Assignment 4 Solutions

Transcription

1 Assgnment 4 Solutons Tmothy Vs February 3, We have n 25, 0.04, P MT 100, and we need to fnd F V. F V P MT (1 + )n ( ) We have F V 8000, n 30, 0.03, and we need to fnd P MT. P MT F V (1 + ) n ( ) We have F V 8000, 0.04, P MT 500, and we need to fnd n. F V P MT (1 + )n ( )n n (8000)(0.04) n 1.64 n ln 1.04 ln 1.64 n ln 1.64 ln

2 We have P MT 7500, r 0.08, t 20 n, and we need to fnd F V. F V P MT (1 + )n ( ) , We have F V 0, 000, t 15, r 0.068, m 2, so that n (15)(2) 30, and , and we need to fnd P MT. P MT F V (1 + ) n , 000 ( ) We have P MT 100, r 0.06, m, so that and n t, and we need to fnd the nterest after each year. We need then to fnd the fnal value after each year. F V P MT (1 + )n 1 ( )t After one year, ths equates to 33.56; after two years to ; after three years to The nterest earned each year s gven by subtractng from that year s fnal value the fnal value of the precedng year and the total contrbutons for that year, whch wll be (m)(p MT ) ()(100) 00 n each case. So, for the frst year, we have For the second year, we have For the thrd year, we have I I I

3 Ths problem s really two parts. We frst use our annuty equaton to determne how much s n t when he makes hs last contrbuton, and then use our compound nterest equaton to determne what that grows to after hs last contrbuton. For the frst part, we have P MT 1000, r 0.064, t n, and we need to fnd F V. F V P MT (1 + )n ) , We use ths value of F V as the value of P n the second part, where we also have r 0.064, t n, and we need to fnd A. A P (1 + ) n 17, ( ) , Here we have P MT 1000, n 30, r 0.064, and we need to fnd F V F V P MT (1 + )n ( ) , Here we have P MT 2000, r 0.066, m, F V 100, 000, so that , and we need to fnd n, where n s the number of months necessary. F V P MT (1 + )n 1 100, ( )n ( ) n (100, 000)(0.0055) n ln ln n ln ln

4 Here we have P MT 2000, n 6, and F V 14, 000, and we need to approxmate. We then set up the followng equatons n the graphng calculator: Y 1 P MT (1 + )n (1 + x)6 1 x Y 2 F V 14, 000 Havng the calculator solve for the ntersect yelds an ntersect where x , correspondng to a rate of 6.14% Here we have n 40, 0.01, P MT 400, and we need to fnd P V. 1 (1 + ) n P V P MT 1 ( ) , Here we have P V 00, n 40, 0.025, and we need to fnd P MT. P MT P V 1 (1 + ) n ( ) Here we have P V 20, 000, , P MT 500 and we need to fnd n. 1 (1 + ) n P V P MT 20, ( ) n ( ) n (20, 000)(0.0175) (1.0175) n 0.3 n ln ln 0.3 n ln 0.3 ln

5 Here we have n 7, , P MT 10, 000, and we need to fnd P V. 1 (1 + ) n P V P MT 1 ( ) 7 10, , Here we have n 36, r , m, P M T 350, so that , and we need to fnd P V and the total nterest pad. 1 (1 + ) n P V P MT (1 + ) , The amount of nterest pad s determned by subtractng the present value (amount borrowed) from the total of the payments (amount pad). I np MT P V (36)(350) 10, , Here we have P V 2500, n 48, 0.05, and we need to fnd P MT. P MT P V 1 (1 + ) n ( ) The nterest pad s determned by subtractng the present value (amount borrowed) from the total of the payments (amount pad). I np MT P V (48)(69.58) To determne f ths s really 0% fnancng we need only check that the payments add up to the purchase prce for the sedan. If the total of the payments equals the purchase prce, we can conclude that ths s really 0% fnancng; f not, we can determne the actual fnance rate. np M T (72)(179), 888 > Obvously, ths advertser should be sued for false advertsng, because the total amount pad s far more than the purchase prce. We now determne both the monthly payment that would be 0% fnancng and we approxmate the nterest rate beng charged on the payments as advertsed. At 0% fnancng, our 5

6 monthly payment s smply We approxmate the nterest rate on a calculator by settng up equatons as follows, notng that r 1 (1 + ) n Y 1 P MT (1 + x ) 72 x Y 2 P V 9330 Havng the calculator solve for the ntersect yelds an ntersect where x 0.195, correspondng to a rate of 11.29% Here we have two optons. Both have n 60, so the better opton wll be, smply, the one wth the lower monthly payment. For the frst opton, 0% fnancng, we have n 60, P V 28, 500, and 0, and we need to fnd P MT. P MT P V n 28, For the second opton, we have n 60, P V 28, 500 6, , 500, and 0.062, and we need to fnd P MT. P MT P V 22, (1 + ) n ( ) 60 The second opton s the one to choose. It saves monthly, or (37.92)(60) over the lfe of the loan We are asked to construct the amortzaton schedule for a debt of 5000 that s to be amortzed n 8 quarterly payments at 2.8% nterest per quarter on the unpad balance. We frst determne the payment amount, wth n 8, P V 5000, P MT P V 1 (1 + ) n ( )

7 Further, the nterest pad for each payment s equal to the nterest rate multpled by the prevous unpad balance. Ths leads to the followng table: Unpad Payment Payment Interest Balance Unpad Number Reducton Balance 0 5, Totals Ths s a two part problem. We frst determne the value of the fund at retrement (after the 20 years of contrbutons) usng the future value technques of secton 3-3. Ths gves us the present value for the amortzaton of the retrement fund durng retrement (the 20 years of payouts). For the frst part, we have P MT 7500, n 20, 0.09, and we need to fnd F V. F V P MT (1 + )n ( ) , For the second part, we now have P V 383, , n 20, 0.09 and we need to fnd P MT. P MT P V 1 (1 + ) n , ( ) 20 42, Here we have P V 1000, n, P MT 90, and we need to approxmate r. Note that r. We set up the followng equatons on the calculator: 1 (1 + ) n Y 1 P MT 7

8 90 1 (1 + r ) r Y 2 P V 1000 Havng the calculator solve for the ntersect yelds an ntersect where x , correspondng to an nterest rate of 14.45%. 8