Chapter 1 Engineering Economic Decisions 1.1 •
•
Lease o Deposit (typically one month worth of deposit) refundable when lease expires. o Monthly lease payment o Monthly maintenance fees o Monthly utility expenses Buy o Closing fees o Down payment o Monthly mortgage payments o Property taxes o Monthly utility fees o Monthly maintenance fees o Repair expenses o Homeowners’ association fee (if applicable)
1.2 •
Option 1: o Total amount at the end of two years: $1,150
•
Option 2: o Loan $500 to a friend for one year and receive $600 o Deposit $500 (left over) in a back at 3% for two years: $500(1.03)(1.03) = $530.45 o Deposit $600 received from your friend at 3% per year for a year: $600(1.03) = $618 Total amount at the end of two years: $530.45 + $618 = $1,148.45 These two options are about the same. But considering the trustworthiness, you could go with Option 2.
Chapter 2 Accounting Information for Engineering Economic Decisions 2.1 (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 2.2 (7), (8), (1), (11), (3), (9) 2.3 (a) • • • •
Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000
(b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15 Market price = $15 + $15 = $30 per share
2.4 (a) Shareholder’s equity in 2021 = $700 - $510 = $190(M) Shareholder’s equity in 2022 = $900 - $640 = $260(M) (b) Net working capital in 2021 = $100 - $60 = $40(M) Net working capital in 2022 = $200 - $90 = $110(M) (c) The income taxes in year 2022: ($2,350 - $1,130-$420-$210) *0.35 = $206.5(M) (d) $383.50 + $420=$803.50 (M) (Cash from Operating activities = Net income + Depreciation)
2.5 (a) ROE (= Net income/Equity) ROA (= Net income + interest expense (1-tax rate)/Average total assets)
Company A 26.03%
Company B 22.29%
17.34%
12.59%
(b) Company A has performed better in terms of profitability. (c) If two companies were merged, the impact on the results of ROE could be positive under the situation where the Company A leads the acquisition using a stock swap instead of issuing new stocks for M&A cost. If Company A uses a stock swap, the stock value wouldn’t be decreased in terms of scarcity. 2.6 Inventory turnover ratio (2021) = Sales/Average inventory balance = $3,776,395 / ($202,794 + $231,313)×0.5 = 17.4 times Inventory turnover ratio (2022) = 15.6 times This ratio shows how many times the inventory of a firm is sold and replaced over a specific period. From the data, Metronix was holding more stocks of inventory than last year; having more inventories on stock is unproductive. 2.7 (b) 2.8 (b) 2.9 (d) 2.10 Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets •
•
Net Income = $8 X Where X = the number of outstanding shares EPS =
Book value =
Total shareholders' equity = $80 X
2.11
•
Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = 6 million shares
•
From the book value per share, we know that the total shareholders’ equity = 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million
•
Debt ratio = $240 million/$720 million = 0.33
(a) Debt ratio (= Total debt/Total assets) = $19,483,000/$38,599,000 = 50.48% (b) Times-interest-earned ratio (= EBIT/Interest expense) = Not defined (c) Current ratio (= Current assets/Current liabilities) = 29,021,000/19,483,000 = 1.49 (d) Quick (acid test) ratio (= (Current assets - Inventories)/Current liabilities)) = (29,021,000-1,301,000)/19,483,000 = 1.42 (d) Inventory-turnover ratio (= Sales/Avg. inventory balance) = 61,494,000/ ((1,301,000+1,051,000)×0.5) = 52.29 (f) Days-sales-outstanding ratio (= Receivables/ (Annual sales/365)) =10,136,000/ (61,494,000/365) = 60.16 (g) Total-assets-turnover ratio (= Sales/Total assets) = 61,494,000/38,599,000 = 1.59 (h) Profit margin on sales (= Net income available to common stockholders/Sales) = 2,635,000/61,494,000 = 4.28% (i) Return on total assets (= (Net income + interest expense (1-tax rate))/Avg. total assets)
= 2,635,000/ ((38,599,000 + 33,652,000)×0.5) = 7.29% (j) Return on common equity (= (Net income available to common stockholders)/Avg. common equity) = 2,635,000/ ((7,766,000 + 5,641,000)×0.5) = 39.31% (k) Price/earnings ratio (= Price per share/Earnings per share) =13.47/ (3,350,000/1,944,000) = 7.82 (l) Book value per share (= (Total stockholders’ equity-Preferred stock)/Shares outstanding) = 7,766,000/1,944,000 = $3.99 To make an informed analysis of the firm’s financial health, we need to calculate the various financial ratios of the firm’s competitors along with the S&P 500. 2.12 Income Statement: A
B
C
D
E
F
$900,000
$585,000
$315,000
$270,000
$108,000
$162,000
Balance Sheet:
$160,000
$120,000
$320,000
$600,000
$900,000
$1,500,000
$450,000
$700,000
$100,000
$700,000
$800,000
•
From Current ratio Total current assets = 2.4 × $250,000 = $600,000 ----------------------------------- ③ Plant and equipment, net = $1,500,000-$600,000=$900,000-----------------------
•
From Quick ratio
Inventory = $600,000 - (1.12 × $250,000) = $320,000 -----------------------------② •
From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) × 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 ------- A
•
From DSO Accounts receivable = 24.3333 × ($1,800,000 ÷365) = $120,000 ------------------ ① Cash = ③-(②+①) = $160,000 -----------------------------------------------------------ⓞ
•
From interest expense of income statement Bond = $450,000 ----------------------------------- ⑥ 250,000 + ⑥ = $700,000 --------------------------⑦
•
From Debt-to-Equity ratio Total Equity ⑩ = $700,000 ÷ 0.875 =$ 800,000 ------------- ⑩ Total assets or Total liabilities and equity = ⑦ + ⑩ = $1,500,000 ------⑤
•
From Return on total assets Net income F = 14%× ($1,350,000) - ($45,000) (0.6) = $162,000
•
From F, D =F ÷ 0.6 = $270,000, E = D× (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000
•
From EPS Stock Outstanding = F ÷ 4.05 = 40,000 shares Common stock = $2.50 × 40,000 = $100,000 -------------------------------⑧ Retained Earnings = ⑩ - ⑧ = $700,000 ------------------------------------
2.13 • • •
•
Accounts receivable = DSO × Sales/365 = 45 days × ($1,200)/365 days) = $147.945 Current assets = (Cash and marketable securities) + (Accounts receivable) + Inventory = $427.945 Long-term debt = (Total assets) – (Current liabilities) – (Common equities) = $427.945 + $280 – (current assets/current ratio) - $500 = ($207.945) – (427.945/3.2) = $74.212 Total assets turnover = Sales/Total assets = $1,200/ ($427.945 + $280) = 1.695 times
2.14
(a) Find Tiger’s accounts receivable. DSO = 91.25 =
AR AR = $50, 000 200, 000 / 365
(b) Determine the amount of current liabilities. CA = Cash + Inventory + AR = $10, 000 + $150, 000 + $50, 000 = $210, 000 $210, 000 Current Ratio = 4.2 = Current Liabilities = $50, 000 Current Liabilities
(c) Calculate the amount of the long-term debt. Total Asset = Current Asset + Fixed Asset = $210, 000 + $90, 000 = $300, 000 $300, 000 = ($50, 000 + Long term debt ) + $200, 000
Long term debt = $50, 000
(d) Calculate the Return on Common Equity. ROE =
net income $15, 000 = = 0.075 7.5% equity $200, 000
2.15 (a) Find Fisher’s accounts receivable. 𝐴𝑅 𝐷𝑆𝑂 = → 𝐴𝑅 = 𝟏𝟒𝟕. 𝟗𝟓𝑴 1,200 365 (b) Calculate the amount of current assets. 𝐶𝐴 = 𝑐𝑎𝑠ℎ + 𝐼𝑛𝑣. +𝐴𝑅 = 100 + 180 + 147.95 = 𝟒𝟐𝟕. 𝟗𝟓𝑴 (c) Determine the amount of current liabilities. 𝐶𝐴 427.95 𝐶𝑅 = 3.2 = = → 𝐶𝐿 = 𝟏𝟑𝟑. 𝟕𝟑𝑴 𝐶𝐿 𝐶𝐿 (d) Determine the amount of total assets. 𝑇𝐴 = 𝐶𝐴 + 𝐹𝐴 = 427.95 + 280 = 𝟕𝟎𝟕. 𝟗𝟓𝑴
(e) Calculate the amount of the long-term debt. 707.95 = 133.73 + 𝐿𝐵 + 500 → 𝐿𝐵 = 𝟕𝟒. 𝟐𝟐𝑴 (f) Calculate the profit margin. 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑝𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑟𝑔𝑖𝑛 = = = 𝟐𝟗. 𝟖𝟑% 𝑠𝑎𝑙𝑒𝑠 1,200 (g) Calculate the Return on Common Equity 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑅𝑂𝐸 = = = 𝟕𝟏. 𝟔% 𝑒𝑞𝑢𝑖𝑡𝑦 500
Short Case Studies with Excel ST2.1 Not provided ST2.2 (a) Working capital = Current assets – Current liabilities Working capital requirements = Changes in current assets – Changes in current liabilities: WC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets. (b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000
ST2.3
(d) Net cash flow: • Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000 • Investing activities = equipment purchase = ($400,000) • Financing activities = borrowed fund = $200,000 • Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 Not provided (Visit the websites and get the most recent financial statements available)
7
Chapter 3: Interest Rate and Economic Equivalence Types of Interest 3.1
I = (iP) N = (0.06)($2, 000)(5) = $600
3.2 •
Simple interest:
$20,000 = $10,000(1 + 0.075N ) (1 + 0.075N ) = 2 N= •
1 = 13.33≈ 14 years 0.075
Compound interest: $20,000 = $10,000(1 + 0.07) N (1 + 0.07) N = 2 N = 10.24 ≈ 11years
3.3 •
Compound interest:
F = $1, 000(1 + 0.065)5 = $1,370.09 •
Simple interest: F = $1, 000(1 + 0.068(5)) = $1, 340 The compound interest option is better.
3.4 •
Simple interest (John):
I = iPN = (0.1)($1,000)(5) = $500 •
Compound interest (Susan):
I = P (1 + i ) N − 1 = $1, 000 (1 + .095)5 − 1 = $574.24 •
Susan’s balance will be greater by $74 (or $74.24 to be exact)
•
Simple interest:
3.5 I = iPN = (0.10)($10, 000)(5) = $5, 000
•
Compound interest:
I = P[(1 + i ) N − 1] = $10, 000(1.6105 − 1) = $6,105 3.6 •
Option 1: Compound interest with 8%:
F = $4,500(1 + 0.085)5 = $5, 000(1.4693) = $6, 766.45 •
Option 2: Simple interest with 9.5%: $4, 500(1 + 0.095 × 5) = $5, 000(1.475) = $6, 637.50
∴ Option 1 is still better. 3.7 End of Year
Principal Repayment
Interest payment
0 1
$4,620.50
$1,200.00
Remaining Balance $15,000.00 $10,379.50
2
$4,990.14
$830.36
$5,389.36
3
$5,389.35
$431.15
$0
Equivalence Concept 3.8
3.9
P = $22, 000( P / F , 5%, 5) = $22, 000(0.7835) = $17, 237.58
F = $30, 000( F / P, 9%, 3) = $30, 000(1.295) = $38,850.87
3.10 F = $100( F / P,10%,10) + $200( F / P,10%,8) = $688
3.11 $1, 000( F / P, i, 2) = $1, 200 $1, 000(1 + i ) 2 = $1, 200
i = 1.2 − 1 i = 9.54%
Single Payments (Use of F/P or P/F Factors) 3.12
i = 10.5% , two-year discount rate is (1 + 0.105) 2 = 1.221 (or 22.1%) 3.13
F = 2 P = P(1 + 0.06) N log 2 = N log 1.06 N = 11.896 years (or 12 years) 3.14 F = $1(1.08)394 = $14, 755, 694, 730, 611
3.15
P = $450, 000( P / F ,5%,5) = 450, 000(0.7835) = $352,575
3.16 F = $250, 000( F / P, 6%,10) = $447, 712
3.17 (a) F = $5, 000( F / P , 7%, 5) = $7, 013 (b) F = $7, 250( F / P , 9%,15) = $26, 408 (c) F = $9, 000( F / P , 6%, 33) = $61,565 (d) F = $12, 000( F / P, 5.5%,8) = $18, 416
3.18 P = $300, 000( P / F ,8%,10) = $138, 958
3.19 (a) P = $25,500( P / F ,12%,8) = $10, 299 (b) P = $58, 000( P / F , 4%,12) = $36, 227 (c) P = $25, 000( P / F , 6%, 9) = $14, 797 (d) P = $35, 000( P / F , 9%, 4) = $24, 795 3.20 (a) P = $12, 000( P / F ,13%, 4) = $7, 360 (b) F = $30, 000( F / P,13%, 5) = $55, 273 3.21 F = 3P = P (1 + 0.08) N log 3 = N log(1.08) N = 14.27 → 15 years 3.22 F = 2 P = P(1 + 0.06) N
•
log 2 = N log(1.06) N = 11.90 years 12 years
•
Rule of 72: 72 / 6 = 12 years
3.23 ($16.50)(100)( F / P, i, 44) = $77.50(204,800) $15,872, 000 ( F / P, i, 44) = = 9, 619.39 $1, 650 i = 23.18%
3.24 (a)
$18( F / P, i, 49) = $190,500 $190,500 = 10,583.33 ( F / P, i, 49) = $18 i = 20.82% (b)
F = $304.8 B( F / P , 20.82%,17) = $7, 592.33B
Uneven Payment Series 3.25
P= 3.26
$2, 000 $800 $1, 000 + + = $3, 230.65 1.11 1.12 1.13
P = $35, 000( P / F ,9%, 4) + $10, 000( P / F ,9%, 2) = $35, 000(0.7084) + $10, 000(0.8417) = $33, 211
3.27 $180, 000 = $20, 000( P / A,9%,5) − $10, 000( P / F ,9%,3) + X ( P / F ,9%, 6)180, 000 − 20, 000(3.8897) + 10, 000(0.7722) = X (0.5963) X = $184,350.16
3.28
P=
$60, 000 $77, 000 $65, 000 $57, 000 45, 000 + + + + = $212,873.89 1.14 1.142 1.143 1.144 1.145
3.29 $1,000 +
X $1,000 $1,500 $1,210 + = + 4 3 2 1.1 1.1 1.1 1.1 X = $2,981
3.30
P=
$15, 000 $23, 000 $36, 000 $48, 000 + + + = $93,564 1.07 2 1.073 1.07 4 1.075
3.31 F = $2, 000( F / P, 6%,10) + $2,500( F / P, 6%,8) + $3, 000( F / P ,6%,6) = $11,822
3.32
P = $3, 000, 000 + $2, 400, 000( P / F ,8%,1) + +$3, 000, 000( P / F ,8%,10) = $20, 734, 618
Or, P = $3, 000, 000 + $2, 400, 000( P / A,8%,5) +$3, 000, 000( P / A,8%,5)( P / F ,8%,5) = $20, 734, 618
3.33 P = $9, 000( P / F ,8%, 2) + $6, 000( P / F ,8%, 5) + $3, 000( P / F ,8%,7) = $13, 550
Equal Payment Series 3.34
A = $250, 000( A / F ,5%,5) = $250, 000(0.1810) = $45, 250 F = $5,000( F / A,5%, 7) = $5, 000(8.1420) = $40, 710 F = $5, 000( F / A, 5%, 7)(1.05) = $5, 000(8.1420)(1.05) = $42, 745.50
(a) (b) 3.35 •
Equal annual payment amount:
A = $20, 000( A / P,10%,3) = $20, 000(0.4021) = $8, 042 •
Loan balance calculation:
End of period 0 1 2 3
Principal Payment $0.00 $6,042.00 $6,646.20 $7,310.82
Interest Payment $0.00 $2,000.00 $1,395.80 $731.18
Interest payment for the second year = $1,395.80
Remaining Balance $20,000.00 $13,958.00 $7,311.80 $0
3.36
F = $500( F / A, 7%,15)(1.07) = $500(25.1290)(1.07) = $13, 444.02 3.37 (a) With deposits made at the end of each year F = $3, 000( F / A, 9%,15) = $88, 083
(b) With deposits made at the beginning of each year F = $3, 000( F / A,9%,15)(1.09) = $96, 010
3.38 F = $10, 000( F / A, 6%,20) = $367,856
3.39 (a) F = $8, 000( F / A,11.75%, 5) = $50, 571 (b) F = $2, 000( F / A, 4.25%,12) = $30, 486 (c) F = $7, 000( F / A, 6.45%, 20) = $270, 309 (d) F = $4, 000( F / A, 7.75%,12) = $74, 793 3.40 (a) A = $45, 000( A / F ,8%,11) = $2, 703 (b) A = $35, 000( A / F , 6%,18) = $1,132 (c) A = $25, 000( A / F , 7%, 6) = $3, 495 (d) A = $12, 000( A / F ,11%,13) = $458 3.41 $50, 000( A / F , 6%,10) = $3, 793.40
3.42 $55, 000 = $3, 000( F / A, 6%,N ) ( F / A, 6%, N ) = 18.33 N = 13years
3.43 $15, 000 = A( F / A,11%,5) A = $2, 408.57
3.44
$10,000 = $1,000( F / P,7%,5) + A( F / A,7%,5) A = $1,536.57 3.45 (a) A = $15, 000( A / P,3.5%, 6) = $2,815.02 (b) A = $7,500( A / P, 7.5%, 7) = $1, 416 (c) A = $2, 500( A / P ,5.25%,5) = $581.43 (d) A = $12, 000( A / P , 6.25%,15) = $1, 255.81 3.46 (a) The capital recovery factor ( A / P, i, N ) for N 35 40
6% 0.0690 0.0665
7% 0.0772 0.0750
To find ( A / P, 6.25%, 38) , first, interpolate for N = 38 : N 38
6% 0.0675
7% 0.0759
Then, interpolate for i = 6.25% ; (A / P, 6.25%, 38) = 0.0696 :
As compared to the value from the interest formula: (A / P, 6.25%, 38) = 0.0694
(b) The equal payment series present-worth factor ( P / A, i, 85) for i
9% 11.1038
10% 9.9970
Then, interpolate for i = 9.25% : ( P / A, 9.25%, 85) = 10.8271
As compared to the value from the interest formula: ( P / A, 9.25%, 85) = 10.8049
3.47 •
Equal annual payment: A = $50, 000( A / P,12%, 3) = $20,817.45
•
Interest payment for the second year: End of Year 0 1 2 3
3.48
Principal Repayment
Interest payment
$14,817.45 $16,595.54 $18,587.01
$6,000 $4,221.91 $2,230.44
A = $15, 000( A / P , 9%,10) = $2, 337.30
3.49 (a) P = $1, 000( P / A, 7.2%,8) = $5,925.29 (b) P = $4, 500( P / A, 9.5%,12) = $31, 427.28 (c) P = $1, 900( P / A,8.25%,13) = $14,812.86 (d) P = $19, 300( P / A, 7.75%,8) = $111, 970.11
Remaining Balance $50,000 $35,182.55 $18,587.01 0
3.50 P = $35,000( P / A,12%,10) = $197,758 Since $200,000 > $197,758, You should not purchase the equipment.
3.51 (a)
P = $9, 600, 000 + $15,100, 000( P / F , 6%,1) +$17,100, 000( P / F , 6%, 2)... + $19, 250, 000( P / F , 6%,5) +$400, 000( P / A, 6%,5) = $89,970,959 (b) P = $7, 600, 000 + $5, 600, 000( P / A, 6%, 4) = $27, 004,591 Since $27, 004,591 > $25,000,000, the prorated payment option is better choice.
Linear Gradient Series 3.52
F = $20, 000( F / A, 6%,5) + $5, 000( F / G , 6%, 5) = $20, 000( F / A, 6%,5) + $5, 000( A / G , 6%,5)( F / A, 6%, 5) = $165,833
3.53
F = $10, 000( F / A,8%,5) − $2, 000( F / G ,8%,5) = $10, 000( F / A,8%,5) − $2, 000( P / G ,8%,5)( F / P, 8%, 5) = $37, 001
3.54 P = $100 + [$100( F / A,9%, 7) + $50( F / A,9%, 6) +$50( F / A,9%,4) + $50( F / A, 9%, 2)]( P / F , 9%, 7) = $991.32
3.55 A = $15, 000 − $1, 000( A / G , 8%, 12) = $10, 404.25
3.56
P = $1, 000( P / A, 6%, 5) + $250( P / G , 6%, 5) = $6,196
Geometric-Gradient Series 3.57
$1, 000, 000 = A( F / A, 6%,30) = A(79.0582) A = $12,649 should be set aside on the account a) $1, 000, 000 = A( P / A, 6%, 20) = A(11.4699) A = $87,185 / year b) $1, 000, 000 = A1 ( P / A1, 3%, 6%, 20) 1 − (1.03) (1.06 ) = A1 0.06 − 0.03 = $68, 674 / year 20
−20
3.58 Using the geometric gradient series present worth factor, we can establish the equivalence between the loan amount $150,000 and the balloon payment series as $150, 000 = A1 ( P / A1 ,10%, 7%,5) = 4.9424 A1
A1 = $30,349.63 Payment series N 1 2 3 4 5 3.59
Payment $30,349.63 $33,384.59 $36,723.05 $40,395.35 $44,434.89
F = $8, 000( P / A1 ,5%, 7%,30)( F / P, 7%,30) = $172,895.56(7.61226) = $1,316,126 3.60 (a) P = $6,000,000( P / A1 ,−10%,12%,7) = $21,372,076 (b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows: An = $60(1 + 0.05) n −1100, 000(1 − 0.1) n −1 = $6, 000, 000(0.945) n −1 = $6, 000, 000(1 − 0.055) n −1
This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%. So,
P = $6, 000, 000( P / A1 , −5.5%,12%,7) = $23,847,897 (c) Computing the present worth of the remaining series ( A4 , A5 , A6 , A7 ) at the end of period 3 gives A4 = 6, 000, 000(1 − 0.055)3 = 5, 063, 451.75 P = $5, 063, 451.75( P / A4 , −5.5%,12%, 4) = $14, 269, 627.82
3.61 20
P = An (1 + i) − n n =1 20
= (2, 000, 000)n(1.06) n −1 (1.06) − n n =1
20 1.06 n = (2, 000, 000 /1.06) n( ) 1.06 n =1 = $396, 226, 415
3.62
(a) The withdrawal series would be Period 11 12 13 14 15
Withdrawal $15,000 $15,000(1.08) $15,000(1.08)(1.08) $15,000(1.08)(1.08)(1.08) $15,000(1.08)(1.08)(1.08)(1.08)
Amount $15,000 $16,200 $17,496 $18,896 $20,407
P10 = $15, 000( P / A1 ,8%,9%,5) = $67,556 Assuming that each deposit is made at the end of each year, then: $67,556 = A( F / A,9%,10) A = $4, 446.54
(b) P10 = $15, 000( P / A1 ,8%, 6%,5) = $73, 476 $73, 476 = A( F / A, 6%,10) A = $5, 574.47
Various Interest Factor Relationships 3.63 (a) (P / F, 8%, 67) = (P / F, 8%,50)(P / F,8%,17) = 0.0058
(P / F, 8%,67) = (1 + 0.08)−67 = 0.0058 i 1 − ( P / F , i, N ) ( P / F , 8%, 42) = ( P / F , 8%,40)( P / F ,8%,2) = 0.0394 0.08 ( A / P, 8%, 42) = = 0.0833 1 − 0.0394
(b) ( A / P, i, N ) =
( A / P, 8%, 42) =
(c) ( P / A, i, N ) =
0.08(1.08) 42 = 0.0833 (1.08) 42 − 1
1 − ( P / F , i, N ) 1 − ( P / F ,8%,100)( P / F ,8%,35) = = 12.4996 i 0.08
(P / A,8%,135) =
(1.08)135 − 1 = 12.4996 0.08(1.08)135
3.64 (a) ( F / P, i, N ) = i ( F / A, i, N ) + 1 (1 + i ) N − 1 +1 i = (1 + i) N − 1 + 1
(1 + i ) N = i
= (1 + i) N (b)
( P / F , i, N ) = 1 − ( P / A, i, N )i (1 + i ) − N = 1 − i =
(1 + i ) N − 1 i (1 + i ) N
(1 + i ) N (1 + i ) N − 1 − (1 + i ) N (1 + i ) N
= (1 + i ) − N (c)
( A / F , i , N ) = ( A / P, i, N ) − i i i (1 + i ) N i (1 + i ) N i[(1 + i ) N − 1] = − i = − (1 + i ) N − 1 (1 + i ) N − 1 (1 + i ) N − 1 (1 + i ) N − 1 i = (1 + i ) N − 1
(d) ( A / P, i , N ) =
i [1 − ( P / F , i, N )]
i (1 + i ) N i = N N (1 + i ) − 1 (1 + i ) 1 − N (1 + i ) N (1 + i ) =
i (1 + i ) N (1 + i ) N − 1
(e) , (f) , (g) Divide the numerator and denominator by (1 + i) N and take the limit N →∞.
Equivalence Calculations 3.65 P = [$100( F / A,12%, 9) + $50( F / A,12%, 7) + $50( F / A,12%, 5)]( P / F ,12%,10) = $740.49
3.66 P (1.08) + $200 = $200( P / F ,8%,1) + $120( P / F ,8%, 2) + $120( P / F ,8%,3) + $300( P / F ,8%, 4) P = $373.92
3.67
A( P / A,15%,5) = $100( P / A,15%,5) + $20( P / A,15%,3)( P / F ,15%, 2) 3.35216 A = $369.74 A = $110.30 3.68 P1 = $200 + $100( P / A,8%,5) + $50( P / F ,8%,1) +$50( P / F ,8%, 4) + $100( P / F ,8%,5) = $750.37 P2 = X ( P / A,8%,5) = $750.37 X = $187.93
3.69
P = $20( P / G ,10%,5) − $20( P / A,10%,12) = $0.96
$40
$60
$80
$20 0
1
2
$20
3
4
5
6
7
8
9 10
11
12
3.70 Establish economic equivalent at N = 8 : C ( F / A,8%,8) − C ( F / A,8%,2)( F / P,8%,3) = $6,000( P / A,8%,2) 10.6366C − (2.08)(1.2597 )C = $6,000(1.7833) 8.0164C = $10,699.80 C = $1,334.73
3.71 The original cash flow series is
N
AN
N
AN
0
0
6
$900
1
$800
7
$920
2 3
$820 $840
8 9
$300 $300
4
$860 10 $300 − $500
5
$880
3.72 $300( F / A,10%,8) + $200( F / A,10%,3) = 2C ( F / P,10%,8) + C ( F / A,10%,7) $4,092.77 = 2C ( 2.1436) + C (9.4872) C = $297.13
3.73
Establishing equivalence at N = 5 $200( F / A,8%,5) − $50( F / P,8%,1) = X ( F / A,8%,5) − ($200 + X )[( F / P,8%, 2) + ( F / P,8%,1)] $1,119.32 = X (5.8666) − ($200 + X )(2.2464) X = $433.29
3.74
Computing equivalence at N = 5 X = $3, 000( F / A,9%,5) + $3, 000( P / A,9%,5) = $29, 623.08
3.75 (b), (d), and (f) 3.76 (b), (d), and (e) 3.77 A1 = ($50 + $50( A / G ,10%,5) − [$50 + $50( P / F ,10%,1)]( A / P,10%,5) = $115.32 A2 = A + A( A / P,10%,5) = 1.2638 A A = $91.25
3.78 (a) 3.79 (b) 3.80 (b) $25, 000 + $30, 000( P / F ,10%, 6) = C ( P / A,10%,12) + $1, 000( P / A,10%, 6)( P / F ,10%, 6) $41,935 = 6.8137C + $2, 458.43 C = $5, 794
Solving for an Unknown Interest Rate of Unknown Interest Periods 3.81 P1 = 30, 723( P / F , i %,5) P2 = A( P / A, i %,10) (1 + i )10 − 1 $50, 000(1 + i ) −5 = $5, 000 10 i (1 + i ) ∴ i = 13.06%
3.82 2 P = P(1 + i )5 21/5 = 1 + i i = 14.87% 3.83
Establishing equivalence at n = 0
$2,000( P / A, i,6) = $2,500( P / A1 ,−25%, i,6) By Excel software, i = 92.36% 3.84 $40, 000 = $15, 000( F / P, i, 5) = $15, 000(1 + i )5 i = 21.67%
3.85 $1, 000, 000 = $2, 000( F / A, 6%, N ) (1 + 0.06) N − 1 0.06 31 = (1 + 0.06) N log 31 = N log1.06 500 =
N = 58.93 ≈ 59years
3.86 Option 1: $100, 000( F / A, 7%, 7)( F / P, 7%,13) = 2, 085, 484.95 Option 2: $100, 000( F / A, 7%,13) = 2, 014, 064.29
$100, 000( F / A, i, 7)( F / P, i,13) = $100,000( F / A, i,13) i = 6.6% 3.87 Assuming that annual renewal fees are paid at the beginning of each year, (a) $15.96 + $15.96(P / A,6%,3) = $58.62 It is better to take the offer because of lower cost to renew.
18
(b)
$57.12 = $15.96 + $15.96( P / A, i,3) i = 7.96% 3.88 If premiums paid at the end of each year, the maximum amount to invest in the prevention program is P = $14,000 ( P / A,12%,5) = $50,467 .
If the premiums paid at the beginning of each year, the solution changes to P = $14, 000 + $14, 000( P / A,12%, 4) = $56,523.
Short Case Studies ST 3.1 (a)
(b)
P = 280, 000( P / A,8%,19) = 2, 689, 007.78
280, 000( P / A, i,19) = 5, 600, 000 − 283, 770 i = 0.00709%
ST 3.2 Establish the following equivalence equation: $140,000 = $32,639 (P / A,i, 9) . The interest rate makes two options equivalent is i = 18.10% by Excel. So, if her rate of return is over 18.10%, it is a good decision.
ST 3.3 (a) PContract = $5, 600, 000 + $7,178, 000( P / F , 6%,1) +$11, 778, 000( P / F , 6%, 2) + +$17, 778, 000( P / F , 6%,9) = $97,102,826.86
(b) PBonus = $5,000,000 + $5,000,000( P / A,6%,5) + $778,000( P / A,6%,9)
= $31,353,535.52 > $23,000,000
It is better stay with the original plan.
Chapter 4 Understanding Money and Its Management Nominal and Effective Interest Rates 4.1 •
Nominal interest rate:
r = 1.3% ×12 = 15.6% •
Effective annual interest rate: ia = (1 + 0.013)12 − 1 = 16.77%
4.2
(a) Monthly interest rate: i = 17.85% ÷12 = 1.4875% Annual effective rate: ia = (1 + 0.014875)12 − 1 = 19.385% (b) $2,500(1 + 0.014875)2 = $2,574.93
4.3 Assuming weekly compounding:
r = 6.89% 0.0689 52 ) − 1 = 0.07128 ia = (1 + 52 4.4 •
Nominal interest rate:
r = 1.3% ×12 = 15.6% •
Effective annual interest rate: ia = (1 + 0.013)12 − 1 = 16.77%
4.5.
$20,000 = $520( P / A, i, 48) ( P / A, i, 48) = 38.4615 Use Excel to calculate i :
i = 0.9431% per month r = 0.9431*12 = 11.32%
4.6.
$16,000 = $517.78( P / A,i,36) ( P / A,i,36) = 30.901155 i = 0.85% per month r = 0.85 × 12 = 10.2%
4.7 •
Interest rate per week Given : P = $550, A = $42, N = 16 weeks
$550 = $42( P / A, i,16) i = 2.46% per week •
Nominal annual interest rate:
r = 2.46% × 52 = 127.95% •
Effective annual interest rate: ia = (1 + 0.0246)52 − 1 = 253.86%
4.8 The effective annual interest rate: ia = e0.087 − 1 = 9.09%
4.9 Interest rate per week:
$450 = $400(1 + i) i = 12.5% per week (a) Nominal interest rate:
r = 12.5% × 52 = 650% (b) Effective annual interest rate
ia = (1+ 0.125)52 −1 = 45,602% 4.10 The effective annual interest rate : ia = e 0.06 − 1 = 6.184%
4.11 24-month lease plan with 40,000 miles over 2 years: P = ($2,399 + $189) + $189( P / A, 0.5%, 23) + ($350 + $0.18 × (40, 000 − 24, 000))( P / F , 0.5%, 24) = $9,869.70
4.12
The three options: a) ia = r = 6.12% 4
0.06 b) ia = 1 + − 1 = 6.136% 4 c) ia = e0.059 −1 = 6.078%
Bank B is the best option. 4.13 12
0.06 ia = 1 + − 1 = 6.168% 12 ia = er − 1 = 0.06168 er = 1.06168 r = 5.985% 4.14 0.05 Bank A: ia = 1 + 365
365
− 1 = 5.127%
Bank B: ia = e0.046 − 1 = 4.707% The difference between two banks after 2 years:
$3,000[ (F / P,5.127%, 2) − (F / P,4.707%, 2)] = $26.44
Compounding More Frequent than Annually 4.15 0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12 a) i = (1 +
4.16
iquarter = e0.09/4 − 1 = 0.022755 (or 2.28%) 4.17 0.06 1 ) − 1 = 0.5% 12 0.06 3 b) i = (1 + ) − 1 = 1.508% 12 0.06 6 c) i = (1 + ) − 1 = 3.038% 12 0.06 12 d) i = (1 + ) − 1 = 6.168% 12 a) i = (1 +
4.18
$25, 000 = $563.44( P / A, i, 48) ( P / A, i, 48) = 44.3703 i = 0.3256% per month
4.19 0.11 1 ) − 1 = 11% 1 0.08 2 b) i = (1 + ) − 1 = 8.16% 2 0.095 4 c) i = (1 + ) − 1 = 9.844% 4 0.075 365 d) i = (1 + ) − 1 = 7.788% 365 a) i = (1 +
4.20
F = PerN = $5,000e(0.06×10) = $9,110.59 4.21
P = Fe− rN = $5,000e−0.06×5 = $3,704.09 4.22
F = PerN = $5, 000e(0.09×5) = $7,841.56 4.23 2 P = Pe 0.06 N ln 2 = 0.06N N = 11.55 years
4.24 (a) Nominal interest rate:
r = 2.32% ×12 = 27.84% (b) Effective annual interest rate: ie = (1 + 0.0232)12 − 1 = 31.68%
(c)
3P = P(1 + 0.0232) N log 3 = N log1.0232 N = 47.90 months ≅ 4 years (d)
4 = e0.0232 N ln(4) = 0.0232 N N = 59.75 months ≅ 5 years 4.25 F = $15, 000(1 +
0.08 8 ) = $15, 000( F / P, 2%,8) 4
= $17,575
4.26
ia = (1 +
0.06 365 ) − 1 = 6.183% 365
F = $15, 000( F / P, 6.183%,12) = $15, 000( F / P,
6 %,12 × 365) 365
= $ 30,815 4.27 (a) 0.082 24 ) = $9,545( F / P, 4.1%, 24) 2 = $25, 037.64
F = $9,545(1 +
(b) 0.06 40 ) = $6,500( F / P,1.5%, 40) 4 = $11, 791.12
F = $6,500(1 +
(c) F = $42,800(1 +
0.09 96 ) = $42, 000( F / P, 0.75%,96) 12
= $87, 693.83
4.28 (a) Quarterly interest rate = 2.25%
3P = P(1 + 0.0225) N log 3 = N log1.0225 N = 49.37quarters = 12.34 years (b) Monthly interest rate = 0.75%
3P = P(1 + 0.0075) N log 3 = N log1.0075 N = 147.03 months = 12.25 years (c)
3 = e0.09 N ln(3) = 0.09 N N = 12.20 years 4.29. (a)
F = $10, 000( F / A, 4%, 20) = $297, 781 (b)
(c)
F = $9, 000( F / A, 2%, 24) = $273, 796.76 F = $5, 000( F / A, 0.75%,168) = $1,672,590.40
4.30 (a)
(b)
(c)
A = $11,000( A / F , 4%, 20) = $369.60 A = $3, 000( A / F ,1.5%, 60) = $31.18 A = $48, 000( A / F ,0.6125%, 60) = $484.46
4.31 (a) Quarterly effective interest rate = 1.5%
F = $10, 000( F / A,1.5%, 60) = $962,147
(b) Quarterly effective interest rate = 1.508% F = $10, 000( F / A,1.508%, 60) = $964, 722
(c) Quarterly effective interest rate = 1.511% F = $10, 000( F / A,1.511%, 60) = $965, 690
4.32
F = $7, 500( F / A, 0.669%, 60) = $551, 479
4.33 (a) Quarterly effective interest rate = 2.25% F = $4,000( F / A,2.25%,40) = $255,145
(b) Quarterly effective interest rate = 2.2669% F = $4,000( F / A,2.2669%,40) = $256,093
(c) Quarterly effective interest rate = 2.2755% F = $4,000( F / A,2.2755%,40) = $256,577
4.34
(d)
Effective interest rate per payment period
i = (1 + 0.01)3 – 1 = 3.03%
0
1
2
3
4
5
6
7
$1,000
4.35 ia = e0.086/4 − 1 = 2.1733% A = $10, 000( A / P , 2.1733%, 20) = $ 621.84
4.36 (a) Monthly effective interest rate = 0.74444% F = $1,500( F / A, 0.74444%, 96) = $209,170
(b) Monthly effective interest rate = 0.75% F = $1, 500( F / A, 0.75%, 96) = $209, 784
(c) Monthly effective interest rate = 0.75282% F = $1, 500( F / A, 0.75282%, 96) = $210, 097
4.37 (b) 4.38 i = e 0.0225 − 1 = 2.2755%
8
9
10
11
12
F = $5, 000( F / A, 2.2755%, 40) = $320, 721
F = $320, 721( F / P, 2.2755%, 20) = $502, 990
4.39 Effective interest rate per month = e 0.0975/12 − 1 = 0.8158% A = $48, 000( A / P, 0.8158%, 60) = $1, 014.90
4.40 Effective interest rate per quarter = e0.0688/ 4 − 1 = 1.7349% P = $2, 500( P / A,1.7349%, 20) = $41, 944
4.41 (a) F = $20, 000( F / A, 4%,10) = $240,122 (b) F = $60, 000( F / A,1.5%, 40) = $3, 256, 074 (c) F = $13, 000( F / A, 0.75%, 72) = $1, 235, 091 4.42
P = Fe− rN = $15,345.36e−0.06×10 = $ 8,421.71 4.43 (a) A = $45,000( A / F ,3.725%,20) = $1,554.80 (b) A = $25,000( A / F ,1.5875%,60) = $252.33 (c) A = $12, 000( A / F , 0.7708%, 60) = $158.06 4.44
id = (1 +
0.05 1 ) − 1 = 0.01369863% 365
F = $3.75( F / A, 0.01369863%,10950) = $95, 299
4.45
$25, 000 = $489.15( P / A, i, 60) i = 0.5416% ia = (1 + 0.005416)12 − 1 = 6.69685%
4.46 $30, 000 = $500( F / A, i, 20) i = 10.4084% i = e r /4 − 1 = e0.25 r − 1 = 0.104084 r = 39.61%
4.47
$15, 000 = $409.61( P / A, i, 42) i = 0.6542% r = 0.006542 ×12 = 7.85%
4.48
A = $70,000( A / F,0.5%,36) = $1,779.54 4.49 (a) P = $5, 000( P / A, 4.5%, 24) = $72, 477 (b) P = $7, 000( P / A, 2%, 40) = $191, 488 (c) P = $3, 500( P / A, 0.5%, 60) = $181, 039 4.50 •
Equivalent future worth of the receipts:
F1 = $1, 500(F / P, 2%, 4) + $2, 500 = $4,123.65
•
Equivalent future worth of deposits:
F2 = A( F / A, 2%,8) + A( F / P, 2%,8) = 9.7546 A ∴ Letting F1 = F2 and solving for A yields A = $422.74
4.51
C ( F / A, 6.168%, 7) + C ( F / P, 0.5%,84) = $1, 600 + $1, 400( F / P, 0.5%,12) + $1, 200( F / P, 0.5%, 24) +$1, 000( F / P, 0.5%,36) = 8.437C + 1.52C = 1, 600 + 1, 486.35 + 1,352.59 + 1,196.68 9.957C = 5, 635.62 ∴ C = $566
4.52 •
Option 1 .06 1 ) − 1 = 1.5% 4 F = $1, 000( F / A,1.5%, 40)( F / P,1.5%, 60) = $132,587 i = (1 +
•
Option 2 .06 4 ) − 1 = 6.136% 4 F = $6, 000( F / A, 6.136%,15) = $141,111 i = (1 +
•
Option 2 – Option 1 = $141,110 – 132,587 = $8,523
•
Select (b)
4.53 Given: r = 7% compounded daily, N = 25 years
•
Since deposits are made at year end, find the effective annual interest rate: ia = (1 + 0.07 / 365)365 − 1 = 7.25%
•
Then, find the total amount accumulated at the end of 25 years: F = $3,250(F / A,7.25%,25) + $150(F / G,7.25%,25) = $3,250(F / A,7.25%,25) + $150(P / G,7.25%,25)(F / P,7.25%,25) = $297,016.95
4.54 •
The balance just before the transfer:
F9 = $22,000(F / P,0.5%,108) + $16,000(F / P,0.5%,72) +$13,500(F / P,0.5%,48) = $77,765.70 Therefore, the remaining balance after the transfer will be $38,882.85. This remaining balance will continue to grow at 6% interest compounded monthly. Then, the balance 6 years after the transfer will be:
F15 = $38,882.85( F / P,0.5%,72) = $55,681.96 •
The funds transferred to another account will earn 8% interest compounded quarterly. The resulting balance six years after the transfer will be:
F15 = $38,882.85( F / P, 2%, 24) = $62,540.63
4.55 Establish the cash flow equivalence at the end of 25 years. Let’s define A as the required quarterly deposit amount. Then we obtain the following: A( F / A,1.5%,100) = $80, 000( P / A, 6.136%,15) 228.8030 A = $770,104 A = $3,365.79
4.56 •
Monthly installment amount:
A = $22, 000( A / P , 0.75%, 60) = $456.68
•
The lump-sum amount for the remaining balance:
P24 = $456.68( P / A,0.75%,36) = $14,361.13 4.57
$225, 000 = $5, 000( P / A, 0.75%, N ) ( P / A, 0.75%, N ) = 45 N = 55 months or 4.58 years
4.58
$20, 000 = $650.52( P / A, i,36) ( P / A, i,36) = 30.7446 i = 0.879% APR = 0.879% × 12 = 10.55%
4.59 Given r = 6% per year compounded monthly, the effective annual rate is 6.168%. Now consider the four options: 1. Renew every three-month at $45 for two years. 2. Renew every year at $160 for two years. 3. Buy a two-year subscription ($279) now. To find the best option, compute the equivalent PW for each option.
o
Poption 1 = $45 + $45( P / A,1.5075%, 7) = $341.83
o
Poption 2 = $160 + $160( P / F , 6.168%,1) = $310.70
o
Poption 3 = $279
Option 3 is the best option. 4.60
To find the amount of quarterly deposit (A), we establish the following equivalence relationship: 4 0.06 ia = 1 + − 1 = 0.06136 4 A( F / A,1.5%, 60) = $60, 000 + $60, 000( P / A, 6.136%,3) A = $219,978 / 96.2147 A = $2, 286.32
4.61 Setting the equivalence relationship at the end of 20 years gives 2
0.06 isemiannual = 1 + − 1 = 3.0225% 4 6% A( F / A, ,80) = $40, 000( P / A,3.0225%, 20) 4 152.71A = $593,862.93 A = $3,888.81 4.62 Given i =
5% = 0.417% per month 12
A = $500,000( A / P,0.417%,120) = $5,303.26 4.63 First compute the equivalent present worth of the energy cost savings during the first operating cycle:
0
1
$70 $70
$70
2
4
June
3
July Aug.
$80 $80 $80 5
6
7
8
9
10
11
12
Dec. Jan. Feb. .
P = $70( P / A, 0.5%,3)( P / F , 0.5%,1) + $80( P / A, 0.5%,3)( P / F , 0.5%, 7) = $436.35
Then, compute the total present worth of the energy cost savings over 5 years. P = $436.35 + $436.35( P / F , 0.5%,12) + $436.35( P / F , 0.5%, 24) +$436.35( P / F , 0.5%,36) + $436.35( P / F , 0.5%, 48) = $1,942.55
Continuous Payments with Continuous Compounding 4.64 Given i = 10%, N = 10 years, and A = $95,000 × 365 = $34,675,000 • Daily payment with daily compounding:
10% ,3650) = $219,170,331.48 365 • Continuous payment and continuous compounding: P = $95,000(P / A,
10
P = Ae− rt dt 0
e(0.10)(10) − 1 = $34,675,000 (0.10)(10) 0.1e = $219,187,803.77
∴ The difference between the two compounding schemes is only $17,472.27.
4.65 Given i = 11%, N = 3, F0 = $500, 000, FN = $40, 000, and 0 ≤ t ≤ 3,
$500,000
$40,000 0
1
2
3
t
460, 000 t 3 3 460, 000 − rt P = (500, 000 − t )e dt 0 3 1 − e −0.11(3) $500, 000 − $40, 000 [ −0.11(3)e −0.11(3) − e −0.11(3) + 1] = 500, 000 − 2 0.11 3(0.11)
f (t ) = 500, 000 −
= $1, 277, 619.39 − $555, 439.67 = $722,179.72
4.66 Given r = 9% , A = $25, 000, N s = 2, N e = 7, 7
P = 25, 000e − rt dt 2
e −0.09(2) − e −0.09(7) = $25, 000 0.09 = $84, 077.34
4.67
yt = 5e−0.25t , yt ut = 275e
ut = $55(1 + 0.09t )
−0.25 t
+ 24.75te −0.25t
20
20
P = 275e−0.25t e −0.12t dt + 24.75te −0.25t e−0.12t dt 0
0
e − 1 24.75 24.75 = 275 + (1 − e −0.37(20) ) − (20e −0.37(20) ) 0.37(20) 2 0.37 0.37e 0.37 = $742.79 + $179.86 = $922.65 0.37(20)
Changing Interest Rates 4.68
F = $10, 000( F / P, 6%,3)( F / P,9%, 4)( F / P,7%,3) = $20,595.62 4.69 Given r1 = 6% compounded quarterly, r2 = 10% compounded quarterly, and r3 = 8% compounded quarterly, indicating that i1 = 1.5% per quarter,
i2 = 2.5% per quarter, and i3 = 2% per quarter. (a) Find P: P = $2,000( P / F ,1.5%,4) + $2,000( P / F ,1.5%,8) + $3,000( P / F ,2.5%,4)( P / F ,1.5%,8) + $2,000( P / F ,2.5%,8)( P / F ,1.5%,8) + $2,000( P / F ,2%,4)( P / F ,2.5%,8)( P / F ,1.5%,8) = $8,875.42
(b) Find F:
F = P( F / P,1.5%,8)( F / P,2.5%,8)( F / P,2%,4) = $13,186 (c) Find A, starting at 1 and ending at 5: F = A + A(F / P, 2%, 4) + A(F / P, 2.5%, 4)(F / P, 2%, 4) + A(F / P, 2.5%, 8)(F / P, 2%, 4) + A(F / P,1.5%, 4)(F / P, 2.5%, 8)(F / P, 2%, 4) = 5.9958A A=
4.70 (a)
$13,186 = $2,199.21 5.9958
P = $300( P / F ,0.5%,12) + $300( P / F ,0.75%,12)( P / F ,0.5%,12) + $500( P / F ,0.75%,24)( P / F ,0.5%,12) + $500( P / F ,0.5%,12)( P / F ,0.75%,24)( P / F ,0.5%,12) = $1,305.26
(b)
$1,305.26 = $300( P / A, i, 2) + $500( P / A, i, 2)( P / F , i, 2) i = 7.818% per year
4.71 Since payments occur annually, you may compute the effective annual interest rate for each year.
i1 = (1 +
0.09 365 ) − 1 = 9.416% , 365
i2 = e 0.09 − 1 = 9.417%
F = $400( F / P,9.416%, 2)( F / P,9.417%, 2) + $250( F / P,9.416%,1)( F / P,9.417%, 2) +$100( F / P,9.417%, 2) + $100( F / P,9.417%,1) + $250 = $1,379.93
4.72 i1 = e0.06 − 1 = 6.18%,
i2 = e0.08 − 1 = 8.33%
F = $1, 000e0.06e0.08 = $1, 000( F / P, 6.18%,1)( F / P,8.33%,1) = $1,150.25
Amortized Loans 4.73 Loan repayment schedule for the selected 6 payments: End of month 0 1 2 13
Interest Payment $0.00 $100.00 $97.46 $68.64
Principal Payment $0.00 $508.44 $510.98 $539.80
Remaining Balance $20,000.00 $19,491.56 $18,980.58 $13,188.31
24 36
$38.20 $3.03
$570.24 $605.41
$7,069.38 $0
4.74 (a) (i) $10,000 ( A / P,0.75%,24) (b) (iii) B12 = A( P / A,0.75%,12) 4.75 Given information: i = 9.45% / 365 = 0.0259% per day , N = 36 months.
•
Effective monthly interest rate, i = (1 + 0.000259)30 − 1 = 0.78% per month
•
Monthly payment, A = $13,000( A / P,0.78%,36) = $415.58 per month
•
Total interest payment, I = $415.58 × 36 − $13, 000 = $1, 960.88
4.76 (a) Using the bank loan at 9.2% compound monthly Purchase price = $22,000, Down payment = $1,800 A = $20,200( A / P,(9.2 / 12)%,48) = $504.59
(b) Using the dealer’s financing, Purchase price = $22,000, Down payment = $2,000, Monthly payment = $505.33, N = 48 end of month payments. Find the effective interest rate: $505.33 = $20, 000( A / P, i, 48) i = 0.8166% per month APR(r ) = 0.8166% × 12 = 9.80%
4.77 Given Data: P = $25,000, r = 9% compounded monthly, N = 36 month, and
i = 0.75% per month. •
Required monthly payment: A = $25, 000( A / P, 0.75%, 36) = $795
•
The remaining balance immediately after the 20th payment:
B20 = $795( P / A,0.75%,16) = $11,944.33 4.78 Given Data: P = $250,000 - $50,000 = $200,000. •
Option 1: N = 15 years × 12 = 180 months APR = 4.25% ∴ A = $200, 000( A / P, 4.25% / 12,180) = $1, 504.56
•
Option 2: N = 30 years × 12 = 360 months APR = 5% ∴ A = $200,000( A / P,5% / 12,360) = $1,073.64 ∴ Difference = $1, 504.56 - $1,073.64 = $430.92
4.79 •
The monthly payment to the bank: Deferring the loan payment for 6 months is equivalent to borrowing $16, 000( F / P, 0.75%, 6) = $16, 733.64
To pay off the bank loan over 36 months, the required monthly payment is A = $16, 733.64( A / P, 0.75%,36) = $532.13 per month
•
The remaining balance after making the 16th payment: $532.13( P / A, 0.75%, 20) = $9,848.67
•
The loan company will pay off this remaining balance and will charge $308.29 per month for 36 months. The effective interest rate for this new arrangement is: $9,848.67 = $308.29( P / A, i,36) ( P / A, i,36) = 31.95 i = 0.66% per month
∴ r = 0.66% × 12 = 7.92% per year
4.80 (a)
A = $300, 000( A / P,
5.5% ,180) = $2, 451.25 12
(b) Remaining balance after the 59th payment:
B59 = $2, 451.25( P / A,
5.5% ,121) = $227, 276.52 12
•
Interest for the 60th payment = (5.5% /12) × B23 = $1,041.68
•
Principal payment for the 60th payment = $2, 451.25 − $1, 041.68 = $1, 409.57
•
9% ,180) = $4, 057.07 12 Total payments over the first 5 years (60 months)
4.81
A = $400, 000( A / P,
$4,057.07 × 60 = $243,424.20
•
Remaining balance at the end of 5 years: B60 = $4,057.07( P / A,0.75%,120) = $320,271.97
• •
Reduction in principal = $400,000 - $320,271.97 = $79,728.03 Total interest payments = $243, 424.20 − $79,728.03=$163,696.17
4.82 The amount to finance = $300,000 - $45,000 = $255,000
A = $255, 000( A / P, 0.5%,360) = $1,528.85
Then, the minimum acceptable monthly salary (S) should be
S=
A $1,528.85 = = $6,115.42 0.25 0.25
4.83 Given Data: purchase price = $150,000, down payment (sunk equity) = $30,000, interest rate = 0.75% per month, N = 360 months, •
Monthly payment: A = $120,000 ( A / P,0.75%,360 ) = $965 .55
•
Balance at the end of 5 years ( 60 months):
B60 = $965.55( P / A,0.75%,300) = $115,056.50 •
Realized equity = sales price – balance remaining – sunk equity: $185,000 - $115,056.60 - $30,000 = $39,943.50 Note: For tax purpose, we do not consider the time value of money on $30,000 down payment made five years ago.
4.84 Given Data: interest rate = 0.75% per month, each individual has the identical remaining balance prior to their 20th payment, that is, $80,000. With equal remaining balances, all will pay the same interest for the 20th mortgage payment. $80,000(0.0075) = $600 4.85 Given Data: loan amount = $130,000, point charged = 3%, N = 360 months, interest rate = 0.75% per month, actual amount loaned = $126,100: •
Monthly repayment: A = $130,000 ( A / P,0.75%,360 ) = $1,046
•
Effective interest rate on this loan
$126,100 = $1,046( P / A, i,360) i = 0.7787% per month ∴ ia = (1 + 0.007787)12 − 1 = 9.755% per year
4.86 (a)
$50,000 = $7,500(P / A,i,5) + $2,500(P / G,i,5) i = 6.914%
(b) P = $50,000 Total payments = $7,500 + $10,000 + … + $17,500 = $62,500 Interest payments = $3,456.87 + … + $1,131.66 = $12,500 End of month 0 1 2 3 4 5
Interest Payment $0.00 $3,456.87 $3,177.34 $2,705.64 $2,028.48 $1,131.66
Principal Payment $0.00 $4,043.13 $6,822.66 $9,794.36 $12,971.52 $16,368.34
Remaining Balance $50,000.00 $45,956.87 $39,134.21 $29,339.85 $16,368.34 $0
4.87 Given Data: r = 7% compounded daily, N = 25 years •
The effective annual interest rate is ia = (1 + 0.07 / 365) 365 − 1 = 7.25%
•
Total amount accumulated at the end of 25 years $75, 000 × 5% = $3, 750
F = $3, 750(F / A, 7.25%, 25) + $150(F / G, 7.25%, 25) = $3, 750(F / A, 7.25%, 25) + $150(P / G, 7.25%, 25)(F / P, 7.25%, 25) = $329, 799.78
4.88 (a) The dealer’s interest rate to calculate the loan repayment schedule. (b)