AMB111F Tut 10 Solutions

Transcription

1 AMB111F Tut 10 Solutions Question One 1.1 (a) Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. When n = 6 yeas, A n = 9500, and = 8, then 9500 = P (1 + 8 )6. This implies P = 9500 = R (1.08) 6 (b) Amount afte n yeas is A n = P (1 + n ) whee = ate and P is the pincipal amount initially invested. When n = 6 yeas, A n = 9500, and = 8, then 9500 = P ( ). This implies P = 9500 = R (1.48) 1.2 Amount afte n yeas is A n = P (1 + the = [(A n /P ) 1/n 1] = [(1450/910) 1/8 1] = Amount afte n yeas is A n = P (1 + the = [(A n /P ) 1/n 1] = [(2P/P) 1/10 1]= Amount afte n yeas is A n = P (1 + the = [(A n /P ) 1/n 1] = [(2P/P) 1/7 1] = Value afte n yeas is V n = P (1 the initial value. Hee P = 00; n = 5 yeas; = 10, thus V 5 = 00(1 10 )5 = R

2 1.6 (a) Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. When = 12 and n = 2 yeas, A 2 = P ( )2 = P (1.254) When =11.6and n = 2 yeas = 8 quates, A 2 = P ( )2 = P (1.257) > P(1.254) So 11.6% p.a. calculated quately is bette. (b) Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. When = 12 and n = 5 yeas, A 5 = P ( )5 = P (1.7623) When =11.6 and n = 5 yeas = 20 quates, A 5 = P ( )20 = P (1.771) > P(1.7623) So 11.6% p.a. calculated quately is bette. Question Two 2.1 Value afte n yeas is V n = P (1 the initial value. Hee P = 2025; n = 10 yeas = 20 half yeas; =10.5/2 pe half yea, thus V 10 = 2025( )20 = R Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. Thus log(a n /P )=nlog(1 + ), implying n = log(an/p ) = log(2250 2/2250) = quates = 8 yeas. log(1+ ) log( ) 2.3 Value afte n yeas is V n = P (1 the oiginal value. Thus log(v n /P )=nlog(1 log(vn/p ) ), implying n = = log(1 ) log( 1 3 P/P) log( ) =8.44 yeas. 2.4 Population afte n yeas is P n = P (1 + the oiginal population figue. Thus log(p n /P )=nlog(1 + ), implying the peiod is n = log(pn/p ) = log(200000/125324) =9.58 yeas. Thus the log(1+ ) log(1+ 5 ) population will exceed in

3 2.5 Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. Thus log(a n /P )=nlog(1 + ), implying peiod is n = log(an/p ) log(1+ ). (a) n = log(800/512) =13.45 quates = 3.36 yeas. log( ) 400 (b) n = log(800/512) =39.9 quates = 3.32 yeas. log( ) Amount afte n yeas is A n = P (1 + the pincipal amount initially invested. Thus (A n /P ) 1/n =(1+ ), implying = [(100/30000) 1/10 1] = Since > 12.5, the appeciation ate has kept up with the aveage inflation ate. Question Thee 3.1 (a) FV = P [(1 + )n 1] whee P is a fixed peiodic deposit. Theefoe Futue value is FV = 2000 [( )10 1] = R (b) Futue value is FV = P (1 + /)[(1 + )n 1] = 2000 ( /)[(1 + 5 )10 1] = R (c) FV = P [(1 + )n 1] whee P is a fixed peiodic deposit. Theefoe Futue value is FV = [( )30 1] = R (d) FV = P deposit. Theefoe Futue value is FV = R (1 + /)[(1 + )n 1] whee P is a fixed peiodic (1 + 12/200)[( )30 1] = 3.2 (a) 5 deposits (b) FV = A = P (1 + /)[(1 + )n 1] = 5000 ( /)[( )4 1] = R Thus she does not have 3

4 enough money fo the deposit since she needs R Inteest = R( ) = R (a) FV = PV(1 + /) n FV implies PV = = 000 = (1+/) n (1+8/) 20 R (b) FV = PV(1 + /) n FV implies PV = = 000 = (1+/) n (1+8/200) 40 R (c) FV = PV(1 + /) n implies PV = R (d) FV = PV(1 + /) n implies PV = R FV (1+/) n = FV (1+/) n = 000 (1+8/400) 80 = 000 (1+8/1200) 240 = 3.4 Fixed deposit = P = FV [(1+ = 12(525000) )n 1] [(1+ 12 )5 1] = R FV = P (1 + /)[(1 + )n 1] implies fixed deposit = P = = R FV (1+/)[(1+ = 12(525000) )n 1] ( )[(1+ )5 1] 3.6 FV = PV(1+/) n FV implies PV = and also FV = P [(1+ (1+/) n )n 1] = [( )30 1] = Theefoe PV = FV = = R (1+/) n (1+8/) 30 Question Fou 4.1 (a) FV = P [(1 + )n 1] whee P is a fixed peiodic deposit. Theefoe Futue value is FV = [( )50 1] = R (b) FV = PV(1 + )n whee PV is a single deposit. Theefoe Futue value is FV = 3750( )50 = R Theefoe (b) yields a bigge etun. 4

5 4.2 The total cost of epayment is just like a futue value in an annuity with pesent value PV = Now PV = P [(1+ )n 1] (1+/) n implies P = PV((1+/) n ) = R [(1+ )n 1] = PV ((1+/)n ) [(1+ = (8/1200)750000((1+8/1200)288 ) )n 1] [( )288 1] = monthly epayments. Thus the total payable = R = R (Note that 24 yeas = 288 months). 4.3 (a) Let PV = R Then the futue value of the machine is FV = PV(1 + /) n = (1 + 13/) 11 = R = Replacement Cost (b) Scap value is SV = PV(1 /) n = (1 9/) 11 = R (c) Now FV - SV = R R = R is the equied value (SF) of the sinking fund. Conside SF = P (1 /)[(1 + )n 1] whee P = fixed instalment. Thus SF P = = R ( = [(1+ )n 1](1+/) 1200[( )132 1](1+14/1200) monthly instalments payable). 4.4 (a) Scap value is SV = PV(1 /) n = (1 10/) 8 = R (b) Let PV = R Then the futue value of the machine is FV = PV(1 + /) n = (1 + 18/) 8 = R = Replacement Cost (c) Now FV - SV = R R = R is the equied value (SF) of the sinking fund. (d) Conside SF = P (1 + /)[(1 + )n 1] whee P = fixed instalment. Thus P = = = SF [(1+ )n 1](1+/) 1200[( )96 1](1+12/1200) R ( = monthly instalments payable). END 5