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HW__7_Solutions

# HW__7_Solutions - 3.1.4 The payments are 310 at time 1 305...

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Unformatted text preview: 3.1.4: The payments are 310 at time 1, 305 at time 2. . . . , 255 at time 12. we can write this present value as 250aﬁl‘02 + 5(Do}ﬁl_m = 2, 643.84+356.16 = 3, 000 3.1.6: Using the retrospective form of the outstanding balance equation, the outstanding balance is the accumulated value of the loan minus the accu— mulated value of the payments made so far. Using the prospective form? the outstanding balance is the PV of the remaining loan payments. I think that for this problem, the prospective form is slightly easier [it doesnit re— quire solving for the original loan amount), so this is the approach we’ll take here. The monthly effective interest rate is % = 0.00T5. Counting time in months, the loan payment at time 1 is 1000, the loan payment at time 2 is 1000(0.08)1, the loan payment at time 3 is 1000(0.08)2, . . ., the loan pay— ment at time 41 is 1000(_ 0.08)“, the loan payment at time :12 is 1000(0.08]41, . . . , and the loan payment at time 60 is 1000(0.08)59. Then the P1", at time £10, of the payments at times 41 through 60, is: 0.340 = 11100111.98)J‘001 + IUDUULQSFW + - -- +1000[0.001511-020 ' D 1 10 = 10011121001314D “'98 1 {1-98 + _ _ _ + 0.98 ‘ ' 1.0075 1.0015 1.0015 1 1— ( 03:" )20 = 1000 0.08 1” (1.0015) 1 1 1 1— ME = 6. 889.11 3.1.7: There 1110. 11 few 111113111 [.0 solve this 11101110111. 0110. 1111}? would be to (-011— Hm‘uc-L 1111 11111011112111.1011 1.0.1110: Time. Payment 1111.. Duo P1111. 11.01.311.111 Outsmuulillg 13111111100. — 1000 10110 1001} 951] = IOUUU — .05) 9112.5 2 1(100[1 — .05}2 T111311 the. outﬁtmlding 11111111100 for 1.110 HOC‘Olld 112]| 13211111101115 forum 11, go.— 011101110 1310310551011. 50 13110.1; 0ng = 1{][1[1[1 — .05)” = 598.?4. This 0111;— Hmulding 11111111100 is the PM” {at 1.11110. 2(1) of the 12151. 11] 11113111101103: 598.?4 = Xaﬁlm =1» X = 914-1. {ht-2. 0 ( fl ‘3 W9 {06m 20’0-00 M x X X Pavmnis % (c) 10,000 2 (2x + qu05% 05” 2 10,000 qlx :: {q'é§(.?% [rewospecﬁue 'Rm) (3) -L I t {1 20,000 - Xamo‘wu + quOW/o (was) 20,000 2 X({(.%o7)+ 9((223760 GEN. 7" canola 5‘: [Sf i ‘ qI (PTO-Spec‘f‘f‘ve ﬁrm) OE! = 930 r {912‘ “39 OEO : 0%; m; : \$62.00 w "L :2 OEJM‘ -=7 C: I. ((3/0 ODS oat, 362.00 1'; : OEmkc 1.. 706M009: 23130 £412?“me Q W ,Ao 4C2‘ﬁ2k3 :l’ﬂ 2K5 Kai “EPIZ‘ 2: HIE/o +(ééoo «:mxo PA =2 K7, IL : (63% (Ni I oar PL 2 we- (gage g W120 Is =1 Oﬁﬁm‘r— S‘VZZOHOoQE 27.” We ; “3,13 : mac! OE; n©BL~P23a 92.20 A mm c \$70.2: I; r:— Ong : @70r21’r(o.05): (<55: m 2 ha a £803“? 08;, "z 08;,» PL! : E7021 430-501 : ig‘MZ L I 0AM 2 l8?\€7'*(0~0§)‘ 0W8 P723: Ire/1‘5 21mm ?¢/“\m/U7) Oggﬁogq’PE\$:O ‘(j Pﬁymevx" 0 1 WHO Z IO 5 WHO q (CHM) S M‘iJO j:an wer 9% Wmo @3790 7/7.?! ((6—51 “H18 ph‘ﬂﬁikad 1243f)qu (4400 (£3.30 OUdS‘fqﬁdhjﬂ Ea lance \$6100 706.00 €¥220 7770. ‘24 Wsz ngwﬂmwma mmmw (Q) (Damp/h , M Q) Wags? MMzﬂw 76k MJM 76k wmg MngJMQ, “ZZ.%77SP Waéﬁﬁg <9wa 71J%77§F’:»Paﬁw% ("r—“37 n f. 193:3”? "1% 22'9775-2Q7‘790/0 ‘5 40—76% m M M to (feta/M76" fut Wﬂ‘WﬂWngw/QﬁwmﬁMQ/q, Mom] l! (2/007 ’L WWW on june L Z007. 100 v +“LOOV;+~-- + 200 V” a 2000,?“ um wt 1,26%?» Mme c (ﬂew ZOOO :: Pam‘so’l‘ﬁ’o “'7 '91 (EH, Arc ﬁg»; W‘*é 762“ Lo 1003?) ~= 20W) ﬁx 0PM (a), m 12» M L0 200 mom (meow-m M m We "boo/\wmm W) 7km W ,q) 100+(24900»7/OO)C V M Ww z» (mama): ,W ﬁg 5:“; 5. W Wham» Aﬁmoe @7762? WWW- MWWWWMM 709% W782» 10m; 20170 : 1000+ Moo; w}? TEE 7k W; a,” {(000.000 (0.06) : €0,000 d m { 300,000010679 Lalo/000 ed? Rig 2 €00,006 (0.05): 9000 605% 3 “DO/DOWW: 7/0000 d M £9 2001000 (0'05): {0/000 00% S TLFV¢éﬂmaQwM Lo {0.000(Da>'?l§% 3" PV W L4) llgﬂmqam : w ae a: Mia a W Pam Wmnﬂlw, Mm, 7% www Pam; 7mmmm m Eq‘PamL a Rm 3» }r2~'\ : )“Vn £:2? gﬂgﬁlqm mL/ﬂLmﬁ MWRW w W tam W a P(I»f““”“).mmmwg 3.241(3): To calculate the interest due in the 1*“ period. we compute the interest on the first \$5. 000 as \$5. 000 4'- 8% = \$400 and the interest on the remainder as \$5. 000 =+= 1055?. = \$500. Then the total interest due for the ﬁrst period \$400+ \$500I = \$000. Since the total payment \$1.500. then the amount of principal repaid in the first period \$1.500 — \$000 = \$000. and the ontstrmding balance after the first payment \$10. 000 — \$000 2 \$0. 400. Continuing in this manner gives the following zunortization table {note that all of the payments are \$1. 500 except for the last one. which is a smaller amount] : Payment Int. Due Prin. Repaid Ontstrmding Balance 0 — — 10000 1 1500 000 000 0400 3 1500 T74 4 1500 701.4 T2154 5 1500 021.54 8? .40 0330.04 0 1500 533.00 00 131 53T0.03 T 1500 43?.00 1002.04 4307.? 8 1500 344.02 1155.38 3152.31 0 1500 10 1500 ' ..' '" i - .'"- 550.80 11 001.41 -' - " '1' '" 0 A @Mﬁcr Mk «WM % W WM: 10,000 : 1009mm ‘ (mw 433qu m 07% Wk“): T70 “MAMA; 7%.301 man/25. w 5w (0,0000% :: (ZS. 0 ‘ 7. 30 Loam \$50,005 Li Mfg; 7,6. 000 \$2000 15,700 0 Sim” ﬁnd Q, 095qu 6,5359! G, 035. CH 250,000 (0.0: 2/3000 ,‘ AW m W W M W W (it %l,03§.0?) w W 7% mm). MW p 9,036.?1. ‘T/LW WM ﬁw ﬁr WW 1'},— 7/§O(OOO 03% 96 'Zow/ (46% W @0293! Sigh : 7/§O,ch>o par—7 ...
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