |5 0 Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest Therefore π −πΏπ‘ = π −0.08π‘ π −0.08 × 0.8 π‘ 5 = 100 | ln π −0.08 × 0.8 0 100 = π −0.08 × 0.8 −0.08 + ln0.8 = 100 π −0.08 × 0.8 −0.08 + ln0.8 π‘ |5 0 5 −1 Rate of payment at time t yrs 100 × 0.8π‘ Rate of payment at a moment is 100 × 0.8π‘ ππ‘ Discounting factor using force of interest Therefore π −πΏπ‘ = π −0.08π‘ π −0.08 × 0.8 π‘ 5 = 100 | ln π −0.08 × 0.8 0 100 = π −0.08 × 0.8 −0.08 + ln0.8 = 100 π −0.08 × 0.8 −0.08 + ln0.8 = 257.42 π‘ |5 0 5 −1 NET PRESENT VALUE There are instances where not all the payments are +ve. In other words payments consist of incomes and outgoes of cash. In this case the present value will be referred to as the net present value. Class question VALUING CASHFLOWS Now we are not always interested in obtaining present value at time, t = 0 There are instances where we require the present value at time say t1 where t1 > 0 of a sum due at time t2 We could do this by discounting the cashflow due at time t2 to t = 0, and then accumulating this to time t1 Graphically You could actually say the following process is really just getting the reciprocal of the accumulation factor from time t1 to t2 You could actually say the following process is really just getting the reciprocal of the accumulation factor from time t1 to t2 You could actually say the following process is really just getting the reciprocal of the accumulation factor from time t1 to t2 In other words π£ π‘2 1 1 = π£ π‘1 π΄ π‘1 , π‘2 We can generalise our earlier expression and say that the present value at time t1 of a discrete cashflow ct at time t (for various value of t), and a continuous payment stream at a rate of ρ(t) per time unit is Explain this formula On a timeline On a timeline We need to accumulate the payments due before the present value date 1/1/2007 and discount payments due after the present value date 1/1/2007 On a timeline We need to accumulate the payments due before the present value date 1/1/2007 and discount payments due after the present value date 1/1/2007 The simplest way to do this is to discount all the payments to time 0 and then accumulate this value to time 2, as follows 100π£ 1 + 130π£ 2 + 150π£ 4 + 160π£ 5 π£ 3 where On a timeline We need to accumulate the payments due before the present value date 1/1/2007 and discount payments due after the present value date 1/1/2007 The simplest way to do this is to discount all the payments to time 0 and then accumulate this value to time 2, as follows 100π£ 1 + 130π£ 2 + 150π£ 4 + 160π£ 5 π£ 3 where 100 0.93 + 130 0.92 + 150 0.84 + 160 0.65 0.84 = 486.38 PAYMENT STREAM WITH FORCE OF INTEREST We have seen how to get the present value of a continuous payment stream as PAYMENT STREAM WITH FORCE OF INTEREST We have seen how to get the present value of a continuous payment stream as Consider a continuous payment stream paid at a rate of ρ(t) from time a to b during which time the force of interest is δ(t). The present value at time a of this payment is π π π‘ ππ‘ × π − π π‘ ππΏ π ππ PAYMENT STREAM WITH FORCE OF INTEREST We have seen how to get the present value of a continuous payment stream as Consider a continuous payment stream paid at a rate of ρ(t) from time a to b during which time the force of interest is δ(t). The present value at time a of this payment is π π π π‘ ππ‘ × π − π π‘ ππΏ π ππ π π‘ π− π π‘ ππΏ π ππ ππ‘ How would the formula look if we were to get the accumulated value at b? Present value as a π π π‘ π π‘ π − π πΏ π ππ ππ‘ How would the formula look if we were to get the accumulated value at b? Present value as a π π π‘ π‘ π − π πΏ π ππ ππ‘ π Accumulated value at b π π π‘ π π π πΏ π‘ π ππ ππ‘ Payment stream starts as 5 and ends a 10, and rate of payment Payment stream starts as 5 and ends a 10, and rate of payment Force of interest Payment stream starts as 5 and ends a 10, and rate of payment Force of interest We therefore have the accumulated value at t = 10 as Payment stream starts as 5 and ends a 10, and rate of payment Force of interest We therefore have the accumulated value at t = 10 as = 86.699 AND LASTLY, INTEREST INCOME An investor who wishes to receive income deposits $1000 in a bank account that pays an effective rate of interest 8% per annum. The interest income is paid to the investor at the end of each year. How much is each payment? 8% * 1000 = $80 Here the investor is not looking to accumulate his cash but instead to receive a regular interest income Note that his capital remains intact and will not be depleted How about if interest is paid continuously with force of interest δ(t)at time t If the initial investment is C then Interest income at time t will be Cδ(t)dt at time t for a very small duration dt And therefore total interest income from time 0 to T AND LASTLY, INTEREST INCOME How about if interest is paid continuously with force of interest δ(t)at time t If the initial investment is C at time 0, then Interest income at time t will be Cδ(t)dt at time t for a very small duration dt And therefore total interest income from time 0 to T If the investor withdraws the capital at time T then the present values of interest and capital are Show that the present values of interest + capital is 2000 [the original amount invested] EXTREMELY IMPORTANT WILL GO THROUGH IN MAKE UP CLASSES Examples you must do Pg 14, Pg 17 Questions you must do 5.11, 5.13, 5.15, 5.16,
