Experiment

Investigation: Body temperature regulation Guiding question: To what point does fur determine an arctic foxes wellbeing in the wild and how much protection against extreme temperatures does it really offer. Introduction: In this experiment I will be attempting to justify my findings that I have gathered over the course of a few different experiments. What I have been gathering information over is the suitability of an arctic foxes fur coat. I have gone about finding this out by simulating a model of an arctic foxes fur coat through many trying and stressful conditions.

The way I did so was by firstly dividing the experiments into three different steps. The first step entailed me attempting to put my model arctic fox in sub zero temperatures, as it would be in real life. What I did however was to put two models in the fridge. The first was a simple test tube covered with cotton, while the second had a space of cool air between the fur and the test tube simulating the space that would be created when a fox puffs out its fur almost like when we get goose pimples.

The second experiment was designed to signal whether it changes anything to the temperature of my two test tubes if there is wind or not. I also did that same experiment of wind with my test tubes both drenched with water, again this would simulate a real life situation such as rain and wind or even snow and wind. Lastly I decided to put my test tubes (or arctic foxes! ) at normal room temperature to determine how much the tests really did influence the temperatures, in this way I could see what exactly was going on in terms of the cooling curve. Hypothesis:

I expect the following experiment to give me a rough enough idea of arctic foxes protective measures and to help me better understand the full extent of their heat retaining abilities. I expect that my experiment has a great deal of flaws yet I also believe that there will be enough raw data to accurately determine a positive result. A positive result would be one that accurately gives me an idea of how to answer my guiding question. If I were to be more specific I would say that in my opinion the test tube with the hot air trapped between the cotton and the actual test tube will make no difference over the one with only cotton around it.

I base this upon the theory that the air temperature between the test tube and the cotton will quickly loose any of its heat and will become redundant. I can prove this by continuing with the experiments I have designed to determine just that. Lastly I also believe that when I will confront the results of my two experiments, the one with only the fan and the other with the fan and the wet cotton I will find that the later will be much more susceptible to the cold and its temperature will decrease much more, due to the water that will constrict its ability to retain heat.

Apparatus: * Two test tubes of equal dimensions * Cotton wool to simulate fur on an arctic fox * Pieces of wood that create a space of air between the tube and the wool * Timer to accurately give me a time p for which to conduct my experiment * Fan to simulate wind * Thermometer to place within test tube which will accurately calculate the temperature of the water held inside it. * Kettle to boil the water before placing it inside the test tube * Selotape to keep the cotton in place Fridge to simulate the sub zero temperatures that arctic foxes must live through Evaluation of apparatus: I think that the apparatus that I used are all extremely good and effective with a few key exceptions. Firstly the cotton is not quite the same material as the fur that an arctic fox uses, meaning that the experiment will not be true to real life. It will only give me a rough idea of the concepts of heat regulation within this particular species of animal, but that will be enough to answer the guiding question.

Another piece of apparatus that does not match exactly my requirements is the fan seeing as it only produces a certain amount of wind and in only one direction meaning that I will have to continually turn my test tube in order to keep the experiment fair and equal. Experiment one: In this experiment I will be placing my test tube’s at room temperature in order to determine before starting the other experiments what the cooling curve would be without it being put through any abnormal conditions.

As I can see through the graphs above the temperature decreases very slowly from the initial 70 degree starting temperature. What happens is that test tube one and two slowly start to disconnect form each other, seeing as test tube one retains heat more effectively. I will now see whether the results will be drastically different with my other experiments. I have also noticed that the test tube 1 has a greater heat retaining ability that may well be due to the layer of air held between it and the test tube. To further prove this theory I will have to continue with my other experiments.

Experiment two: In this experiment I have put my two test tubes inside the fridge in order to simulate the sub zero air conditions. In my hypothesis I said that I thought there would be no difference between test tube one and two. In the first experiment it was proved otherwise, yet in the following experiment there will be a greater temperature change so the differences in temperature will become more evident as time goes on. As you can see in the following graph there is a much more significant difference between test tube one and test tube two.

This shows that so far the space of hot air between the cotton wool and the test tube is beginning to work much more effectively. This is completely against what I initially said in my hypothesis, where I clearly stated I thought It would make no difference. However if you look carefully at this gathering of information you are able to see that the results start to separate from each other much more significantly than the first experiment this shows that the more drastic the temperature the more it helps to have that layer of hot air to protect and keep the ‘arctic fox’ warm. Experiment three:

This is the most important experiment, in my opinion to determine to what extent an arctic foxes fur helps keep out the cold. This is because I will be keeping the test tubes at room temperature while placing a fan in front of it to simulate wind and the cold wind that comes with that. It is not as drastic as my previous experiment but it is just as important. I expect to see results that are of a higher temperature than the last experiment, but I also expect the difference between the two test tubes to increase seeing as test tube 2 is much better equipt against this kind of heat and temperature difference.

Eng ineeri ng Economy Third Edition Leland T. Blank, P. E. Department of Industrial Engineering Assistant Dean of Engineering Texas A & M University Anthony J. Tarquin, P. E. Department of Civil Engineering Assistant Dean of Engineering The University of Texas at EI Paso McGraw-Hill Book Company New York S1. Louis San Francisco Auckland Bogota Caracas Colorado Springs Hamburg Lisbon London Madrid Mexico Milan Montreal New Delhi Oklahoma City Panama Paris San Juan Silo Paulo Singapore Sydney Tokyo Toronto 4 Level One 1. Define and recognize in a problem statement the economy symbols P, F, A, n, and i. 1. 6 Define cash flow, state what is meant by end-of-period convention, and construct a cash-flow diagram, given a statement describing the amount and times of the cash flows. Study Guide 1. 1 Basic Terminology Before we begin to develop the terminology and fundamental concepts upon which engineering economy is based, it would be appropriate to define what is meant by engineering economy. In the simplest terms, engineering economy is a collection of mathematical techniques which simplify economic comparisons.

With these techniques, a rational, meaningful approach to evaluating the economic aspects of different methods of accomplishing a given objective can be developed. Engineering economy is, therefore, a decision assistance tool by which one method will be chosen as the most economical one. In order for you to be able to apply the techniques, however, it is necessary for you to understand the basic terminology and fundamental concepts that form the foundation for engineering-economy studies.

Some of these terms and concepts are described below. An alternative is a stand-alone solution for a give situation. We are faced with alternatives in virtually everything we do, from selecting the method of transportation we use to get to work every day to deciding between buying a house or renting one. Similarly, in engineering practice, there are always seveffl ways of accomplishing a given task, and it is necessary to be able to compare them in a rational manner so that the most economical alternative can be selected.

The alternatives in engineering considerations usually involve such items as purchase cost (first cost), the anticipated life of the asset, the yearly costs of maintaining the asset (annual maintenance and operating cost), the anticipated resale value (salvage value), and the interest rate (rate of return). After the facts and all the relevant estimates have been collected, an engineering-economy analysis can be conducted to determine which is best from an economic point of view.

However, it should be pointed out that the procedures developed in this book will enable you to make accurate economic decisions only about those alternatives which have been recognized as alternatives; these procedures will not help you identify what the alternatives are. That is, if alternatives ,4, B, C, D, and E have been identified as the only possible methods to solve a Particular problem when method F, which was never recognized as an alternative, is really the most attractive method, the wrong decision is certain to be made because alternative F could never be chosen, no matter what analytical techniques are used.

Thus, the importance of alternative identification in the decision-making process cannot be overemphasized, because it is only when this aspect of the process has been thoroughly completed that the analysis techniques presented in this book can be of greatest value. In order to be able to compare different methods for accomplishing a given objective, it is necessary to have an evaluation criterion that can be used as a basis Terminology and Cash-Flow Diagrams 5 for judging the alternatives. That is, the evaluation criterion is that which is used to answer the question “How will I know which one is best? Whether we are aware of it or not, this question is asked of us many times each day. For example, when we drive to work, we subconsciously think that we are taking the “best” route. But how did we define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, depending upon which criterion is used to identify the best, a different route might be selected each time! (Many arguments could have been avoided if the decision makers had simply stated the criteria they were using in determining the best). In economic analysis, dollars are generally used as the basis for comparison.

Thus, when there are several ways of accomplishing a given objective, the method that has the lowest overall cost is usually selected. However, in most cases the alternatives involve intangible factors, such as the effect of a process change on employee morale, which cannot readily be expressed in terms of dollars. When the alternatives available have approximately the same equivalent cost, the nonquantifiable, or intangible, factors may be used as the basis for selecting the best alternative, For items of an alternative which can be quantified in terms of dollars, it is important to recognize the concept of the time value of money.

It is often said that money makes money. The statement is indeed true, for if we elect to invest money today (for example, in a bank or savings and loan association), by tomorrow we will have accumulated more money than we had originally invested. This change in the amount of money over a given time period is called the time value of money; it is the most important concept in engineering economy. You should also realize that if a person or company finds it necessary to borrow money today, by tomorrow more money than the original loan will be owed. This fact is also explained by the time value of money.

The manifestation of the time value of money is termed interest, which is a measure of the increase between the original sum borrowed or invested and the final amount owed or accrued. Thus, if you invested money at some time in the past, the interest would be Interest = total amount accumulated – original investment (1. 1) On the other hand, if you borrowed would be Interest money at some time in the past, the interest (1. 2) = present amount owed – original loan In either case, there is an increase in the amount of money that was originally invested or borrowed, and the increase over the original amount is the interest.

The original investment or loan is referred to as principal. Probs. 1. 1 to 1. 4 1. 2 Interest Calculations When interest is expressed as a percentage of the original amount per unit time, the result is an interest rate. This rate is calculated as follows: . Percent interest rate = interest accrued per unit time 00% .. I x 1 0 origma amount (1. 3) 6 Level One By far the most common time period used for expressing interest rates is 1 year. However, since interest rates are often expressed over periods of time shorter than 1 year (i. e. 1% per month), the time unit used in expressing an interest rate must also be identified and is termed an interest period. The following two examples illustrate the computation of interest rate. Example 1. 1 The Get-Rich-Quick (GRQ) Company invested $100,000 on May 1 and withdrew a total of $106,000 exactly one year later. Compute (a) the interest gained from the original investment and (b) the interest rate from the investment. Solution (a) Using Eq. (1. 1), Interest = 106,000 – 100,000 = $6000 (b) Equation (1. 3) is used to obtain Percent interest rate = 6000 per year 100,000 x 100% = 6% per year

Comment For borrowed money, computations are similar to those shown above except that interest is computed by Eq. (1. 2). For example, if GRQ borrowed $100,000 now and repaid $110,000 in 1 year, using Eq. (1. 2) we find that interest is $10,000, and the interest rate from Eq. (1. 3) is 10% per year. Example 1. 2 Joe Bilder plans to borrow $20,000 for 1 year at 15% interest. Compute (a) the interest and (b) the total amount due after 1 year. Solution (a) Equation (1. 3) may be solved for the interest accrued to obtain Interest = 20,000(0. 15) = $3000 (b) Total amount due is the sum of principal and interest or Total due Comment = 0,000 + 3000 = $23,000 Note that in part (b) above, the total amount due may also be computed as Total due = principal(l + interest rate) = 20,000(1. 15) = $23,000 In each example the interest period was 1 year and the interest was calculated at the end of one period. When more than one yearly interest period is involved (for example, if we had wanted to know the amount of interest Joe Bilder would owe on Terminology and Cash-Flow Diagrams 7 the above loan after 3 years), it becomes necessary to determine whether the interest . payable on a simple or compound basis. The concepts of simple and compound interest are discussed in Sec. . 4. Additional Examples 1. 12 and 1. 13 Probs. 1. 5 to 1. 7 1. 3 Equivalence The time value of money and interest rate utilized together generate the concept of equivalence, which means that different sums of money at different times can be equal in economic value. For example, if the interest rate is 12% per year, $100 today (i. e. , at present) would be equivalent to $112 one year from today, since mount accrued = 100 =$112 Thus, if someone offered you a gift of $100 today or $112 one year from today, it would make no difference which offer you accepted, since in either case you would have $112 one year from today.

The two sums of money are therefore equivalent to each other when the interest rate is 12% per year. At either a higher or a lower interest rate, however, $100 today is not equivalent to $112 one year from today. In addition to considering future equivalence, one can apply the same concepts for determining equivalence in previous years. Thus, $100 now would be equivalent to 100/1. 12 = $89. 29 one year ago if the interest rate is 12% per year. From these examples, it should be clear that $89. 29 last year, $100 now, and 112 one year from now are equivalent when the interest rate is 12% per year.

The fact that these sums are equivalent can be established by computing the interest rate as follows: 112 100 = 1. 12, or 12% per year and 8~~~9 = 1. 12, or 12% per year The concept of equivalence can be further illustrated by considering different loan-repayment schemes. Each scheme represents repayment of a $5000 loan in 5 years at 15%-per-year interest. Table 1. 1 presents the details for the four repayment methods described below. (The methods for determining the amount of the payments are presented in Chaps. 2 and 3. ) • Plan 1 a interest or principal is recovered until the fifth year.

Interest accumulates each year on the total of principal and all accumulated interest. • Plan 2 The accrued interest is paid each year and the principal is recovered at the end of 5 years. • Plan 3 The accrued interest and 20% of the principal, that is, $1000, is paid each year. Since the remaining loan balance decreases each year, the accrued interest decreases each year. + 100(0. 12) = 100(1 + 0. 12) = 100(1. 12) 8 Level One Table 1. 1 Different repayment schedules of $5,000 at 15% for 5 years (1) End of year (2) = 0. 15(5) Interest for year (3) = (2) + (5) Total owed at end of year (4) Payment per plan (3) – (4) Balance after payment (5) Plan 1 0 1 2 3 4 5 Plan 2 0 1 2 3 4 5 Plan 3 0 1 2 3 4 5 Plan 4 0 1 2 3 4 5 $ 750. 00 862. 50 991. 88 1,140. 66 1,311. 76 5,750. 00 6,612. 50 7,604. 38 8,745. 04 10,056. 80 0 0 0 0 10,056. 80 $10,056. 80 $ $5,000. 00 5,750. 00 6,612. 50 7,604. 38 8,745. 04 0 $750. 00 750. 00 750. 00 750. 00 750. 00 $5,750. 00 5,750. 00 5,750. 00 5,750. 00 5,750. 00 $ 750. 00 750. 00 750. 00 750. 00 5,750. 00 $8,750. 00 $5,000. 00 5,000. 00 5,000. 00 5,000. 00 5,000. 00 0 $750. 00 600. 00 450. 00 300. 00 150. 00 $5,750. 00 4,600. 00 3,450. 00 2,300. 00 1,150. 00 $1,750. 00 1,600. 00 1,450. 0 1,300. 00 1,150. 00 $7,250. 00 5,000. 00 4,000. 00 3,000. 00 2,000. 00 1,000. 00 0 $750. 00 638. 76 510. 84 363. 73 194. 57 $5,750. 00 4,897. 18 3,916. 44 2,788. 59 1,491. 58 $1,491. 58 1,491. 58 1,491. 58 1,491. 58 1,491. 58 $7,457. 90 $5,000. 00 4,258. 42 3,405. 60 2,424. 86 1,297. 01 0 • Plan 4 Equal payments are made each year with a portion going toward princi- pal recovery and the remainder covering the accrued interest. Since the loan balance decreases at a rate which is slower than in plan 3 because of the equal end-of-year payments, the interest decreases, but at a rate slower than in plan 3. te that the total amount repaid in each case would be different, even though each repayment scheme would require exactly 5 years to repay the loan. The difference in the total amounts repaid can of course be explained by the time value of money, since the amount of the payments is different for each plan. With respect to equivalence, the table shows that when the interest rate is 15% per year, $5000 at time 0 is equivalent to $10,056. 80 at the end of year 5 (plan 1), or $750 per year for 4 years and $5750 at the end of year 5 (plan 2), or the decreasing amounts shown in years 1 through 5 (plan 3), or $1,491. 8 per year for 5 years (plan 4). Using the formulas developed in Chaps. 2 and 3, we could easily show that if the payments in Terminology and Cash-Flow Diagrams 9 each plan (column 4) were reinvested at 15% per year when received, the total amount of money available at the end of year 5 would be $10,056. 80 from each repayment plan. Additional Examples 1. 14 and 1. 15 Probs. 1. 8 and 1. 9 1. 4 Simple and Compound Interest The concepts of interest and interest rate were introduced in Sees. 1. 1 and 1. 2 and ed in Sec. 1. 3 to calculate for one interest period past and future sums of money equivalent to a present sum (principal).

When more than one interest period is involved, the terms simple and compound interest must be considered. Simple interest is calculated using the principal only, ignoring any interest that was accrued in preceding interest periods. The total interest can be computed using the relation Interest = (principal)(number of periods)(interest rate) = Pni (1. 4) Example 1. 3 If you borrow $1000 for 3 years at 14%-per-year simple interest, how much money will you owe at the end of 3 years? Solution The interest for each of the 3 years is = Interest per year 1000(0. 14) = $140 Total interest for 3 years from Eq. (1. 4) is Total interest = 1000(3)(0. 4)= $420 Finally, the amount due after 3 years is 1000 + 420 Comment = $1420 The $140 interest accrued in the first year and the $140 accrued in the second year did not earn interest. The interest due was calculated on the principal only. The results of this loan are tabulated in Table 1. 2. The end-of-year figure of zero represents th~ present, that is, when the money is borrowed. Note that no payment is made by the borrower until the end of year 3. Thus, the amount owed each year increases uniformly by $140, since interest is figured only on the principal of $1000. Table 1. 2 Simple-interest (1) (2) computation (3) (4) (2) + (3) Amount owed (5) End of year 0 1 2 Amount borrowed $1,000 Interest Amount paid 3 $140 140 140 $1,140 1,280 1,420 $ 0 0 1,420 10 Level One In calculations of compound interest, the interest for an interest period is calculated on the principal plus the total amount of interest accumulated in previous periods. Thus, compound interest means “interest on top of interest” (i. e. , it reflects the effect of the time value of money on the interest too). Example 1. 4 If you borrow $1000 at 14%-per-year compound interest, instead of simple interest as in the preceding example, compute the total amount due after a 3-year period.

Solution The interest and total amount due for each year is computed as follows: Interest, year 1 = 1000(0. 14) = $140 Total amount due after year 1 = 1000 + 140 = $1140 Interest, year 2 = 1140(0. 14) = $159. 60 Total amount due after year 2 = 1140 + 159. 60 = $1299. 60 Interest, year 3 = 1299. 60(0. 14)= $181. 94 Total amount due after year 3 = 1299. 60 + 181. 94 = $1481. 54 Comment The details are shown in Table 1. 3. The repayment scheme is the same as that for the simple-interest example; that is, no amount is repaid until the principal plus all interest is due at the end of year 3.

The time value of money is especially recognized in compound interest. Thus, with compound interest, the original $1000 would accumulate an extra $1481. 54 – $1420 = $61. 54 compared with simple interest in the 3-year period. If $61. 54 does not seem like a significant difference, remember that the beginning amount here was only $1000. Make these same calculations for an initial amount of $10 million, and then look at the size of the difference! The power of compounding can further be illustrated through another interesting exercise called “Pay Now, Play Later”. It can be shown (by using the equations that will be developed in Chap. ) that at an interest rate of 12% per year, approximately $1,000,000 will be accumulated at the end of a 40-year time period by either of the Table 1. 3 Compound-interest (1) (2) computation (3) (4) = (2) + (3) (5) End of year 0 1 2 3 Amount borrowed $1,000 Interest Amount owed $1,140. 00 1,299. 60 1,481. 54 Amount paid $140. 00 159. 60 181. 94 $ 0 0 1,481. 54 Terminology and Cash-Flow Diagrams 11 – llowing investment schemes: • Plan 1 Invest $2610 each year for the first 6 years and then nothing for the next 34 years, or • Plan 2 Invest nothing for the first 6 years, and then $2600 each year for the next 34 years!! ‘ote that the total investment in plan 1 is $15,660 while the total required in plan _ to accumulate the same amount of money is nearly six times greater at $88,400. Both the power of compounding and the wisdom of planning for your retirement at he earliest possible time should be quite evident from this example. An interesting observation pertaining to compound-interest calculations in-olves the estimation of the length of time required for a single initial investment to double in value. The so-called rule of 72 can be used to estimate this time.

The rule i based on the fact that the time required for an initial lump-sum investment to double in value when interest is compounded is approximately equal to 72 divided by the interest rate that applies. For example, at an interest rate of 5% per year, it would take approximately 14. 4 years (i. e. , 72/5 = 14. 4) for an initial sum of money to double in value. (The actual time required is 14. 3 years, as will be shown in Chap. 2. ) In Table 1. 4, the times estimated from the rule of 72 are compared to the actual times required for doubling at various interest rates and, as you can see, very good estimates are obtained.

Conversely, the interest rate that would be required in order for money to double in a specified period of time could be estimated by dividing 72 by the specified time period. Thus, in order for money to double in a time period of 12 years, an interest rate of approximately 6% per year would be required (i. e. , 72/12 = 6). It should be obvious that for simple-interest situations, the “rule of 100” would apply, except that the answers obtained will always be exact. In Chap. 2, formulas are developed which simplify compound-interest calculations. The same concepts are involved when the interest period is less than a year.

A discussion of this case is deferred until Chap. 3, however. Since real-world calculations almost always involve compound interest, the interest rates specified herein refer to compound interest rates unless specified otherwise. Additional Example 1. 16 Probs. 1. 10 to 1. 26 Table 1. 4 Doubling time estimated actual time from rule of 72 versus Doubling lime, no. of periods Interest rate, % per period 1 Estimated from rule 72 Actual 70 35. 3 14. 3 7. 5 2 5 10 20 40 36 14. 4 7. 2 3. 6 1. 8 3. 9 2. 0 12 Level One 1. 5 Symbols and Their Meaning The mathematical symbols: relations sed in engmeenng economy employ the following P = value or sum of money at a time denoted as the present; dollars, pesos, etc. F A n i = value or sum of money at some future time; dollars, pesos, etc. = a series of consecutive, equal, end-of-period month, dollars per year, etc. amounts of money; dollars per = number of interest periods; months, years, etc. = interest rate per interest period; percent per month, percent per year, etc. The symbols P and F represent single-time occurrence values: A occurs at each interest period for a specified number of periods with the same value.

It should be understood that a present sum P represents a single sum of money at some time prior to a future sum or uniform series amount and therefore does not necessarily have to be located at time t = O. Example 1. 11 shows a P value at a time other than t = O. The units of the symbols aid in clarifying their meaning. The present sum P and future sum F are expressed in dollars; A is referred to in dollars per interest period. It is important to note here that in order for a series to be represented by the symbol A, it must be uniform (i. e. the dollar value must be the same for each period) and the uniform dollar amounts must extend through consecutive interest periods. Both conditions must exist before the dollar value can be represented by A. Since n is commonly expressed in years or months, A is usually expressed in units of dollars per year or dollars per month, respectively. The compound-interest rate i is expressed in percent per interest period, for example, 5% per year. Except where noted otherwise, this rate applies throughout the entire n years or n interest periods. The i value is often the minimum attractive rate of return (MARR).

All engineering-economy problems must involve at least four of the symbols listed above, with at least three of the values known. The following four examples illustrate the use of the symbols. Example 1. 5. If you borrow $2000 now and must repay the loan plus interest at a rate of 12% per year in 5 years, what is the total amount you must pay? List the values of P, F, n, and i. Solution In this situation P and F, but not A, are involved, since all transactions are single payments. The values are as follows: P = $2000 Example 1. 6 i = 12% per year n = 5 years

If you borrow $2000 now at 17% per year for 5 years and must repay the loan in equal yearly payments, what will you be required to pay? Determine the value of the symbols involved. Terminology and Cash-Flow Diagrams 13 ~- ution = S2000 = ? per year for 5 years = 17% per year = 5 years – ere is no F value involved. – 1 In both examples, the P value of $2000 is a receipt and F or A is a disbursement. equally correct to use these symbols in reverse roles, as in the examples below. Example 1. 7 T you deposit $500 into an account on May 1, 1988, which pays interest at 17% per year, hat annual amount can you withdraw for the following 10 years?

List the symbol values. Solution p = $500 A =? per year i = 17% per year n= 10 years Comment The value for the $500 disbursement P and receipt A are given the same symbol names as before, but they are considered in a different context. Thus, a P value may be a receipt (Examples 1. 5 and 1. 6) or a disbursement (this example). Example 1. 8 If you deposit $100 into an account each year for 7 years at an interest rate of 16% per year, what single amount will you be able to withdraw after 7 years? Define the symbols and their roles.

Solution In this example, the equal annual deposits are in a series A and the withdrawal is a future sum, or F value. There is no P value here. A = $100 per year for 7 years F =? i = 16% per year n = 7 years Additional Example 1. 17 Probs. 1. 27 to 1. 29 14 Level One 1. 6 Cash-Flow Diagrams Every person or company has cash receipts (income) and cash disbursements (costs) which occur over a particular time p. These receipts and disbursements in a given time interval are referred to as cash flow, with positive cash flows usually representing receipts and negative cash flows representing disbursements.

At any point in time, the net cash flow would be represented as Net cash flow = receipts – disbursements (1. 5) Since cash flow normally takes place at frequent and varying time intervals within an interest period, a simplifying assumption is made that all cash flow occurs at the end of the interest period. This is known as the end-of-period convention. Thus, when several receipts and disbursements occur within a given interest period, the net cash flow is assumed to occur at the end of the interest period.

However, it should be understood that although the dollar amounts of F or A are always considered to occur at the end of the interest period, this does not mean that the end of the period is December 31. In the situation of Example 1. 7, since investment took place on May 1, 1988, the withdrawals will take place on May 1, 1989 and each succeeding May 1 for 10 years (the last withdrawal will be on May 1, 1998, not 1999). Thus, end of the period means one time period from the date of the transaction (whether it be receipt or disbursement).

In the next chapter you will learn how to determine the equivalent relations between P, F, and A values at different times. A cash-flow diagram is simply a graphical representation of cash flows drawn on a time scale. The diagram should represent the statement of the problem and should include what is given and what is to be found. That is, after the cash-flow diagram has been drawn, an outside observer should be able to work the problem by looking at only the diagram. Time is considered to be the present and time 1 is the end of time period 1. (We will assume that the periods are in years until Chap. . ) The time scale of Fig. 1. 1 is set up for 5 years. Since it is assumed that cash flows occur only at the end of the year, we will be concerned only with the times marked 0, 1, 2, … , 5. The direction of the arrows on the cash-flow diagram is important to problem solution. Therefore, in this text, a vertical arrow pointing up will indicate a positive cash flow. Conversely, an arrow pointing down will indicate a negative cash flow. The cash-flow diagram in Fig. 1. 2 illustrates a receipt (income) at the end of year 1 and a disbursement at the end of year 2.

It is important that you thoroughly understand the meaning and construction of the cash-flow diagram, since it is a valuable tool in problem solution. The three examples below illustrate the construction of cash-flow diagrams. ° Figure 1. 1 A typical cash-flow time scale. Year 1 Year 5 r=;:;; r+;:;. I 1 2 Time o I I 3 4 I 5 Terminology and Cash-Flow Diagrams 15 + Figure 1. 2 Example of positive and negative cash flows. 2 3 Time Example 1. 9 Consider the situation presented in Example 1. 5, where P = $2000 is borrowed and F is to be found after 5 years.

Construct the cash-flow diagram for this case, assuming an interest rate of 12% per year. Solution Figure 1. 3 presents the cash-flow diagram. Comment While it is not necessary to use an exact scale on the cash-flow axes, you will probably avoid errors later on if you make a neat diagram. Note also that the present sum P is a receipt at year 0 and the future sum F is a disbursement at the end of year 5. Example 1. 10 If you start now and make five deposits of $1000 per year (A) in a 17%-per-year account, how much money will be accumulated (and can be withdrawn) immediately after you have made the last deposit?

Construct the cash-flow diagram. Solution The cash flows are shown in Fig. 1. 4. Since you have decided to start now, the first deposit is at year 0 and the [lith Comment deposit and withdrawal occur at the end of year 4. Note that in this example, the amount accumulated after the fifth deposit is to be computed; thus, the future amount is represented by a question mark (i. e. , F = ? ) Figure 1. 3. Cash-flow diagram for Example 1. 9. + P = $2. 000 i = 12% o 2 3 4 5 Year F= ? 16 Figure 1. 4 Cashflow diagram for Example 1. 10. Level One F= ? i = 17″10 2 0 3 4 Year A=$1. 000 Example 1. 11

Assume that you want to deposit an amount P into an account 2 years from now in order to be able to withdraw $400 per year for 5 years starting 3 years from now. Assume that the interest rate is 151% per year. Construct the cash-flow diagram. Figure 1. 5 presents the cash flows, where P is to be found. Note that the diagram shows what was given and what is to be found and that a P value is not necessarily located at time t = O. Solution Additional Examples 1. 18 to 1. 20 Probs. 1. 30 to 1. 46 Additional Examples Example 1. 12 Calculate the interest and total amount accrued after 1 year if $2000 is invested at an interest rate of 15% per year.

Solution Interest earned = 2000(0. 15) = $300 Total amount accrued = 2000 + 2000(0. 15) = 2000(1 + 0. 15) = $2300 Figure 1. 5 Cashflow diagram for Example 1. 11. A = $400 o 2 3 4 5 6 7 Year p=? Terminology and Cash-Flow Diagrams 17 Example 1. 13 a) Calculate the amount of money that must have been deposited 1 year ago for you to have $lOQO now at an interest rate of 5% per year. b) Calculate the interest that was earned in the same time period. Solution a) Total amount accrued = original deposit + (original deposit)(interest rate). If X = original deposit, then 1000 = X + X(0. 5) = X(l + 0. 05) 1000 = 1. 05X 1000 X=-=952. 38 1. 05 Original deposit = $952. 38 (b) By using Eq. (1. 1), we have Interest = 1000 – 952. 38 = $47. 62 Example 1. 14 Calculate the amount of money that must have been deposited 1 year ago for the investment to earn $100 in interest in 1 year, if the interest rate is 6% Per year. Solution Let a = a = = total amount accrued and b = original deposit. Interest Since a Interest Interest b b + b (interest rate), interest can be expressed as + b (interest rate) b =b = b (interest rate) $100 = b(0. 06) b = 100 = $1666. 67 0. 06 Example 1. 5 Make the calculations necessary to show which of the statements below are true and which are false, if the interest rate is 5% per year: (a) $98 now is equivalent to $105. 60 one year from now. (b) $200 one year past is equivalent to $205 now. (c) $3000 now is equivalent to $3150 one year from now. (d) $3000 now is equivalent to $2887. 14 one year ago. (e) Interest accumulated in 1 year on an investment of $2000 is $100. Solution (a) Total amount accrued = 98(1. 05) = $102. 90 =P $105. 60; therefore false. Another way to solve this is as follows: Required investment = 105. 60/1. 05 = $100. 57 =P $9? Therefore false. b) Required investment = 205. 00/1. 05 = $195. 24 =p $200; therefore false. 18 Level One (e) Total amount accrued = 3000(1. 05) = $3150; therefore true. (d) Total amount accrued = 2887. 14(1. 05) = $3031. 50 “# $3000; therefore false. (e) Interest = 2000(0. 05) = $100; therefore true. Example 1. 16 Calculate the total amount due after 2 years if $2500 is borrowed now and the compoundinterest rate is 8% per year. Solution The results are presented in the table to obtain a total amount due of $2916. (1) (2) (3) (4) = (2) + (3) (5) End of year Amount borrowed $2,500 Interest Amount owed Amount paid o 1 2 Example 1. 17 $200 216 2,700 2,916 $0 2,916 Assume that 6% per year, starting next withdrawing Solution P = you plan to make a lump-sum deposit of $5000 now into an account that pays and you plan to withdraw an equal end-of-year amount of $1000 for 5 years year. At the end of the sixth year, you plan to close your account by the remaining money. Define the engineering-economy symbols involved. $5000 A = $1000 per year for 5 years F = ? at end of year 6 i = 6% per year n = 5 years for A Figure 1. 6 Cashflow diagram for Example 1. 18. $650 $625 $600 $575 $ 550 $525 $500 $625 t -7 -6 -5 -4 -3 -2 -1 t o Year P = $2,500 Terminology and Cash-Flow

Diagrams 19 Example 1. 1B The Hot-Air Company invested $2500 in a new air compressor 7 years ago. Annual income “-om the compressor was $750. During the first year, $100 was spent on maintenance, _ cost that increased each year by $25. The company plans to sell the compressor for salvage at the end of next year for $150. Construct the cash-flow diagram for the piece f equipment. The income and cost for years – 7 through 1 (next year) are tabulated low with net cash flow computed using Eq. (1. 5). The cash flows are diagrammed . Fig. 1. 6. Solution End of year Net cash flow Income Cost -7 -6 -5 -4 -3 -2 -1 0 1 Example 0 750 750 750 750 750 750 750 750 + 150 $2,500 100 125 150 175 200 225 250 275 $-2,500 650 625 600 575 550 525 500 625 1. 19 Suppose that you want to make a deposit into your account now such that you can withdraw an equal annual amount of Ai = $200 per year for the first 5 years starting 1 year after your deposit and a different annual amount of A2 = $300 per year for the following 3 years. How would the cash-flow diagram appear if i is 14! % per year? Solution The cash flows would appear as shown in Fig. 1. 7. Comment The first withdrawal (positive cash flow) occurs at the end of year 1, exactly one year after P is deposited.

Figure 1. 7 Cash-flow diagram for two different A values, Example 1. 19. A2 = $300 A, = $200 0 1 2 3 4 i = 14+% 5 6 7 8 Year p=? 20 Level One p=? j = 12% per year Figure 1. 8 Cash-flow diagram for Example 1. 20. F2 1996 1995 A = $50 A = $150 = $50 F, = $900 Example 1. 20 If you buy a new television set in 1996 for $900,. maintain it for 3 years at a cost of $50 per year, and then sell it for $200, diagram your cash flows and label each arrow as P, F, or A with its respective dollar value so that you can find the single amount in 1995 that would be equivalent to all of the cash flows shown.

Assume an interest rate of 12% per year. Solution Comment Figure 1. 8 presents the cash-flow diagram. The two $50 negative cash flows form a series of two equal end-of-year values. As long as the dollar values are equal and in two or more consecutive periods, they can be represented by A, regardless of where they begin or end. However, the $150 positive cash flow in 1999 is a single-occurrence value in the future and is therefore labeled an F value. It is possible, however, to view all of the individual cash flows as F values. The diagram could be drawn as shown in Fig. . 9. In general, however, if two or more equal end-of-period amounts occur consecutively, by the definition in Sec. 105 they should be labeled A values because, as is described in Chap. 2, the use of A values when possible simplifies calculations considerably. Thus, the interpretation pictured by the diagram of Fig. 1. 9 is discouraged and will not generally be used further in this text. p=? j = 12% per year F. = $150 1. 9 A cash flow for Example 1. 20 considering all values as future sums. Figure 1996 1995 1997 1998 1999 F2 = $50 F3 = $50 F, = $900

The environment can affect the degree to which a gene may be expressed In an Individual. For example, solo type and weather conditions may affect plant growth. Seeds deliver energy for germination and Minimal growth until plants begin to photosynthesis and create their own glucose for further development. Aim/Purpose To demonstrate the effect of environment on phenotype to formulate a hypothesis about the effect of environment on phenotype To design and conduct a controlled experiment to demonstrate the effect of one environmental factor on a plant phenotype. Hypothesis-

Radish plants that is grown in sunlight, room light and in the shade where there is absence of light, those that in the shade will display isolating where there Is pale stems and leaves, elongated stems and leaves reducing In size whilst those grown In light will be green, sturdier and shorter. The plants that are grown in the room light will be In the middle of the other two (light and dark). Apparatus and materials- 6 plant pots Beaker to measure amounts of water for the watering of plants Pack of radish seeds 30 CM ruler that has mm written as well 3 trays Soil enough for 6 pots Risk Assessment-

The bean seedlings may have contagious diseases. Therefore gloves should be worn. Leather shoes is most recommended to be worn as handling glass may break and water might make you slip Method/Procedure- 1. Put solo Into each of the pots up to where the crease Is, making sure It Is even In all pots 2. Plant seeds by following Instructions on the back of the packet 3. Germinate radish seeds with equal amounts 4. Two seedlings were left as a control: it was watered and taken care of normally under shade. 5. The other four seedlings were placed in either light covered area, or one in dark covered area 6.

Water occasionally and wait for observable phonological results. Note Don’t overflow the pots with water Just till moist. 7. Take notes of Plant growth and observations seen. Keep a record of all data being done Results- Observations: Dates: Task: 27. 2. 14 Germination of plants 13. 14 Watered: mall 11 . 3. 14 Watered: mall 14. 3. 14 Measured/taken pictures 18. 3. 14 Light: Stems turning purple starting from the bottom to the top of the stem Leaves vibrant green Thicker stems Control Pale looking Control g’s stem from the bottom is more thicker than control 1 Control 2 is more shriveled up while control 1 is still slightly straight

Leaves have shrunk Dark Stem of dark 2 is more of white than dark 2 where it is like a pale green/grey Discussion- *trends happening Conclusion- The phenotype expressed in the light ones display green pigment for the environment influenced the need of chlorophyll for photosynthesis. The ones in the dark turned albino, in the absence of light, photosynthesis cannot take place. When these albino plants were put in the sun, over the course of 2 days they altered to a green color again. From Day 7 to Day 10 plants raised in shade displayed isolating whereas those raised in light did not Observations of plants grown in each

There were some factors that have affected some of the result of the experiment. The position of the plants should have been closer together not that far apart. The plants placed in the dark under the bench were the furthest from where the plants in the light and room light were assigned. The amount of water being put in the pots and the consistent watering made the trays flood with water making the pots more than Just moist. The water in the trays wasn’t emptied so it was there for quite a while and was it got emptied a bit later on in the experiment.

The dependent variables in the experiment were the growth and the progress of the Lana and this depends on the amount of water, amount of sunlight being taken into place. The control that was kept the same for each plant was the amount of water being put into the pots. The accuracy of the experiment wasn’t to professional expectations but it was taken into considerations of the factors that can affect it. There could be more to make this experiment better and also with the results that were taken. The reliability of the experiment was reliable as it was compared to another student’s similar works done in their time in year 12.

But what could have made it more liable is that the experiment could have been repeated another time since the plants start sprouting after 5-7 days after being planted. The plants in sunlight grew normally but those of the same species grown in the dark where there is absence of light elongated much faster to increase their chance of finding light. They were pale in color due to the lack of chlorophyll. The plants that were place in the normal room light grew but the stem wasn’t as thick as the one in the light. The energy reserves of the seed were used up therefore it leads to the plant dying in comparison to the one exposed to light.

Other factors in the environment, such as salinity and pH of soil, also affect the phenotype of a plant that would otherwise appear normal. The aim/purpose and the hypothesis were met and the results were similar. “Radish plants that is grown in sunlight, room light and in the shade where there is absence of light, those that in the shade will display isolating where there is pale stems and leaves, elongated stems and leaves reducing in size whilst those grown in light will be green, sturdier and shorter. The plants that are grown in the room light will be in the middle of the other two (light and dark). “

In addition, there have been many catastrophic events in the world caused by lack of sleep or fatigue in certain situations. These include the disaster of Coherency, the Challenger explosion In 1986 and the Exxon Valued oil spill, many claim, were caused by lack of sleep and fatigue. The employees at Coherency were overworked, working 13 hours or more. The pilots of the Challenger had a significant lack of sleep and the oil spill caused by the workers working over 22 hours per day. Furthermore, these catastrophic events are all that could be affected or prevented with a quick reaction time.

Therefore, could the fatigue and lack of sleep have led to slower reaction times causing these events to occur or was it another variable that the fatigue caused? I wonder if there is a relationship between amount of sleep and reaction time. Plan: To find out the relationship between amount of sleep and reaction time, if there is one, I will be using two different type of investigation. Firstly, I will be doing some of the investigation myself by taking a sample from my year group and having them complete a test.

In addition to that, I will also be getting results from other studies, Investigations and reports on the same experiment as the results would be more reliable and varied. For my own Investigation, since I am comparing, It Is difficult to have certain independent or dependent variables. However, since I am seeing how amount of sleep affects reaction time, as my independent variable, I will use the amount of sleep measured in hours and I will be using the reaction time measured in seconds. I will obtain the amount of sleep by asking the people within the sample and I will get the reaction time by having the sample take a free online test.

For this investigation, I will be using convenience sampling since the investigation has a very small time limit and I will not have any resources available to investigate on a larger sample or to do sample which is outside of this sample. Therefore these results are quite unreliable since It Is only done on a very small group and It Is only convenience sampling so It only takes Into account a very small group of people in a certain location. This Is why I will also be using other Investigations, papers and studies to get more reliable and accurate information.

I will have a sample of 15 people from etc. To find more reliable, accurate and more varied data. Once I have collected the results, I will firstly create a table out of the raw information to simply take down the results. Then, I will create a processed data able to make it easier to create graphs. Then I will work out averages for the reaction times and amount of sleep making them into box and whisker plots. Then, I will create a scatter graph to compare both my variables. I believe that the results will be that the amount of sleep has a very large influence over one’s reaction time.

This selection was purely reliability and credibility of the source as the 2 most credible and reliable sources were chosen for the data and information. The first source was the paper “Effect of Total Sleep Deprivation on Reaction Time and Waking EGG Activity in Man” by the American Sleep Disorders Association and Sleep Research Society, written by l. Lorenz, J. Ramose, C. Race, M. A. Guava and M. Coors-Caber. The second paper was by Mitch Leslie for the Stanford report with the name “Sleep impacts reaction time as much as alcohol. Both of these sources gave information such as averages, graphs and conclusions with some analysis but did not give the raw data. Therefore, some of this processed data could be misleading and could be biased. In addition, the data that these sources give are quite different compared to the data given from my investigations. This would mean that my results are probably unreliable as it was conducted on a very small group, none of the variables were changed so there was not much variety and the results were mostly inaccurate.

In addition, there are some missing values especially for the time some people went to the estimated hours of sleep they had gotten. I will treat this as all the other values although if they seem like outliers on a graph, it would be because they are inaccurate. To analyses, I will group the data and information by source treating them differently. Overall, I believe that I have enough data to analyses and to create a inclusion since I have the data I have collected and the information from the other studies.

Research question: Does the concentration (mol DMS) of sulfur dioxide in wine rise or fall when exposed to alarm for different time periods (O, 75,150,225, 300 minutes)? Purpose: Many adults enjoy the consumption of wine but are not aware of the different preservatives and chemicals that are added to the drink. Sulfur Dioxide, which is added to many food products including wine because it acts as a redundant, is Well known as a poisonous and allergenic substance (Echo-consult, n. D), making it a somewhat harmful ingredient.

The purpose of this experiment Is to determine how the amount of sulfur dioxide In white wine Is affected by the exposure to the alarm over different time periods and whether this will negatively or positively affect the human body. The boiling point of sulfur dioxide is -ICC, therefore when it is above this temperature it is expected to evaporate. According to Rutledge Estates, an Australian wine companyџtriangle Estates, 2011), the concentration of sulfur dioxide In wine reduces when subjected to aeration, this loss of sulfur dioxide increases over time.

This reduction in sulfur dioxide can be beneficial for people with allergies but can be harmful for the wine as oxidation causes a loss of the fruity flavor, browning, and the development of allowedly or nutty flavors (Threadlike, 2013). Hypothesis: As the wine Is left out for longer the sulfur dioxide content falls. Variables: Variable Dependent The amount of Sulfur doodle Independent Amount of time Controlled The equipment used The rinsing techniques The measurements of each Sodium hydroxide (ml) White wine (ml) Sulfuric acid (1 Mol) Starch indicator (2 ml) Set Up: mall flask Stopwatch Pipette

These results did not conform to the hypothesis; this difference could be a result of the percentage error in the equipment, and the systematic error. The results for each trial only varied slightly therefore the narrow time limit could have affected the results, there may have been larger difference of SIS if the experiment was conducted over a larger period of time. If this was the problem then it can be seen that over short periods of time the concentration of SIS does not change very much.

This means that wine will take longer to oxides and will have antibacterial properties for a longer time, making it retain its flavor and quality for a longer time. Percentage errors in instruments: Percentage error for Pipette- Percentage error com of pipettes wine) X 100 x 100 ?0. 2% Percentage error for Burette titration 1- Percentage error =(Uncertainty com of average titration) X 100 =(0. 049. 60) XIII Table of percentage errors in instruments Instrument Titration Uncertainty Percentage error (%) В±o. Ml 0. % В±o. Ml 0. 4% 5 Evaluation: Weakness Reason Improvement The timing of each interval. The laboratory was lock at certain times, this was not accounted for in the planning stage and made the timing very inaccurate. Plan the experiment at the beginning of the day accounting for all breaks and laboratory trading hours. The number of trials There was not enough time to do enough trials to get three concordant results for each test Repeat experiment until there are at least three concordant results Seeing the end point

It was hard to tell what color the endpoint should be as the previous titration would change color after a period of time Have a color chart that is permanent which the color of the titer can be compared to Contamination of wine The beakers holding the wine were exposed to the air and there may have been gases in the air which contaminated the wine. Place the wine in a place which is not exposed to things which could contaminate it. Rinsing technique The equipment was rinsed multiple times but foreign chemicals could still have been present Repeat all of the rising steps twice to ensure they are not contaminated

Measuring inaccuracies Seeing how much iodine was still in the burette was difficult because the lines were very close together. Spend more time with maximum concentration on viewing the measurements on the burette. Bibliography: Threadlike, M 2013 Wine Aeration and Its Adverse Effects, Iowa State University, accessed 25 November 2013, . Shannon, C 2011 Is aerating wine Just hot air? , Rutledge Estates, accessed November 2013, . The use of Sulfur Dioxide in Must and Wine n. D. , Echo-consult, Pdf, accessed Xavier, L n. D. Titration, CICADAS, accessed 25 November 2013, .

Prior to our procedure, we measured the temperatures of each pond area. We used the low-temperature climate (bird inaccuracy pond) to compare to the higher-temperature climate (Lake Calhoun holding pond and Lake Harriet duck area. After completing our experiment by surveying various sections of each three experimental sites, we gathered our information using a stream study. We surveyed the areas four different times for maximum proficiency. After recording each sample study four times for each area, we added up their water quality rating total index count and divided it by four, generating an average for our results. Conclusion Our groups question was: How does temperature change have an affect on the water quality of aquatic macro invertebrates?

After concluding our stream study, our group was able to determine that as temperature increased, the water quality index for aquatic macro invertebrates decreased. With the results that the stream study gave us, we were able to conclude that the area with the lowest temperature of 22 degrees Celsius (bird sanctuary) had an excellent quality rating of over 22. The Lake Calhoun holding pond (24 degrees Celsius) had a water quality rating of six, which fell under poor quality.

As for the Lake Harriet Duck Area, we were able to generate a good quality rating of 22. Thus concluding that as the temperature increased, the water quality index decreased. We believe that as temperature is higher, the pond areas are more accessible by other animals/humans, creating the water quality to be less. Some limitations we occurred during our experiment were assuring from three different environments so the surrounding environments could have had an effect on the water quality rating.

For example, the bird sanctuary pond is not accessible by people for recreational use, whereas both lake Harriet and lake Calhoun allow recreational use. This could have altered the results of our experiment. Another reason that our data could have been skewed was because recently Lake Calhoun had experienced an E-Coli problem, which is a reason the water quality was so low.

Individual Research Project-Part 1 Godfrey Boyd American Intercontinental University Abstract There are theoretical and philosophical differences between “formal research” and “business proposals”. The advantages and disadvantages related to both can be a benefit depending upon the problem studied. Research is not fully understood in business. This paper will attempt to highlight reasons for limited utilization of research in business. Small businesses are less likely to conduct marketing research. Amy E.

Knaup, an economist with the Office of Employment statistics, Bureau of Labor Statistics reported in 2005, 44% of all new business failed in the first 2 years. Individual Research Project-Preliminary Research Design In business, formal research and business proposals have significant differences and some similarities. It is very important to understand the purposes and goals of each. What are some commonalities? Research and business proposals have similar components. Backround information must be identified for both formal research and the business proposal. (American Journal of Small Business, Vol.

IX, No. 4, Spring 1985. ) When someone conducts formal research or a business proposal, they have similar goals. Research and business proposals are conducted to find solutions, or solve problems. What are some differences? A formal research proposal differs significantly. Both proposals may appear similar. The purpose of formal research is mainly to find data to solve problems or help businesses make improved decisions on daily operations, or use of the best product. Apple’s iPod was a huge financial success in the early to middle 2000s. In 2008, iPod sales dropped significantly. (Business Week).

Apple introduced and upgraded the iPod after a significant market study indicated that the economy and practicality of use were indicators of decreased iPod sales. The business proposal will focus more on the product and make it more successful. The attitude toward market research from small businesses is also a possible reason for businesses failing in less than two years. (Barnes Journal of Small Busienss Management). The formal research problem can focus on problems that sociological, and present clear concepts. Theory is also a hallmark of formal research. The problem studied and evaluated can be compared to previous research.

Business proposals are more time sensitive. One of the goals of the business proposal is to research and find a solution to why a product is not doing well in a particular market. Time is of essence in business in order for the business to be successful. The formal research may not have a time constraint but the goal is to find a solution or solve a problem. What are some capabilities of one in contrast with the capabilities of the other? Formal research has the capability to pinpoint data through correct analysis. Hypotheses are validated through appropriate statistical analysis.

Business proposals often suggest new products or services with the goal being to make money for the business. The formal research study has the capability to approach the issue theoretically and suggest a practical solution in just what to do to make money. A formal research study will provide valuable conclusions and insight. Formal research highlights complexities, context and emphasizes research. Rigor is the result of carefully collected data and strong accurate investigations. Formal research has to have testability. This means that data was collected correctly and sample sizes were adequate.

A capability of a business proposal is the proposal can be done in relatively shorter time period. The business proposal will focus on a product or a business related problem that needs a solution in order to make money. Method Participant (subject) characteristics Sampling procedures Sample size, power, and precision Measures and covariates Research design Experimental manipulations or interventions Results Recruitment Statistics and data analysis Ancillary analyses Participant flow Intervention or manipulation fidelity Baseline data Adverse events Discussion References