Alifia Farradita Fedli / 119410006 1.2. The Accumulation and Amount Function [PDF]

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Alifia Farradita Fedli / 119410006 1.2. The accumulation and amount function 1.

consider the amount function A(t) = t2+2t+3. a) find the corresponding accumulation function a(t). Answer : A(t) = t2+2t+3 and A(0) = 3 = k So that A (t) A(t ) 1 2 a(t) = = = (t +2t+3). A (0) 3 k b) verify that a(t) satisfies the three properties of an accumulation function. Answer : (1) a(0) = 1 >>> a(0) = (2) a(t) =

1 2 (t +2t+3). 3

1 1 [(0)2 + (0)+ 3] = (3) = 1. 3 3

1 (2t+2) > 0 , for t≥ 0 3 so that a(t) is a an increasing function. (3) a(t) is continuous if : lim a( x ) = lim ¿ = lim ¿ x→ t−¿ a(x)¿ x→ t+ ¿a (x)¿ x→ t a’(t) =





a(t) >> example : t = -1 2 a(-1) = 3 lim a( x )



>> example : t = -1 lim 1 lim a(x ) = x →−1 (t2+2t+3). x→−1 3 1 = [(-1)2+2(-1)+3] 3 2 = 3 lim ¿

x→ t

x→ t−¿ a(x)¿

>> example : t = -1 lim ¿ = lim x→−1−¿a (x)¿

1 x→−1−¿ ¿ 3

¿

(t2+2t+3).

1 [(-1)2+2(-1)+3] 3 2 = 3 =



lim

¿

x→ t+ ¿a (x)¿

>> example : t = -1 lim ¿ = lim 1 ¿ (t2+2t+3). x→−1+¿ a(x)¿ x→−1+¿ ¿ 3

1 = [(-1)2+2(-1)+3] 3 2 = 3 lim ¿ = lim ¿ = a(t) = 2 , so a(t) is a a( x ) = Because lim x→ t−¿ a(x)¿ x→ t+ ¿a (x)¿ x→ t 3 continuous function. c) find In Answer : In = A(n) – A(n-1) = [n2 + 2n + 3] – [(n-1)2 + 2(n-1) + 3] = n2 + 2n + 3 – [n2 – 2n + 1 + 2n – 2 + 3] = n2 + 2n + 3 – n2 + 2n – 1 - 2n + 2 – 3 = 2n – 1. 2. a) prove that A(n) – A(0) = I1+I2+....+In. Answer : I1+I2+....+In = A(1) – A(0) + A(2) – A(1) +..........+ A(n) – A(n-1) , n ≥ 1 = A(n) – A(0). b) verbally interpret the result obtained in (a). Answer : RHS adalah jumlah bunga bunga yang diperoleh selama setiap periode n. LHS adalah kenaikan dana selama n periode, yang seluruhnya terkait dengan bunga yang diperoleh. 3. For the $5000 investment given Example 1.1, find the amount of interest earned during the second year of investment. i.e. between times t = 3 and t = 4. Answer :  From the example 1.1, we know that A(2) = 11.130,00. A(3) = 11.575,20. A(4) = 12.153,96.  Let K be the accumulated value of the $5000. K canbe determined by ratio and proportion. A ( 4 )− A(3) K 12.153,96−11.575,20 → K = 5000 = = $260. 5000 12.153,96 . A (2) 4. It is known that a(t) is of the form at2 + b. If $100 invested at time 0 accumulates to $172 at time 3,find the accumulated value at time 10 of $100 invested at time 5. Answer : Diket :  a(t) = at2 + b.  $100, at time 0.

 $172, at time 3. Dik : find the accumulated value at time 10 of $100 invested at time 5 ? ≫From a(t) = at2 + b, we can subsitution t = 0 dan t = 3, and elimination of the equation. a (0) = b = 100 a(3) = 9a + b = 172 ≫From the equation above, it is known that the values of a=8 and b=100. Thus, a (5) = 25a + b = 25(8) + 100 = 300. a(10) = 100a + b = 100(8) + 100 = 900. And the answer is 100

a(10) 900 = 100 = $300. 300 a(5)

1.3. The Effective rate of interest 5. Asumme that A(t) = 100 + 5t. a) Find i5. Answer : A (5) – A (4) 100+5 (5 ) – [100+5 ( 4 ) ] 125−120 1 I5 = = = = . 120 24 A ( 4) 100+5 (4) b) Find i10. Answer : A (10) – A (9) 100+5 (10 ) – [100+5 ( 9 ) ] 150−145 1 = = = . 145 29 A (9) 100+5 (9) t 6. Assume that A(t) = 100 (1.1) . a) Find i5. Answer : A (5) – A (4) 100(1.1)5 – 100(1.1)4 161.05−146.41 14.64 I5= = = = = 0.1. 146.41 146.41 A ( 4) 100 (1.1)4 b) Find i10. A (10) – A (9) 100[ (1.1 ) 10 – ( 1.1 ) 9] 11−9.9 I10 = = = = 0.1 9.9 A (9) 100[ ( 1.1 ) 9] 7. Show that A(n) = (1 + in) A(n – 1),where n is positive integer. Answer : A (n)– A( n−1) From formula : In = A (n−1) So that , A(n) – A (n−1) = In [ A(n−1)¿ A(n) = In [ A(n−1)¿ + A(n−1) A(n) = (in + 1) A(n – 1). I10 =

8. If A(4) = 1000 and in = 0,1n, where n is a positive integer, find A(7). Answer :

A(7) = A(4)(1+i5)(1+i6)(1+i7) = 1000 (1.05)(1.06)(1.07) = $1190.91. 1.4. Simple Interest 9. a) At what rate of simple interest will $500 accumulate to $615 in 2 ½ years? Answer : Diket : - k = $500 - AC = $615 - t = 2.5 Dit : i ? AC = k. (1+ it) 615 = 500 (1+ i2.5) 615 = 500 + 1250i 1250i = 115 115 i= = 0.092 or 9.2% 1250 b) In how many years will $500 accumulate to $630 at 7.8% simple interest? Answer : Diket : - k = $500 - AC = $630 - i = 7.8% = 0.078 Dit : t ? AC = k. (1+ it) 630 = 500 (1+0.078t) 630 = 500 + 39t 39t = 130 130 1 t= = 3 years. 39 3 10. At the certain rate of simple interest $1000 will accumulate to $1110 after a certain period of time. Find the accumulated value of $500 at a rate of simple interest three fourths as great over twice as long a period of time. Answer : Diket : - AC1 = $1110. k1 = $1000 k2 = $500 3 i2 = i1 4 t2 = 2t1 Dit : AC2 ? >>AC1 = k1. (1+ i1t1) 1110 = 1000 (1+ i1t1) 1110 = 1000 + 1000 i1t1 1110 i1t1 = = 0.11. 1000 >> AC2 = k2. (1+ i2t2)

3 i1.2t1) 4 3 = 500 [1+ (0.11)] 2 = $582.5 = 500 (1+

11. Simple interest of i = 4% is being credited to a fund. In which period is this equivalent to an effective rate of 2 ½ % ? Answer : Diket : - i1 = 4% = 0.04 - i2 = 2.5% = 0.025. Dit : n? i 0.04 in = and 0.025 = 1+ i(n−1) 1+ 0.04( n−1) so that 0.025 + 0.001(n−1) = 0.04, and n =16. 12. A deposit of $1000 is invested at simple interest at time t = 0. The rate of simple interest during year t is equal to 0.01t for t = 1,2,3,4, and 5. Find the total accummulated value of this investment at time t = 5. Answer : AC = k1 (1+0.01t1) + k2 (1+0.01t2) + k3 (1+0.01t3) + k4 (1+0.01t4) + k5 (1+0.01t5) >> Because k1 = k2 = k3 = k4 = k5 = 1000, so AC = k [(1+0.01t1) + (1+0.01t2) + (1+0.01t3) + (1+0.01t4) + (1+0.01t5)] = 1000 [(1+0.01(1)) + (1+0.01(2)) + (1+0.01(3)) + (1+0.01(4)) + (1+0.01(5))] = 1000 (1.01 + 1.02 + 1.03 + 1.04 + 1.05) = 1000 (1.15) = $1150 1.5. Compound Interest 13. It is known that $600 invested for two years will earn $264 in interest. Find the accumulated value of $2000 invested at the same rate of compound interest for three years. Answer : AC1 = k. (1+ i)t 600 + 264 = 600 (1 + i)2 864 = (1 + i)2 600 864 1+i =√ 600 i = 1.2 – 1 i = 0.2. So that AC2 = k. (1+ i)t = 2000 (1 + 0.2)3

= $3456 14. Show that the ratio of the accumulated value of 1 invested at rate I for n periods, to the accumulated value of 1 invested at rate j for n periods. i > j, is equal to the accumulated value of 1 invested for n periods at rate r. Find an expression for r as a function of i and j. Answer : >>a(n)i = (1+ i)n >>a(n)j = (1+ j)n >>a(n)r = (1+ r)n (1+i)n 1+i (1+ r)n = and 1+ r = 1+ j (1+ j)n So that, 1+i ( 1+i ) −(1+ j) i− j r= -1= = 1+ j 1+ j 1+ j 15. At a certain rate of compound interest, 1 will increase to 2 in a years, 2 will increase to 3 in b years, and 3 will increase to 15 in c years. If 6 will increase to 10 in n years, express n as a function of a, b, and c. Answer : From the information given : (1 + i)a = 2 >> (1 + i)a = 2 >> (1 + i) log 2 = a 3 3 2(1 + i)b = 3 >> (1 + i)b = >> (1 + i) log = b 2 2 c c (1 + i) 3(1 + i) = 15 >>(1 + i) = 3 >> log 5 = c 10 10 6(1 + i)n = 10 >>(1 + i)n = >> (1 + i) log =n 6 6 Find n:

10 6 = (1 + i) log 10 - (1 + i) log 6

n = (1 + i) log

= (1 + i) log 5 + (1 + i) log 2 – [(1 + i) log

3 (1 + i) + log 4] 2

= c + a - b – 2a n=c–b–a 16. An amount of money is invested for one year at a rate of interest of 3% per quarter. Let D(k) be the difference between the amount of interest carned on a compound interest basis and on a simple interest basis for quarter k, where k = 1,2,3,4. Find the ratio of D(4) to D(3). Answer : D(4) : simple = 0.03. Compound = (1.03)4 - (1.03)3 D(3) : simple = 0.03. Compound = (1.03)3 - (1.03)2 Find the ratio of D(4) to D(3) : 4 3 D(4) ( 1.03 ) − (1.03 ) −0.03 . = = 1.523. D(3) ( 1.03 )3−( 1.03 )2−0.03

1.6 Present Value 17. The two sets of granparents for a newborn baby wish to invest enough money immediately to pay $10,000 per year for four years toward college costs starting age 18. Grandparents A agree to fund the first two payments. If the effective rate of interest is 6% per annual. find the difference between the contributions of Granparents A and B. Answer : A : 10000 [(1.06)-18 + (1.06)-19] = 6808.57 A : 10000 [(1.06)-20 + (1.06)-20] = 6059.60 Difference : 6808.57- 6059.60 = $748.97. 18. The sum of the present value of 1 paid at the end of n periods and 1 paid at the end of 2n periods is 1. Find ( 1 + i )2n. Answer : PV1 + PV2 = 1 (1+i)-n + (1+i)-2n = 1 1 1 >> =1 −n + ( 1+i ) ( 1+i )−2n >> (1+i)2n - (1+i)n -1 = 0 −b ± √b 2−4 ac 1± √ 1+ 4 1± √ 5 (1+i)n = = = 2a 2 2 Because value of (1+i) must ≤1 , so that 1+ √5 (1+i)n = 2 Finally, 1+ √5 2 3+ √ 5 (1+i)2n = ( ) = . 2 2 19. It is known that an investment of $500 will increase to $4000 at the end of 30 years. Find the sum of the present values of three payments of $10,000 each which will occur at the end of 20,40,an 60 years. Answer : 4000 = 500(1+i)30 (1+i)30 = 8 10000(v20+ v40 + v60) = 10000(8-2/3 + 8-4/3 + 8-2) 1 1 1 = 10000 ( + + ) 4 16 64 = $3281.25 1.7. The Effective rate of discount 20. a) Find d5 if the rate of simple interest is 10%. Answer : The formula : a(t) = 1 + it , i = 0.1.

I 5 a ( 5 )−a (4 ) 1.5−1.4 0.1 1 = = = = A5 1.5 1.5 15 a(5) b) Find d5 if the rate of simple discount is 10%. Answer : The formula : a-1(t) = 1 - it = 1 - dt, d = 0.1 d5 =

1 1 − −1 −1 I 5 a ( 5 )−a (4 ) ( 1−0.5 ) −( 1−0.4 ) 0.5 0.6 1 d5 = = = = = . −1 1 A5 6 a(5) ( 1−0.5 ) 0.5 21. Find the effective rate of discount at which a payment of $200 immediately and $300 one year from today will accumulate to $600 two years from today. Answer : 200 + 300(1 – d) = 600(1 – d)2 2 + 3 – 3d – 6 + 12d – 6d2 = 0 6d2 - 9d + 1 = 0 Use the quadratic formula to find the solution. −b ± √b 2−4 ac −(−9) ± √ (−9 ) 2−4.(6.1) 9 ± √ 57 d = = = 2a 12 2.6 Because value of d must ≤1 , so that 9− √ 57 d= = 0.1208 = 12.08% 12 22. the amount of interest earned on A for one year is $336, while the equivalent amount of discount is $300. Find A. Answer : 336 Amount of interest : iA = 336 → i = A 300 Amount of discount : dA = 300 → d= A d 336 300 / A i= and = 1−d A 1−300/ A So that 336 (A – 300) = 300A 36A = 100,800 A = $2800 23. Find the present value of $5000 to be paid at the end of 25 months at a rate of discount of 8% convertible quarterly : a) Assuming compound discount throughout. Answer : 25 Dik : t = 25 months. Because it is convertible quartely, so t = . 3 0.08 i = 0.08 . Because it is convertible quartely, so i = = 0.02. 4 the exact answer is 5000v25/3 = 5000(1 – d)t = 5000(1 – 0.02)25/3 = $4225.27.

b) Assuming simple discount during the final fractional period. Answer : 1 5000v8 (1- d) = 5000 (1 – 0.02) 8[1- ¿)(0.02)] = $4225.46. 3 24. Show that

(i−d) 2 d3 = (1−d) 2 1−v

Answer : LHS = RHS d3 i3 v 3 = 2 = i3v = i2d.. and (1−d) 2 v ( i−d )2 ( id )2 2  RHS: = = i d. 1−v d 25. If i and d are equivalent rates of simple interest and simple discount over t periods, show that. i – d = idt. Answer : Simple interest : a(t) = 1 + it Simple discount : a-1(t) = 1 – dt 

LHS:

1 1 – dt 1 – dt + it - idt2 = 1 it – dt = idt2 i – d = idt 1 + it =