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Words 771

Pages 4

Bond Valuation with Annual payments

Jackson Corporation’s bonds have 12 years remaining to maturity. Interest is paid annually, the bonds have a $1,000 par value, and the coupon interest rate is 8%. The bonds have a yield to maturity of 9%. What is the current market price of these bonds?

P = f*r*[1-(1+i)^-n]/i+C*(i^-n)

F = par value = 1000

C = maturity value = 1000 r = coupon rate per coupon payment period = .08 i = effective interest rate per coupon payment period = .09 n = number of coupon payments remaining = 12

=1000*0.08*[1-(1+1.09)^-12]/0.09+1000*(1.09^-12)

= 928.3927

= $928.39

5-2

Yield to Maturity for Annual payments

Wilson Wonder’s bonds have 12 years remaining to maturity. Interest is paid annually, the bonds have a $1,000 par value, and the coupon interest rate is 10%. The bonds sell at a price of $850. What is their yield to maturity?

TTM = 12 years

Par = $1000

n = 12; PMT = 100; PV = -850; i = 12.4751% c = 10% (100) price = $850 YTM =

PMT = 100 i = solve

Yield to Maturity = 12.48%

Time to maturity = 12 yrs.

Par value = $1,000

Coupon interest rate = 10%

Price of bond = $850

Value of bond = t=1nParvalue*coupon interest rate1+YTMt+Par value1+YTMn

5-6

Maturity Risk Premium

The real risk-free rate is 3%, and inflation is expected to be 3% for the next 2 years. A 2-year Treasury security yields 6.3%. What is the maturity risk premium for the 2-year security?

R = r*+IP+MRP

6.3 = 3+3+MRP

6.4 = 6 + MRP

MRP = 6.3 – 6

MRP = 0.3%

5-7

Bond Valuation with Semiannual payments

Renfro Rentals has issued bonds that have a 10% coupon rate, payable semiannually. The bonds mature in 8 years, have a face value of $1,000, and a yield to maturity of 8.5%. What is the price of the bonds?

FV 1,000 PMT 50 N 16 R 4.25% Present Value = $1,085.80

$1,085.80

5-13

Yield to Maturity and Current Yield

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...Week 6 homework 1. (TCO 5) Which component of an Intelligent Storage System provides the interface between the storage system and the host? a. The Front End b. The Cache Memory c. The Back End d. The Physical Disks 2. (TCO 5) Which component of an Intelligent Storage System is made of semiconductor memory where data is placed temporarily? a. The Front End b. The Cache Memory c. The Back End d. The Physical Disks 3. (TCO 5) What is the smallest unit of cache that can be allocated? a. book b. chapter c. section d. page 4. (TCO 5) What is used by cache memory to indicate whether data in cache has been written to disk or not? a. A dirty bit flag b. A grimy bit flag c. An uncommitted flag d. A committed flag 5. (TCO 5) What term is give used to describe an operation where the required data was found in cache? d. Read loss e. Read miss f. Read find d. Read hit 6. (TCO 5) Select two protocols that are used as the transport mechanism when implementing IP SAN storage systems. g. IPX/SPX b. iSCSI c. FCIP d. SNA 7. (TCO 5) Which protocol is a host-based encapsulation of SCSI I/O over IP using an Ethernet NIC card or an HBA? h. TCP/IP i. FC j. FCIP d. iSCSI 8. (TCO 5) Which protocol is extensively used in disaster-recovery implementations where data is duplicated on disk or tape to an alternate site? k. TCP/IP l. FC c. FCIP d. iSCSI ...

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...Week 3 homework Exercise 7.1, problem 10a The total number of pairs we can either include or not is 4^2-4=16. Any reflexive relation is a subset of this set of 12 elements; we know there are 2^12 such subsets. Problem 10b The number of decisions we can make for any symmetric relation is 4+ (16-4)/2=4+6=10. The number of possible symmetric relations is 210. Exercise 7.2, Problem 15a a) Draw the digraph G1 (V1, E1) where V1 {a, b, c, d, e, f } and E1 {(a, b), (a, d), (b, c), (b, e), (d, b), (d, e), (e, c), (e, f), (f, d)}. Exercise 7.3, Problem 1 Draw the Hasse diagram for the poset ⊆, where{1, 2, 3, 4}. (1,1)<(1,2)<(1,3)<(1,4)<(2,1)<(2,2)<...<(4,3)<(4,4 ). o------o------o------o------o------o--- ... ---o------o (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) ... (4,3) (4,4) 3 1 2 4 3 1 2 4 Problem 6a . For A {a, b, c, d, e}, the Hasse diagram for the poset (A, R) is shown in Fig. 7.23. (a) Determine the relation matrix for R. a b c d e a 1 1 1 0 0 b 1 1 0 1 0 c 1 0 1 1 0 d 0 1 1 1 0 e 1 0 0 0 1 Exercise 7.4, Problem 1a Determine whether each of the following collections of sets is a partition for the given set A. If the collection is not a partition, explain why it fails to be. a) A {1, 2, 3, 4, 5, 6, 7, 8}; A1 {4, 5, 6}, A2 {1, 8}, A3 {2, 3, 7}. The Question A is a partition. Problem 2a There are only three ways, corresponding to the partition A_1 = {1,2,8}, A_2 = {3,4}, A_3 =......

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...WEEK 3 Homework * * * What is the relationship between a firm’s life cycle stage and its ability to accurately forecast sales? Why is this the case, and how do capital suppliers respond? Why is so much emphasis placed on forecasting sales and cash balances in building proformas and projections? * * * Professor, class, * * There is a tight relationship between a firm's life cycle stage and its ability to accurately forecast sales. The earlier a venture is at its life cycle stage the more difficult it will have to accurately forecast its sales. It is usually easier to forecast a firm being at the later stages of life cycle such as maturity compared to early stage firms that have not operated yet. Firms at the early life cycle stage do not have any historical sales and records that are the crucial element to guide the future performances. Capital suppliers emphasize a lot on forecasting sales that is an efficient way to see future performances and results of the ventures they are about to finance * * * * * * * * * Chapter 6: Discussion Question: #4 p. 223 Why is it usually easier to forecast sales for seasoned firms in contrast with early-stage ventures? We usually forecast sales efficiently based on an horizontal analysis in other words, based on sales from previous years. The difficulty in forecasting sales for early stage ventures rely on the fact the ventures do not have any past......

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...Homework Week 3 September 22, 2013 DeVry University Online Operant conditioning sometimes referred to as instrumental conditioning is a method of learning that occurs through rewards and punishments for behavior. Through operant conditioning, an association is made between a behavior and a consequence for that behavior. A good example of Operant conditioning is: employees finishing projects to receive praise or promotions (Cherry, 2013). Observational learning occurs when one observes another’s behaviors, which enables one to reenact those behaviors. In the workplace observational learning is used in a few ways. For example: whenever a manager started to train an employee they used instructional videos, so that the new employees could observe the job functions they were to undertake. Next, normally the manager would take the new employees into the actual work area that they would occupy and allow them to observe someone performing the job task. After that, the manager would pair a new employee with an employee that performed their job title well enough to train another employee. This allowed the new employee to observe the other employee, which enabled the new employee to learn though observing the other employees behaviors (Whitaker, 2012). The social learning theory, a system of learning most commonly associated with behaviorist Albert Bandura, is most commonly applied in educational settings. You can also apply this theory, which argues that people learn from each......

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...MAT540 Week 3 Homework Click Link Below To Buy: http://hwcampus.com/shop/mat540-week-3-homework/ MAT540 Week 3 Homework Week 3 Page 1 of 3 MAT540 Week 3 Homework Chapter 14 1. The Hoylake Rescue Squad receives an emergency call every 1, 2, 3, 4, 5, or 6 hours, according to the following probability distribution. The squad is on duty 24 hours per day, 7 days per week: Time Between Emergency Calls (hr.) Probability 1 0.15 2 0.10 3 0.20 4 0.25 5 0.20 6 0.10 1.00 a. Simulate the emergency calls for 3 days (note that this will require a “running” , or cumulative, hourly clock), using the random number table. b. Compute the average time between calls and compare this value with the expected value of the time between calls from the probability distribution. Why are the result different? 2. The time between arrivals of cars at the Petroco Services Station is defined by the following probability distribution: Time Between Emergency Calls (hr.) Probability 1 0.35 2 0.25 3 0.20 4 0.20 1.00 MAT540 Homework Week 3 Page 2 of 3 a. Simulate the arrival of cars at the service station for 20 arrivals and compute the average time between arrivals. b. Simulate the arrival of cars at the service station for 1 hour, using a different stream of random numbers from those used in (a) and compute the average time between arrivals. c. Compare the results obtained in (a) and (b). 3. The Dynaco Manufacturing Company produces a product in a process......

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...MAT540 Week 3 Homework Click Link Below To Buy: http://hwcampus.com/shop/mat540-week-3-homework/ MAT540 Week 3 Homework Week 3 Page 1 of 3 MAT540 Week 3 Homework Chapter 14 1. The Hoylake Rescue Squad receives an emergency call every 1, 2, 3, 4, 5, or 6 hours, according to the following probability distribution. The squad is on duty 24 hours per day, 7 days per week: Time Between Emergency Calls (hr.) Probability 1 0.15 2 0.10 3 0.20 4 0.25 5 0.20 6 0.10 1.00 a. Simulate the emergency calls for 3 days (note that this will require a “running” , or cumulative, hourly clock), using the random number table. b. Compute the average time between calls and compare this value with the expected value of the time between calls from the probability distribution. Why are the result different? 2. The time between arrivals of cars at the Petroco Services Station is defined by the following probability distribution: Time Between Emergency Calls (hr.) Probability 1 0.35 2 0.25 3 0.20 4 0.20 1.00 MAT540 Homework Week 3 Page 2 of 3 a. Simulate the arrival of cars at the service station for 20 arrivals and compute the average time between arrivals. b. Simulate the arrival of cars at the service station for 1 hour, using a different stream of random numbers from those used in (a) and compute the average time between arrivals. c. Compare the results obtained in (a) and (b). 3. The Dynaco Manufacturing Company produces a product in a process......

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...MAT540 Week 3 Homework Click Link Below To Buy: http://hwcampus.com/shop/mat540-week-3-homework/ MAT540 Week 3 Homework Week 3 Page 1 of 3 MAT540 Week 3 Homework Chapter 14 1. The Hoylake Rescue Squad receives an emergency call every 1, 2, 3, 4, 5, or 6 hours, according to the following probability distribution. The squad is on duty 24 hours per day, 7 days per week: Time Between Emergency Calls (hr.) Probability 1 0.15 2 0.10 3 0.20 4 0.25 5 0.20 6 0.10 1.00 a. Simulate the emergency calls for 3 days (note that this will require a “running” , or cumulative, hourly clock), using the random number table. b. Compute the average time between calls and compare this value with the expected value of the time between calls from the probability distribution. Why are the result different? 2. The time between arrivals of cars at the Petroco Services Station is defined by the following probability distribution: Time Between Emergency Calls (hr.) Probability 1 0.35 2 0.25 3 0.20 4 0.20 1.00 MAT540 Homework Week 3 Page 2 of 3 a. Simulate the arrival of cars at the service station for 20 arrivals and compute the average time between arrivals. b. Simulate the arrival of cars at the service station for 1 hour, using a different stream of random numbers from those used in (a) and compute the average time between arrivals. c. Compare the results obtained in (a) and (b). 3. The Dynaco Manufacturing Company produces a product in a process......

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Words: 520 - Pages: 3

...dividend. A similar stock is selling on the market for $70. Burnwood must pay flotation costs of 5% of the issue price. What is the cost of the preferred stock? Dps = 6%*$60 = $3.60 Pps = 70 F = 5% Rps = Dps / Pps(1 – F) Rps = $3.60 / $70(1 – 0.05) Rps = $3.60 / $70(0.95) Rps = $3.60 / 66.5 Rps = 0.0541 or 5.41% 9-5 - Summerdahl Resort’s common stock is currently trading at $36 a share. The stock is expected to pay a dividend of $3.00 a share at the end of the year (D1 = $3.00), and the dividend is expected to grow at a constant rate of 5% a year. What is its cost of common equity? Rs = (D1/P0) + Expected g Rs = (3.00/36) + 5% Rs = 8.33% + 5% Rs = 13.33 % 9-6 - Booher Book Stores has a beta of 0.8. The yield on a 3-month T-bill is 4% and the yield on a 10-year T-bond is 6%. The market risk premium is 5.5%, and the return on an average stock in the market last year was 15%. What is the estimated cost of common equity using the CAPM? Rs = Rrf + (RPm)Bi Rs = 6.0% + (5.5%*0.8) Rs = 6.0% + 4.4% Rs = 10.4% 9-7 - Shi Importer’s balance sheet shows $300 million in debt, $50 million in preferred stock, and $250 million in total common equity. Shi’s tax rate is 40%, rd = 6%, rps = 5.8%, and rs = 12%. If Shi has a target capital structure of 30% debt, 5% preferred stock, and 65% common stock, what is its WACC? WACC = WdRd(1 – T) + WpsRps + WsRs WACC = (0.3*6.0)(1-.40) + (0.05*5.8) + (0.65*12.0) WACC = 1.08% + 0.29% + 7.8% WACC = 9.17%...

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...FI515 Week 7 Homework Problems pp. 681-682 16-1 Cash Management Williams & Sons last year reported sales of $10 million and an inventory turnover ratio of 2. The company is now adopting a new inventory system. If the new system is able to reduce the firm’s inventory level and increase the firm’s inventory turnover ratio to 5 while maintaining the same level of sales, how much cash will be freed up? Sales=$10,000,000 Inventory turnover ratio (old) 2 Inventory Turnover ratio (new) 5 Freed up Cash ? (old)=10,000,000/2=$5,000,000 (new)=10,000,000/5=$2,000,000 Freed up cash=old inventory-new inventory $5,000,000-2,000,000 =$3,000,000 Medwig 16-2 Receivables Investment Corporation has a DSO of 17 days. The company averages $3,500 in credit sales each day. What is the company’s average accounts receivable? DSO= Accounts Receivables/Credit Sales(365) 17= Accounts Receivables/$3,500 Account Receivables= 17*$3,500=$59,500 16-3 Cost of Trade Credit What is the nominal and effective cost of trade credit under the credit terms of 3/15, net 30? Nominal cost of trade formula= discount percentage 100-discount percentage*365days credit is outstanding-discount period =397*36530-15 =0.03093*24.33 =0.75263 =75.26% Effective cost of trade formula= Periodic rate=0.03/0.97=0.3093 Periods/year=365/(30-15)=24.33 EAR=(1+periodic rate)n-1 =(1.03093)24.33-1=109.84% 16-4 Cost of Trade Credit ...

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...Rose Sandeen Week 3 Homework Chapter 8 You are given the following equations for the aggregate demand (AD) and short-run aggregate supply (SAS) curves: AD: Y = 1.25Aᴩ + 2.5Mˢ/P SAS: Y = 11,250 - 20W + 1,000P where Y is real GDP, Aᴩ is the amount of autonomous planned spending that is independent of the interest rate, Mˢ is the nominal money supply, P is the price level, and W is the nominal wage rate. Assume that Aᴩ equals 5,000, Mˢ equals 2,000, W equals 50, and natural real GDP, Yᴺ, equals 11,250. Use the values for the amounts of autonomous planned spending that is independent of the interest rate and the nominal money supply to derive the equation for the aggregate demand curve. Compute the amount of aggregate demand when the price level equals 2.0, 1.25, 1.0, 0.8, and 0.5. Graph the aggregate demand curve. Answer: The equation for aggregate demand is Y = 6,250 + 5,000/P. See below graph for aggregate demand curve. Price Level |2.0 |1.25 |1.0 |0.8 |0.5 | |Aggregate Demand |8,750 |10,250 |11,250 |12,500 |16,250 | | Derive the equation for the short-run aggregate supply curve, given that the nominal wage rate equals 50. Compute the amount of short-run aggregate supply when the price level equals 2.0, 1.25, 1.0, 0.8, and 0.5. Graph the short-run aggregate supply curve. Answer: The equation for the short-run aggregate supply curve is Y = 10,250 + 1,000P. See below graph for short-run aggregate supply curve. Price Level |2.0 |1.25 |1.0 |0.8 |0.5 |......

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...Easy Problems 1-5 (3-1) Days Sales Outstanding Greene Sisters has a DSO of 20 days. The company’s average daily sales are $20,000. What is the level of its accounts receivable? Assume there are 365 days in a year. DSO= 20; Average Daily Sales= 20,000 20 = A/R / 20,000 20 x 20,000= A/R 400,000 = A/R (3-2) Debt Ratio Vigo Vacations has an equity multiplier of 2.5. The company’s assets are financed with some combination of long-term debt and common equity. What is the company’s debtratio? Equity Multiplier= 2.5 Equity Ratio= 1/EM 1 / 2.5= .40 Debt Ratio + Equity Ratio= 1 Therefore Debt Ratio= 1-Equity Ratio= 1- 0.40 = .60% (3-3) Market/Book Ratio Winston Washers’s stock price is $75 per share. Winston has $10 billion in total assets. Its balance sheet shows $1 billion in current liabilities, $3 billion in long-term debt, and $6 billion in common equity. It has 800 million shares of common stock outstanding. What is Winston’s market/book ratio? Market Value per share= 75; Common Equity= 6,000,000; Number of shares outstanding= 800 million Market-to-book ratio = market value per share/(common equity/number of shares outstanding) Market-to-book ratio = $75/(6,000,000/800,000,000) = $75/7.5 = 10 market-to-book ratio (3-4) Price/Earnings Ratio A company has an EPS of $1.50, a cash flow per share of $3.00, and a price/cash flow ratio of 8.0. What is its P/E ratio? Price per share = $8 x $3 = $24 P/E = $24 / 1.5 = 16 P/E ratio (3-5)......

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...3-1 Days Sales Outstanding Accounts Receivable Average Credit Sales Solution: ? = 20,000x20 = $400,000 (Accounts Receivable) 400,000 = 20 20,000 3-2 Debt Ratio Total Liabilities / Total Assets Solution: 60% 3-3 Market/Book Ratio Market price per share / Book value per share Book value per share = Common equity / shares outstanding Book value per share = $6,000,000,000 / 800,000,000 = 7.5 Solution: $75 / 7.5 = 10 3-4 PE Ratio Price/Earnings Ratio = Price per share / Earnings per share 8*3.00 = 24 24/$1.50 = 16 3-5 ROE ROE = ROA x Equity Multiplier ROA = Profit Margin x Total assets turnover ROA = .03 x 2 = .06 ROE = .06 x 2.0 = .12 = 12% 3-6 Du Pont Analysis Total Assets Turnover = 5 ROA = Profit Margin x Total Assets Turnover 10% = 2% x ? 10% / 2% = 5 Equity Multiplier = 1.5 ROE = Profit Margin x Total Assets Turnover x Equity Multiplier 15% = 2% x 5 x ? 15% / .1 = 1.5 3-7 Current and Quick Ratios Quick Ratio = Current Assets – Inventories / Current Liabilities Current Liabilities = $3,000,000 / 1.5 Current Liabilities = $2,000,000 1.0 = $3,000,000 - ? / $2,000,000 Inventories = $1,000,000 4-1 FV of Single Amount FV = PV (1+ I)^N FV = $10,000 (1+.10)^5 FV = $10,000 (1.10)^5 FV = $16,105.10 4-2 PV of Single Amount 1292.10 PV = FVn / (1+I)^N PV = $5,000 / (1+ .07)^20 PV = $5,000 / (1.07)^20 PV = $1292.10 4-6 FV of Ordinary Annuity FVAn = PMT { (1+I)^n / I – 1/I } FVAn = $300 [ (1+.07)^5 / .07 – 1/.07] FVAn = $300 [...

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...& Sons last year reported sales of $10 million and an inventory turnover ratio of 2. The company is now adopting a new inventory system. If the new system is able to reduce the firm’s inventory level and increase the firm’s inventory turnover ratio to 5 while maintaining the same level of sales, how much cash will be freed up? Old inventory-2 10,000,000/2= 5,000,000 New inventory-5 10,000,000/5=2,000,000 Old inventory-new inventory=freed cash 5,000,000-2,000,000= $3,000,000 (16-2) Receivables Investment Medwig Corporation has a DSO of 17 days. The company averages $3,500 in credit sales each day. What is the company’s average accounts receivable? 17x 3,500= $59,500 (16-3) Cost of Trade Credit What is the nominal and effective cost of trade credit under the credit terms of 3/15, net 30? Nominal- 397 x 36530-15= 0.03093 x 24.33= 0.75263= 75.26% Effective- 0.03/0.97=0.3093= Periods/years=365/(30-15)=24.33 0.3093 x 24.33-1=109.84% (16-4) Cost of Trade Credit A large retailer obtains merchandise under the credit terms of 1/15, net 45, but routinely takes 60 days to pay its bills. (Because the retailer is an important customer, suppliers allow the firm to stretch its credit terms.) What is the retailer’s effective cost of trade credit? 0.01/0.99=0.01010 365/(60-15)=8.11 EAR=(1+0.01010) x 8.11-1= 8.49% (16-5) Accounts Payable A chain of appliance stores, APP Corporation, purchases inventory with a net price of $500,000 each day. The......

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...========================================================== FIN515: Week-1 Assignment ========================================================== MINI-CASE (p. 45): Consulting Michelle DellaTorre To explain the U.S. Financial System to Ms. DellaTorre, I would provide her with the following answers to the questions given to me by my boss: 1. Why is corporate finance important to all managers? Corporate finance is a basic component of how a business is run. All managers should keep this in mind to direct funds to the optimal division or product in a company. In addition, managers should understand how their company is financed and whether it has a risk of bankruptcy. Conversely, managers should understand if the equity in the business is undervalued and has the potential to grow. 2. Describe the organizational forms a company might have as it evolves from a start-up to a major corporation. List the advantages and disadvantages of each form. Sole Proprietorship: A type of business entity that is owned and run by one individual and in which there is no legal distinction between the owner and the business. The owner receives all profits (subject to taxation specific to the business) and has unlimited responsibility for all losses and debts. Every asset of the business is owned by the proprietor and all debts of the business are the proprietor's. This means that the owner has no less liability than if they were acting as an individual instead of as a business. Some......

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