Free Essay

Reliey Solutions

In:

Submitted By november20
Words 1110
Pages 5
CHAPTER 5

Answers to Problems

1(a). Given a three security series and a price change from period t to t+1, the percentage change in the series would be 42.85 percent.

Period t Period t+1 A $ 60 $ 80 B 20 35 C 18 25 Sum $ 98 $140

Divisor 3 3

Average 32.67 46.67

1(b). Period t Stock Price/Share # of Shares Market Value A $60 1,000,000 $ 60,000,000 B 20 10,000,000 200,000,000 C 18 30,000,000 540,000,000 Total $800,000,000

Period t+1 Stock Price/Share # of Shares Market Value A $80 1,000,000 $ 80,000,000 B 35 10,000,000 350,000,000 C 25 30,000,000 750,000,000 Total $1,180,000,000

1(c). The percentage change for the price-weighted series is a simple average of the differences in price from one period to the next. Equal weights are applied to each price change.

The percentage change for the value-weighted series is a weighted average of the differences in price from one period t to t+1. These weights are the relative market values for each stock. Thus, Stock C carries the greatest weight followed by B and then A. Because Stock B had the greatest percentage increase (75%) and the second largest weight, the percentage change would be larger for this series than the price-weighted series.
2(a). Period t Stock Price/Share # of Shares Market Value A $60 16.67 $ 1,000,000 B 20 50.00 1,000,000 C 18 55.56 1,000,000 Total $3,000,000

Period t+1 Stock Price/Share # of Shares Market Value A $80 16.67 $ 1,333.60 B 35 50.00 1,750.00 C 25 55.56 1,389.00 Total $4,472.60

2(b).

The answers are the same (slight difference due to rounding). This is what you would expect since Part A represents the percentage change of an equal-weighted series and Part B applies an equal weight to the separate stocks in calculating the arithmetic average.

2(c). Geometric average is the nth root of the product of n items.

The geometric average is less than the arithmetic average. This is because variability of return has a greater affect on the arithmetic average than the geometric average.

3. Student Exercise

4(a).

Day 1 Company Price/Share

A 12

B 23

C 52

Day 2

(Before Split) (After Split) Company Price/Share Price/Share

A 10 10

B 22 44

C 55 55

Day 3

(Before Split) (After Split) Company Price/Share Price/Share

A 14 14

B 46 46

C 52 26

Day 4

Company Price/Share

A 13

B 47

C 25

Day 5

Company Price/Share

A 12

B 45

C 26

4(b). Since the index is a price-weighted average, the higher priced stocks carry more weight. But when a split occurs, the new divisor ensures that the new value for the series is the same as it would have been without the split. Hence, the main effect of a split is just a repositioning of the relative weight that a particular stock carries in determining the index. For example, a 10% price change for company B would carry more weight in determining the percent change in the index in Day 3 after the reverse split that increased its price, than its weight on Day 2.

4(c). Student Exercise

5(a). Base = ($12 x 500) + ($23 x 350) + ($52 x 250) = $6,000 + $8,050 + $13,000 = $27,050

Day 1 = ($12 x 500) + ($23 x 350) + ($52 x 250) = $6,000 + $8,050 + $13,000 = $27,050

Index1 = ($27,050/$27,050) x 10 = 10

Day 2 = ($10 x 500) + ($22 x 350) + ($55 x 250) = $5,000 + $7,700 + $13,750 = $26,450

Index2 = ($26,450/$27,050) x 10 = 9.778

Day 3 = ($14 x 500) + ($46 x 175) + ($52 x 250) = $7,000 + $8,050 + $13,000 = $28,050

Index3 = ($28,050/$27,050) x 10 = 10.370

Day 4 = ($13 x 500) + ($47 x 175) + ($25 x 500) = $6,500 + $8,225 + $12,500 = $27,225

Index4 = ($27,225/$27,050) x 10 = 10.065

Day 5 = ($12 x 500) + ($45 x 175) + ($26 x 500) = $6,000 + $7,875 + $13,000 = $26,875

Index5 = ($26,875/$27,050) x 10 = 9.935

5(b). The market values are unchanged due to splits and thus stock splits have no effect. The index, however, is weighted by the relative market values.

6. Price-weighted index (PWI)2008 = (20 + 80+ 40)/3 = 46.67

To accounted for stock split, a new divisor must be calculated: (20 + 40 + 40)/X = 46.67 X = 2.143 (new divisor after stock split)

Price-weighted index2003 = (32 + 45 + 42)/2.143 = 55.53

VWI2002 = 20(100,000,000) + 80(2,000,000) + 40(25,000,000) = 2,000,000,000 + 160,000,000 + 1,000,000,000 = 3,160,000,000

assuming a base value of 100 and 1998 as base period, then (3,160,000,000/3,160,000,000) x 100 = 100

VWI2003 = 32(100,000,000) + 45(4,000,000) + 42(25,000,000) = 3,200,000,000 + 180,000,000 + 1,050,000,000 = 4,430,000,000

assuming a base value of 100 and 2002 as period, then (4,430,000,000/3,160,000,000) x 100 = 1.4019 x 100 = 140.19

6(a). Percentage change in PWI = (55.53 - 46.67)/46.67 = 18.99%

Percentage change in VWI = (140.19 - 100)/100 = 40.19%

6(b). The percentage change in VWI was much greater than the change in the PWI because the stock with the largest market value (K) had the greater percentage gain in price (60% increase).

6(c). December 31, 2002 Stock Price/Share # of Shares Market Value

K $20 50.0 $1,000.00

M 80 12.5 1,000.00 R 40 25.0 1,000.00 Total $3,000.00

December 31, 2003

Stock Price/Share # of Shares Market Value

K $32 50.0 $1,600.00

M 45 25.0* 1,125.00 R 42 25.0 1,050.00 Total $3,775.00

(*Stock-split two-for-one during the year.)

Unweighted averages are not impacted by large changes in stocks prices (i.e. price-weighted series) or in market values (i.e. value-weighted series).

-----------------------
[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Similar Documents

Free Essay

Science

...Detailed Lesson Plan (Grade 7) 1. Objectives After providing the necessary materials, each student; 1.1 investigates the different types of solutions: -unsaturated -saturated -supersaturated 1.2 performs an actual activity about solubility. 1.3 values the common solutions that can be found at home and can be used in daily living. 1. Learning Tasks 2.1 Topic: Solutions 2.2 Concept: The unsaturated solution has a less amount of solute to be dissolved. The saturated solution can hold no more solute to be dissolved. The supersaturated solution cannot hold more solute. 2.3 Materials 2.3.1 Textbook/Other Reference -Science Grade 7: Matter (K-12 Curriculum), pages 1-16 - http://www.infoplease.com/encyclopedia/science/solution-heat-solution.html 2.3.2 Instructional Materials - video clips, materials enumerated for the experiment and activity sheet. 2. Methodology 3.1 Daily Activities 3.1.1 Prayer/ Greetings Christ Jesus Whom we open our eyes, may you be there; When we open our ears, may you be there; When we open our mouths, may you be there; When we open our diaries, may you be there. Help us to see with your eyes; Help us to hear with your ears; Help us to speak your truth in love; Help us to make time for you… for others… for ourselves. Amen. 3.1.2 Checking of Attendance 3.1.3 Checking of Assignment 3.2 Preparatory Activities 3.2.1 Review Teacher: Class...

Words: 1391 - Pages: 6

Free Essay

Lab Report

...Osmosis Introduction: Tissues are made up of cells that have the same structure and function. In this experiment osmosis will be measured in a piece of tissue. Water potential can be determined by the changes in dimensions of the tissues when it is placed in solutions of different concentrations. Research Question: What is the effect of sucrose solution on potato cells Hypothesis: Osmosis will increase when the solutions are less concentrated with sucrose Variables: Dependant: Osmosis Independent: Length Materials: 12 large test tubes or specimen tubes with bungs, marker pen, potato chip cutter, scalpel, fine forceps, 3 watch glasses or petri dishes, ruler, 6 labeled beakers one containing 50cm^3 of distilled water and other four containing 50cm^3 of 0.2 mol dm^-3, 0.4 mol dm^-3, 0.6 mol dm^-3 and 0.8 mol dm^-3. Method: 1- Using the potato chip cutter, cut 24 chips from the potato. Cut the chips into 5cm lengths using the scalpel. Be as accurate as possible. Place 4 chips into each of six, labeled test tubes, one test tube for each of the different sucrose solutions. Pour in enough of each respective solution to cover the potato tissue. Put a bung in each one of the test tubes and label this series “potato”. 2- Repeat the method using the apple fruit or turnip root, putting four chips of tissue 5cm long into each of the second series of test tubes. Label these tubes “apple” or “turnip”. 3- Leave the tubes for at least an hour 4- After at least...

Words: 604 - Pages: 3

Free Essay

Environement Polltuion

...FUNDAMENTALS OF SURFACE MODES: ¥ COLLOIDS ¥ means ÒglueÓ in Greek ¥ was coined in 1861 by Thomas Graham. ¥ ¥ ¥ ¥ ¥ WHAT IS A COLLOID? usually consists of two phases; one continuous phase in which the other phase is dispersed. Size of particles: larger than the size of molecules and small enough for the dispersed phase to stay suspended for a longer period of time. ¥ No strict boundaries for the size limits. OFFICIAL DEFINITION ¥ In 1903 Wolfgang Ostwald formulated the official definition of a colloid: ¥ a system containing entities having at least one length scale in between 1nm and 1µm. ¥ For smaller particles there is no distinct boundaries between the phases and the system is considered a solution; ¥ for larger entities the particles will fall to the bottom due to the gravitational force, and the phases are separated. 1 2 3 MESOSCOPIC PHYSICS ¥ The particle size is in the so-called mesoscopic range in between the macroscopic and microscopic limits. LARGE INTERFACIAL AREA ¥ One very important quality of the colloids is the large interfacial area between the dispersed and the continuous phases. WHAT EFFECTS HAS THIS? ¥ This means that interface effects and hence the electromagnetic surface modes, are very important for the properties of the colloids. ¥ It costs energy to create this much surface and the particles would clump together if this isnÕt prevented. ¥ Usually the particles are charged and hence repel each other. 5 6 4 Four states...

Words: 2895 - Pages: 12

Premium Essay

Poop

...encouraged to ask question/s that is related to this module and that you don’t know the answer to and want it to be answered. The question/s can be answered by anyone in class. The instructor will allow sometime for other students to answer the question/s before contributing. You can post your questions in MODULE 2 forum under the Discussion Forums. Complete Assignment #2. The assignments are posted in the ASSIGNMENTS area of the website. Even though the homework assignments are not to be turned in and graded, you are strongly encouraged to do them to understand the material and to prepare for quizzes and tests. A complete solution to this assignment will be posted on the course website under ASSIGNMENTS on Thursday (Sept 4th). You are encouraged to review the solutions and compare your work to the solution, ensuring that you understand the reasons the solutions appear as they do. You are also encouraged...

Words: 485 - Pages: 2

Free Essay

Speech 203

... * Has no secretary The most important problem would be catching the flight for the meeting in Chicago since it’s an emergency and work related. Solutions: Getting work done for the day on the air plane on the way to the meeting. Call the secretary to find out at least some of the info for the meeting. Item#2 Problems: * Internal politicking in the company * Finding someone to replace F.T. Dickenson * Dealing with eliminating his overtime hours The two most important problems would the politicking, because someone else may feel the same way and decide to leave the company too. Also, finding someone to replace him and do his unfinished work in such a short period of time. Solution: Search for a new employee ASAP. Item#3 Problems: * Workers threatening to walk out over a co-worker * 10 votes to dismiss Foreman Edward George The workers are the most important asset to the company, so their interest should be first. Second, you have to figure out what to do about the votes to dismiss Ed George. Solutions: Talk to Ed George about his problems with his co-workers. Hold a meeting and get even more workers involved in the voting process, the take another vote. Item#4 Problems: * Overloading which can result in interruption of electrical power. * Not being reachable for Southern Power Solution: Contact Southern Power ASAP. See if they could come out and help with the problem. Item#5 Problem: * Balancing your work and his until Wednesday ...

Words: 922 - Pages: 4

Free Essay

Snnsnjs

...Name Date Class COLLIGATIVE PROPERTIES OF SOLUTIONS Section Review Objectives • Identify the three colligative properties of solutions • Describe why the vapor pressure, freezing point, and boiling point of a solution differ from those properties of the pure solvent. Vocabulary • colligative properties • freezing-point depression • boiling-point elevation Part A Completion Use this completion exercise to check your understanding of the concepts and terms that are introduced in this section. Each blank can be completed with a term, short phrase, or number. In a solution, the effects of a nonvolatile _______ on the properties of the solvent are called _______. They include _______ point and vapor pressure _______, and boiling point _______. In each case, the magnitude of the effect is _______ proportional to the number of solute molecules or ions present in the _______. Colligative properties are a function of the number of solute _______ in solution. For example, one mole of sodium chloride produces _______ as many particles in solution as one mole of sucrose and, thus, will depress the freezing point of water _______ as much. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Part B True-False Classify each of these statements as always true, AT; sometimes true, ST; or never true, NT. 11. When added to 1000 g of water, 2 moles of a solute will increase the boiling point by...

Words: 483 - Pages: 2

Free Essay

Potato Osmolarity

... it increases the water movement out of the plant by a process called transpiration. When there is a shortage of water,  the guard cells well get smaller and close the stomata, and the transpiration process will move slower. The osmolarity would be tested with solute concentrations ranging from 0.0 M to 0.6 M.  The concentrations increased 0.1 M each time and the solute used was sucrose. The goal of this experiment was to determine which concentration had the least effect on the potato after being incubated. This information helped us estimate the osmolarity of the potato tuber tissue. Our group hypothesis is that the osmolarity of the potato will have the concentration with the 0.1 M solution. This hypothesis is based off the predicted outcome that smaller molarity concentrations will have the least effect on the potato.  The more solute added to a solution decreases the concentration of water in most cases (Kosinski).  The decrease in water concentration would then lead to a lower weight of the potato tuber once it has finished incubating. Methods & Materials: For experiment A you will need the following items: 1 large potato tuber  Forceps petri dish razor blade DI water metric ruler 7 250 mL beakers/ or disposable cups   balance that weighs to the nearest 0.01g   sucrose...

Words: 1866 - Pages: 8

Premium Essay

Effect Of Solute Concentration On Egg Osmosis

...The data solute concentration of the egg was found by taking the average percent change in mass for each percent solute concentration used in the lab and putting the data into the graph(Mass vs. Solute Concentration Over 3 Days). According to the data from the class, the solute concentration inside an egg is 25.1%. The egg would be in an isotonic solution at 25.1% solute concentration. This is when the concentration outside the egg is equal with the concentration of water inside the egg. The point of equilibrium is the point at which the trendline crosses the x-axis. At this point, the egg would no longer gain or lose mass. If the concentrations are the same inside and outside the egg, diffusion will not be able to take place. All averages...

Words: 318 - Pages: 2

Premium Essay

Carcinus Maenas Lab

...Analysis of Carcinus maenas ingestion rates Materials and methods Part 1 - Control experiment Two mussels (mytilus edulis) were opened and the mussels inside cut into 4 pieces, producing 8 pieces of roughly equal size in total. A balance was then tared with a weighing boat placed on it. Each piece of mussel flesh was placed individually and in turn into the weighing boat to be weighed. The pieces were then placed into a plastic bucket half filled with filtered seawater. The weight, time of weighing and time of immersion into the bucket were all noted down for each piece of mussel. The position of each piece of flesh inside the bucket was also taken down so as not to mix them up. One of the eight pieces was left in the bucket for the duration of the experiment until being removed and weighed at the end and was written down as “the last”. Another piece was removed frequently throughout the experiment and was given the name “the one”. The last 6 pieces were each removed once and weighed at regular intervals to provide a range of submersion times. The experiment was run for roughly 20 minutes. The method for removing and weighing the pieces of mussel was kept consistent throughout the experiment. Each piece was removed, blotted clear of any excess water and weighed. If the piece of mussel was “the one”, it was returned to the bucket of seawater, if not, it was kept aside and moist for the crab feeding (Part 2). Once all of the pieces were removed and weighed (multiple times for “the...

Words: 933 - Pages: 4

Premium Essay

Abrasion Lab Report

... The purpose of the experiment was to see how abrasion works and how it affects rocks and minerals with the use of sugar cubes. I measured the mass of all 5 sugar cubes together then placed it in a jar with a lid on and shook it 20 times, as I was shaking the jar my lab partner recorded the mass that was taking with the number of times I shake the jar. After which we drew the shape of the sugar cubes when it was poured out on a paper, then we calculated the percent change. This procedure was repeated 5 times and with each trial the shaking time was increased by 20. After all the calculations was done, we graphed the data. After each trial the size and mass of the sugar cubes reduced, this was due to the pressure put on the jar as we were shaking it. Supposing little pressure was put on the jar, the outcome would have been different and would have shown little abrasion effects on minerals and rock so therefore this made me conclude that, for abrasion to have a greater impact on minerals and rocks the pressure of the water or wind must be strong to push the minerals against each other. Another factor that contributes to how abrasion works could be the amount of space the minerals and rocks have because during the experiment, the sugar cubes pumped into each other and the walls of the jar and since there wasn’t enough space for movement the sugar cubes weathered and reduce in size with the mass decreasing which I believe wouldn’t occur if there were to be enough space for the sugar...

Words: 462 - Pages: 2

Premium Essay

Polyimides

...Polyimides are a synthetic polymeric resin of a class resistant to high temperatures, wear, and corrosion, used primarily as a coating or film on a substrate substance. Polyimides are also polymers that usually consist of aromatic rings coupled by imide linkages-that is, linkages in which two carbonyl groups are attached to the same nitrogen atom. Polyimides cover the whole range of high performance polymers. They are a sophisticated family of materials which have applications in highly technical end use fields from aerospace to microelectronics. In this paper I will give an introduction to polyimides, and then I will talk about the formation of Polyimides, and finally the effects of solvents in Polyimides. Polyimides possess very unique key properties such as thermoxidative stability, high modulus, high mechanical strength, excellent electrical properties, and superior chemical resistance. Because of these merits, general difficulty in processing polyimides and their high demand cost did not deter advanced exploration of new compositions and new processing methods which points at value-added niche markets in advanced technology applications. A very important consideration in the successful synthesis of Polyimides is a design/creation of proper composition and proper decisions of synthesis methods. Ultimately, the latter is determined by the chemical and physical properties of monomers and polymers and intermediates. When a diamine and a dianhydride are added into a dipolar...

Words: 554 - Pages: 3

Premium Essay

Egg Drop Project Research Paper

...The experience of the egg drop project was one that will help you realize that not every way that you try will work, and just because you fail once doesn't mean you can't try again. I believe my egg broke possibly because when I was testing mine I was working with a larger egg. So when it came to the drop it wasn't supported by the walls as much as it should have been. I believe also because the straws on the bottom didn't really reach the egg since it was a smaller egg than what I was practicing with. I feel if I had also dropped it a couple more times or had made it a tiny bit smaller it might have concealed the egg better , and hopefully wouldn't have cracked than.I believe also that when it came to the test it didn't land straight on...

Words: 285 - Pages: 2

Premium Essay

Essays

...dissolved in H2S c. CO2 dissolved in ammonia d. Mg(NO3)2 dissolved in water 2. At sea level, the partial pressure of O2 is 0.21 atm. However, the Dead Sea is not at sea level -- it’s at a much lower altitude, where the atmospheric pressure is 806 mmHg and the temperature is 20˚C. Use Henry’s Law and Table 13.2 to calculate the molar concentration of O2 in the Dead Sea. 3. Suppose we make a solution by dissolving 42.5 g of I2 in 1.50 mol of CCl4. a. What is the molality of this solution? b. Suppose we wanted to use our 42.5 g of I2 to make a 0.500 m solution. What volume of CCl4 would we need? The density of CCl4 is 1.589 g/mL. 4. We dissolve 200.0 g of NaCl in 684.5 g of water and find that the density of the solution is 1.15 g/mL. Calculate: a. The % by mass of NaCl b. The mole fraction of NaCl c. The molarity of NaCl d. The molality of NaCl 5. Calculate the number of moles of solute in each solution. a. A solution of Br2 dissolved in 50.0 g of...

Words: 741 - Pages: 3

Free Essay

Solutions to Grooming Teens for Adulthood

...Solutions to Grooming Teens for Adulthood Reasoning and Problem Solving CST 1 November 30, 2009 To solve a problem one must often dive beneath the surface of the reflected obvious to reveal the rest of the issue hidden in the depths below. In Task one for this course the question of what is the best way to prepare teen’s for a successful adulthood has been addressed through several viewpoints and approaches. Just as there are multiple approaches in rearing children, there does not appear to be a single solution to the problem. Upon investigation it quickly becomes apparent that various groups can look at the same issue and will ultimately form different solutions that reflect their own skew on the problem. A closer look at example solution’s utilizing life skills through school settings, community resources, and Socratic home environments will demonstrate this concept. As an educator I believe that preparation is gleaned through understanding and understanding is gleaned from education. My solution would involve taking an active approach in educating the future educator by preparing the young to facilitate life skills for themselves and their own children one day. Equipping children with life applications of what they potentially will face as an adult is much like training a soldier for battle. Our county does not expect our military personnel to enlist and not receive training for what they will expect to encounter. Our children should not...

Words: 3075 - Pages: 13

Free Essay

Blowfly

...netting, as otherwise they will get a larger dose than those on the bottom of the container. You should also use the same procedure for each pot. Note that since there are only 11 pots, 1 pot will be the control and 5 pots will be used for each pesticide. Ensure that you use the following concentrations of rotenone (0.03, 0.1, 0.3, 1 and 3x for Rotenone) and primicarb (3 x 10-8, 3 x 10-7, 3 x 10-6, 3 x 10-5 and 3 x 10-4M for pirimicarb). It is important that you start with the control, then the most dilute solution of one of the pesticides. Work your way up to the top concentration of the pesticide, then thoroughly wash the spray top with distilled water. Then you should use the spray top to dispense control solution in order to clean the inside of the spray. Now you can move onto the most dilute solution of the 2nd pesticide and work your way up to the top concentration. When dispensing pesticide solutions it is essential that you use a new clean pipette tip for each solution to avoid cross contamination of...

Words: 266 - Pages: 2