Power sizing exponent

Economic Analysis Homework 1

Solve the following questions. Cite in IEEE if your database is a google search.

Cost Estimating and Estimating Models

Engineering economic analysis involves present and future economic factors; thus, it is critical to obtain reliable estimates of future costs, benefits and other economic parameters. Estimates can be rough estimates, semi detailed estimates, or detailed estimates, depending on the needs for the estimates. A characteristic of cost estimates is that errors in estimating are typically nonsymmetric because costs are more likely to be underestimated than overestimated. Difficulties in developing cost estimates arise from such conditions as one-of-a-kind estimates, resource availability, and estimator expertise. Generally the quality of a cost estimate increases as the resources allocated to developing the estimate increase. The benefits expected from improving a cost estimate should outweigh the cost of devoting additional resources to the estimate improvement.

Several models are available for developing cost (or benefit) estimates.

1. The per-unit model is a simple but useful model in which a cost estimate is made for a single unit, then the total cost estimate results from multiplying the estimated cost per unit times the number of units.

2. The segmenting model partitions the total estimation task into segments. Each segment is estimated, then the segment estimates are combined for the total cost estimate.

3. Cost indexes can be used to account for historical changes in costs. The widely reported Consumer Price Index (CPI) is an example. Cost index data are available from a variety of sources. Suppose A is a time point in the past and B is the current time. Let IVA denote the index value at time A and IVB denote the current index value for the cost estimate of interest. To estimate the current cost based on the cost at time A, use the equation:

Cost at time B = (Cost at time A) (IVB / IVA).

4. The power-sizing model accounts explicitly for economies of scale. For example, the cost of constructing a six-story building will typically be less than double the construction cost of a comparable three-story building. To estimate the cost of B based on the cost of comparable item A, use the equation

Cost of B = (Cost of A) [ ("Size" of B) / ("Size" of A) ]

where x is the appropriate power-sizing exponent, available from a variety of sources. An economy of scale is indicated by an exponent less than 1.0 An exponent of 1.0 indicates no economy of scale, and an exponent greater than 1.0 indicates a diseconomy of scale.

"Size" is used here in a general sense to indicate physical size, capacity, or some other appropriate comparison unit.

1. The cost to provide workplace safety training to a new employee is estimated at $600 for a particular manufacturer. Use the per-unit cost estimating model to estimate the annual workplace safety training cost if an estimated 30 new employees are hired annually.

2. Given the data below, use the segmenting model to estimate a manufacturer's total annual cost.

i. Estimated annual production volume = 400,000 units

ii. Estimated direct labor cost per unit = $3.75

iii. Estimated indirect labor cost per unit = $0.92

iv. Estimated raw material and purchased part cost per unit = $32.50

v. All other costs are independent of production volume and are estimated at $800,000 per year.

3. Given the data below, use the segmenting model to estimate a manufacturer's average cost per unit produced. Note that this is different from the previous question.

i. Estimated annual production volume = 400,000 units

ii. Estimated direct labor cost per unit = $3.75

iii. Estimated indirect labor cost per unit = $0.92

iv. Estimated raw material and purchased part cost per unit = $32.50

v. All other costs are independent of production volume and are estimated at $800,000 per year.

4. Use the cost index method to estimate the current construction cost for a building equivalent to one constructed in 1980 at a cost of $2.7 million. According to the Engineering News-Record, the 1980 average Building Cost Index was 1941. Suppose that the current year Building Cost Index has been estimated at 3620.

Source:http://enr.construction.com/features/conEco/costIndexes/bldIndexHist.asp

5. Use the power-sizing model to estimate the cost of a piece of equipment that has 75% more capacity than a similar piece of equipment that cost $1,000. The appropriate power sizing exponent for this type of equipment is 0.725.

6. Use a combined cost index and the power-sizing cost estimating model to estimate the current cost of a piece of equipment that has 50% more capacity than a similar piece of equipment that cost $30,000 five years ago. The appropriate power-sizing exponent for this type of equipment is 0.8, and the ratio of the cost indexes (current to 5 years ago) is 1.24. (Note that this is more complex than the previous questions.)

B. Cash Flow Diagram

Cash flow diagrams visually represent income and expenses over some time interval. The diagram consists of a horizontal line with markers at a series of time intervals. At appropriate times, expenses and costs are shown.

Note that it is customary to take cash flows during a year at the end of the year, or EOY (end-of-year). There are certain cash flows for which this is not appropriate and must be handled differently. The most common would be rent, which is normally taken at the beginning of a cash period. There are other pre-paid flows, which are handled similarly.

For example, consider a truck that is going to be purchased for $55,000. It will cost $9,500 each year to operate including fuel and maintenance. It will need to have its engine rebuilt in 6 years for a cost of $22,000 and it will be sold at year 9 for $6,000.

http://global.oup.com/us/companion.websites/fdscontent/uscompanion/us/static/companion.websites/9780199339273/cha2/image1.png

1. Given the cash flow diagram below, answer the questions by clicking on the correct answer. Note that there are several questions; as you correctly answer each, you go to the next question. (Note that these questions will take more time than previous questions did.)

http://global.oup.com/us/companion.websites/fdscontent/uscompanion/us/static/companion.websites/9780199339273/cha2/image2.png

i. What is the initial cost of this new machine?

ii. What is the rebuild cost?

iii. What is the salvage value for the machine

iv. In what single year is the total combined value of the machine positive?

Single payment simple interest formulas

Simple interest, as opposed to compound interest, is rare. With an investment that pays simple interest, the amount of interest accumulated each period depends solely on the amount invested, not on prior interest earned and left in the account.

The following single payment equation applies to simple interest: F = P (1 + I * n)

Example: If $100 is invested at 6% interest (compound interest) for four years, the amount accumulated at the end of four years is:

F = P (1 + i)n= $100 (F/P,6%,4)

= $100 (1.262) = $126.20

If, however, interest is simple rather than compound, then the amount accumulated at the end of four years is only:

F = P (1 + in) = $100 [ 1 + (0.06)(4) ]

= $100 (1.24) = $124.00

1. How much must be invested now at 6% simple interest to accumulate $1,000 at the end of 5 years?

a. P = $1,000 / [ 1 + (0.06)(5) ] = $1,000 / 1.30 = $769

b. F = $1,000 [ 1 + (0.06)(5) ] = $1,000 (1.30) = $1,300

c. P = $1,000 / 5 = $200

d. P = $1,000 (P/F,6%,5) = $1,000 (0.7473) = $747

2. Suppose that $100 is invested at 5% simple interest for 20 years. How much will be in the account after 20 years?

a. P = $100 / [ 1 + (0.05)(20) ] = $100 / 2 = $50

b. F = $100 (20)(0.05) = $100

c. F = $100 [ 1 + (5)(5) ] = $100 (26) = $2,600

d. F = $100 [ 1 + (0.05)(20) ] = $100 (2) = $200

Single Payment Compound Interest Formulas (annual)
Interest and Equivalence
Given a present dollar amount P, interest rate i% per year, compounded annually, and a future amount F that occurs n years after the present, the relationship between these terms is F = P (1 + i)n

In equations, the interest rate i must be in decimal form, not percent.

Example: If $100 is invested at 6% interest per year, compounded annually, then the future value of this investment after 4 years is

F = P (1 + i)n= $100 (1 + 0.06) 4

= $100 (1.06)4= $100 (1.2625) = $126.25

Solving the above equation for P yields: P = F (1 + i)-n

The factors F/P and P/F are available in interest tables, simplifying somewhat the calculations. The common notation for these factors is

(F/P,i%,n) = (1 + i)n

(P/F,i%,n) = (1 + i)– n

1. Suppose that $1,000 is invested for six years at an interest rate of 10% per year, compounded annually. How much will be in the account at the end of six years?

a. F = $1,000 + $600 = $1,600

b. F = $1,000 (1.06)10= $1,791

c. F = $1,000 (1.10)6= $1,772

d. F = $1,000 (1.06)-10= $558

2. Suppose that an investor wishes to deposit an amount now so that in 30 years $1,000,000 will be in an account that pays 10% interest per year, compounded annually. What amount must be deposited now?

a. P = $1,000,000 (1.10)-30= $57,309

b. P = $1,000,000 / 30 = $33,333

c. P = $1,000,000 (0.10) = $100,000

d. P = $1,000,000 (1.10)30= $17,449,402

3. This question deals with the use of interest tables instead of equations. Use the interest tables to answer the question.

Suppose that $100 is invested for five years at an interest rate of 8% per year, compounded annually. How much will be in the account at the end of five years?

a. F = $100 (P/F,8%,5) = $100 (0.6806) = $68.06

b. F = $100 (F/P,8%,5) = $100 (1.469) = $146.90

c. P = $100 (P/A,8%,5) = $100 (3.993) = $399.30

d. F = $100 (F/A,8%,5) = $100 (5.867) = $586.70

4. This question deals with the use of interest tables instead of equations. Use the interest tables to answer the question.

Suppose that an investor wishes to deposit an amount now so that in 20 years there will be $50,000 in an account that pays 7% interest, compounded annually. What amount must be deposited now?

a. F = $50,000 (F/P,7%,20) = $50,000 (3.870) = $193,500

b. P = $50,000 (P/F,20%,7) = $50,000 (0.2791) = $13,955

c. P = $50,000 (P/F,7%,20) = $50,000 (0.2584) = $12,920

d. P = $50,000 (P/F,7%,20) = $50,000 (0.2415) = $12,075

Single Payment Compound Interest Formulas (other periods)
Interest and Equivalence
Single payment compound interest formulas (other periods)
If the interest period and compounding period are not stated, then the interest rate is understood to be annual with annual compounding. Examples:

"12% interest" means that the interest rate is 12% per year, compounded annually.

"12% interest compounded monthly" means that the interest rate is 12% per year (not 12% per month), compounded monthly. Thus, the interest rate is 1% (12% / 12) per month.

"1% interest per month compounded monthly" is unambiguous.

When the compounding period is not annual, problems must be solved in terms of the compounding period, not years.

Example: If $100 is invested at 6% interest, compounded monthly, then the future value of this investment after 4 years is:

F = P (1 + i)n= $100 (1 + 0.005)48

= $100 (1.005)48= $100 (1.2705) = $127.05

Note that the interest rate used above is (6% / 12) = 0.5% per month = 0.005 per month, and that the number of periods used is 48 (months), not 4 (years).

1. Use interest tables. Suppose that $1,000 is invested for 4 years at an interest rate of 12%, compounded quarterly. How much will be in the account at the end of 4 years?

a. F = $1,000 (F/P,12%,4) = $1,000 (1.574) = $1,574

b. F = $1,000 (F/P,12%,16) = $1,000 (6.130) = $6,130

c. F = $1,000 (F/P,3%,4) = $1,000 (1.126) = $1,126

d. F = $1,000 (F/P,3%,16) = $1,000 (1.605) = $1,605

2. Use interest tables. How much must be invested now at 6% interest, compounded monthly, to accumulate $1,000 at the end of five years?

a. P = $1,000 (P/F,6%,60) = $1,000 (0.0303) = $30.30

b. P = $1,000 (P/F,0.5%,60) = $1,000 (0.7414) = $741.37

c. P = $1,000 (P/F,0.5%,5) = $1,000 (0.9754) = $975.40

d. P = $1,000 (P/F,6%,5) = $1,000 (0.7473) = $747.30

Single Payment Compound Interest (solving for i or n)
Interest and Equivalence
The single payment compound interest formula

F = P (1 + i)n

or single payment interest table factors can be used to solve for unknown i or n.

Example: A $100 investment now in an account that pays compound interest annually will be worth $250 at a point exactly 31 years from now. What annual interest rate does this account pay?

Solving the equation for i:

250 = 100 (1 + i)31

1 + i = (250/100)1/31= 1.0299

yields an answer of 3%.

Or, using interest tables, note that i = 3%, because (F/P,3%,31) = 2.500.

1. How many years are required for an investment to double in value at 10% interest?

a. n = 2 years, because (F/A,10%,2)>2 and (F/A,10%,1) < 2

b. n = 3 years, because (P/G,10%,3)>2 and (P/G,10%,2) < 2

c. n = 8 years, because (F/P,10%,8)>2 and (F/P,10%,7) < 2

d. n = 10 years, because interest is 10% per year

2. At what annual compound interest rate is $100 today equivalent to $370, seventeen years from now?

a. i = 6%

b. i = 7%

c. i = 8%

d. i = 9%