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1. What are the common types of Engineering Economic Decisions? Give an example of each. Briefly describe the fundamental principles of Engineering Economics. The five main types of engineering economic decisions are 1. Equipment or Process Selection: the choice of material will dictate the manufacturing process for the body panels in the automobile so there are many factors will affect the ultimate choice of the material and engineers, should consider all major cost elements, such as the cost of machinery and equipment, tooling, labor, and material. 2. Equipment replacement: the example for this category, a company may purchase 10 large presses, they expecting them to produce stamped metal parts for 10 years, so after 5 years, it may become necessary to produce the parts in plastic, which would require retiring the presses early and purchasing plastic molding machines. At the end, the company will find a making the purchased machines become obsolete earlier then expected. 3. New Product or Product Expansion: this category increase company revenues by making a new product or product expansion. One common type of expansion decision includes decisions about expenditures aimed at increasing the output of existing production or distribution facilities. Other type of expenditure decision includes considering expenditures necessary to produce a new product or to expand into a new geographic area. 4. Cost reduction: this class of engineering decision considering the cost reduction, which is a project that attempts to lower a firm’s operating costs. 5. Improvement in Service or Quality: there are many examples that mentions in the previous categories that related to economic decisions help to improving services or quality of product. There are fundamental principles to follow in engineering economics, which are: • • • • Principle 1: A dollar earned today is worth more than a dollar earned in the future. Principle 2: All that counts are the differences among alternatives. Principle 3: Marginal revenue must exceed marginal cost. Principle 4: Additional risk is not taken without the expected additional return. The four fundamental principles that much be applied in all engineering economic decisions are 1. 2. 3. 4. The time value of money Differential cost and revenue Marginal cost and revenue The trade-off between risk and reward Formula 𝑷𝑽 = 𝑪𝟏 (𝟏 + 𝒓)𝒏 𝑷𝒓𝒆𝒔𝒆𝒏𝒕 𝑽𝒂𝒍𝒖𝒆 = Where; C1 = Cash Flow at period 1 r = Rate of return 6% n = Number of periods 𝑪𝒂𝒔𝒉 𝑭𝒍𝒐𝒘 (𝟏 + 𝑹𝒂𝒕𝒆)𝑻𝒆𝒓𝒎 Period (n) Cash Flow (C) 0 0 1 -4,000.00 2 -2,500.00 3 3,500.00 4 -4,500.00 5 -2,000.00 6 8,686.55 7 3,000.00 = 𝑪𝟏 (𝟏 + 𝒓)𝒏 0 Present Value 0 = −4,000 (1 + 6%)1 -3,773.58 = −2,500 (1 + 6%)2 -2,224.99 = 3,500 (1 + 6%)3 2,938.67 = −4,500 (1 + 6%)4 -3,564.42 = −2,000 (1 + 6%)5 -1,494.52 = 𝟖, 𝟔𝟖𝟔. 𝟓𝟓 (𝟏 + 𝟔%)𝟔 6,123.67 = 3,000 (1 + 6%)7 1,995.17 Total = 0 I used other method to know the minimum of positive cash flow required in year 6; 𝑷𝑽 = 𝐶. 𝑃𝑉𝐹 𝑷𝑽𝑭 = 𝟏 (𝟏 + 𝒓)𝒏 Period (n) Cash Flow (C) 0 0 1 -4,000.00 2 𝑷𝑽𝑭 = 𝟏 (𝟏 + 𝒓)𝒏 Present Value = C. PVF 0 0 1 -3,773.58 -2,500.00 = (1+6%)2 = 0.8900 1 -2,224.99 3 3,500.00 = (1+6%)3 = 0.8396 1 2,938.67 4 -4,500.00 = (1+6%)4 = 0.7921 1 -3,564.42 5 -2,000.00 = (1+6%)5 = 0.7473 1 -1,494.52 6 8,686.55 = (𝟏+𝟔%)𝟔 = 0.7050 𝟏 6,123.67 7 3,000.00 = (1+6%)7 = 0.6651 1 1,995.17 NPV 0 = (1+6%)1 = 0.9434 3 – What report is used to describe a company’s financial position at the end of a reporting period? Briefly describe its contents. Balance sheet depicts the financial position of a company for a particular date; it shows the overall financial condition or health of the company, and sometimes called its “ Statement of financial position” repots three main categories of items, which are: assets, liabilities and stockholders’ equity. The relationship about balance sheet is: Assets = Liabilities + Shareholder’ Equity The Asset: contains Current Assets including cash and bank balance, inventories, receivables, short term investments and any prepaid expenses. Non-Current assets include land, property, plant and equipment net of depreciation, intangible assets like patents net of amortization, goodwill, and long term investments. Liabilities: contains current liabilities, which includes payables, any expenses payable, deferred revenue, short term debt and current portion of long term debt. Non-current liabilities include long-term debts. Shareholder’s Equity: section includes common stock, preferred stock and retained earnings. 4 – You have just purchased 1000 shares of stock at $70 per share. Your analysis indicates that the stock price will increase 10% per year. How much will your investment be worth in 5 years? When will the market price have doubled? Assume no dividend payments for this calculation. Given: P = $70,000 N = 5 years i = 10% per year The interest-earing process repeats, and after N periods, the total accumulated value (balance) F (the future sum) will grow to : 𝑭 = 𝑷(𝟏 + 𝒊)𝒏 𝐹 = 70,000(1 + 10%)5 𝐹 = 112,735.7 Investment in 5 years $112,735.70 Figure 1 - A cash flow diagram for 5 years 𝑭 = 𝑷(𝟏 + 𝒊)𝒏 140,000 = 70,000(1 + 10%)𝑛 𝑛 = 7.27 𝑦𝑒𝑎𝑟𝑠 The market price will be doubled in 7.27 years 5 – Suppose you make an annual contribution of $5000 to your investment account at the end of each year for 5 years. If the account earns 10% annually, how much can be withdraw early in the 11th year. Given: PMT = $5000 r = 5 years i = 10% per year (𝟏 + 𝖎)𝒏 − 𝟏 𝑭 = 𝑷[ ] 𝖎 𝐹 = 5,000 [ (1 + 10%)5 − 1 ] 10% 𝐹 = 30,526 After 5 years: PMT = $30,526 r = 5 years i = 10% per year 𝑭 = 𝑷(𝟏 + 𝒊)𝒏 𝐹 = 30,526(1 + 10%)5 𝐹 = 49,162 6 – What will be the amount accumulated by each of these present investments? P a b c d 5000 7250 9000 12000 Term ( r ) Rate (i) 7 15 30 8 7% 8% 6% 5% Compounding 1 4 12 Continuously a 𝐹 = 𝑃(1 + 𝑖)𝑛 𝐹 = 5000(1 + 7%)7 𝐹 = 8,028.91 b 𝐹 = 𝑃(1 + 𝑖)𝑛 𝐹 = 7250(1 + 8%)15 𝐹 = 23,787.47 c 𝐹 = 𝑃(1 + 𝑖)𝑛 𝐹 = 9,000(1 + 6%)30 𝐹 = 54,203.18 d 𝐹 = 𝑃 × 𝑒 𝑖×𝑛 𝐹 = 12,000 × 𝑒 5%×8 𝐹 = 17,901.90 F $8,028.91 $23,787.47 $54,203.18 $17,901.90 7 – If a bank advertises a savings account that pays a 7% nominal interest rate compounded continuously, what is the effective annual percentage rate? Nominal rate (r) = 7% 𝒊 = 𝒆𝒓 − 𝟏 𝑖 = 2.718280.07 − 1 𝑖 = 7.251% so, the effective annual interest rate for a nominal interest rate of 7% compounded continuously is 7.251% 8 – If the interest rate is 8% compounded continuously, what is the required quarterly payment to repay a loan of $10,000 in five years? The formula of effective interest rate is given below: 𝒓 𝒊𝒂 = 𝒆𝒌 − 𝟏 Here, r = 0.080 (8%) and K = 4 𝑖𝑎 = 𝑒 8% 4 −1 𝑖𝑎 = 2.02% Thus, the quarterly effective interest rate is approximately 2.02%. Now we will find the number of compounding year as follows: 𝑵 = 𝑲(𝑵𝒐 𝒐𝒇 𝒚𝒆𝒂𝒓𝒔) 𝑁 =4×5 𝑁 = 20 years In last step, use (i) and (n) in appropriate equivalence formula that is shown below: 𝐴 = 𝑃(𝐴⁄𝑃 , 𝑖, 𝑁) Where, P refer to the present worth of the sum of money in this situation, A refers to the annual amount placed in the account, (i) refer to the interest rate and N refer to the number of terms that the money is for 𝐴 = 10,000(𝐴⁄𝑃 , 2.02%, 20) 𝐴 = 10,000(0.06703) 𝐴 = 670.3 Hence, the required annual payment made over 5 years with given 8% interest rate, which is compounded continuously, is $670.3 9 – You are considering two types of machines for a manufacturing process: Machine A: PW(13%)A = $75,200 – ($6,000 + $2,400) (P / A, 13%, 6) + $20,000 (P/F, 13%, 6) = - $98,973 Years Machine A Investment Cost Salvage Tax Net CashFlow Present Value 0 1 2 3 4 5 6 $(6,000) $(6,000) $(6,000) $(6,000) $(6,000) $(2,400) $(8,400) $(2,400) $(8,400) $(2,400) $(8,400) $(2,400) $(8,400) $(2,400) $(8,400) $(6,000) $20,000 $(2,400) $11,600 $(75,000) $(75,000) $(98,973) Machine B: PW(13%)B = -$40,000 – $11,000 (P / A, 13%, 6) = - $83,973.05 Machine B Investment Cost Salvage Tax Net Cash Flow 0 $(40,000) Present Value $(83,973.05) $(40,000) 1 2 3 4 5 7 $(11,000) $(11,000) $(11,000) $(11,000) $(11,000) $$(11,000) $$(11,000) $$(11,000) $$(11,000) $$(11,000) $(11,000) $$$(11,000) 𝑁𝑒𝑡 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = ∑ 𝐶𝑎𝑠ℎ𝑓𝑙𝑜𝑤𝑠 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑉𝑎𝑙𝑢𝑒 = ∑ { Machine B is a better choice 𝑁𝑒𝑡 𝑃𝑒𝑟𝑖𝑜𝑑 𝐶𝑎𝑠ℎ𝐹𝑙𝑜𝑤 } − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 (1 + 𝑅)𝑇 3 TIMES= (1+8%)N 1LOG 3=NLOG 1.08 N=14.27 YEARS. By using Excel file: Initial Investment Revenue Capital Exp Net Cash Flows Payback Period IRR Assumption Rate of Return NPV 2016 -10000000 -10000000 2.11 57.72% 15% $15,592,371 4250000 -250000 4000000 2017 2018 2019 2020 5100000 10100000 11500000 11000000 -100000 -100000 -500000 5000000 10000000 11000000 11000000 Extra information required would be how many average or minimum number of sales units does McDonalds make per day, the pricing of their products, and the financials for the last few years to determine the consistency or growth of sales. This would help determine the position of the firm if the salary for all employees would be increased. Assumptions made -That the Big Mac would still be sold at the same rate and same number of sales units as before with the increased cost of 68 cents per unit. -The assumption that the labor force would be the same number -The assumption that clients would tolerate the incremental cost of 68 cents per Big Mac Opinion The above assumption cannot be true as doubling the salary for all employees’ means doubling the expenditure on salaries which makes a large portion of the recurrent expenditure. The statement also cannot be true as there are assumptions that cannot just be made to happen. The variables could shift in any direction for instance decline in sales would negatively affect the firm If the minimum wage were tripled the statement would still not be true as the assumptions would be indeterminable for certain.
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Running head ENGINEERING ECONOMIC DECISIONS

Engineering Economic Decisions
Student’s Name
Institutional Affiliation

1

ENGINEERING ECONOMIC DECISIONS

2

Engineering Economic Decisions
What are the common types of Engineering Economic Decisions? Give an example of each.
Briefly describe the fundamental principles of Engineering Economics.
Engineering economic decisions comprises of five primary categories.
Product expansion or new products increases the revenue of the company by producing
a new product or by expanding it. A usual kind of expansion decision made is such as decisions
regarding expenditure that focuses on raising the output of subsisting distribution and production
facilities. Another sort of expenditure decision made includes contemplating the disbursement
obligatory to manufacture a new product or that required enlarging the existing one to a new
geographical place.
Cost reduction is a type of engineering economic decision that considers the lowering of
a company’s operating cost. For example, the firm has 500 marketers; the management
employed this significant number of employees thinking that a large number will result in high
sales. Their salary amount to $250,000 per month. Reducing this number to 150 competent
marketers and doubling their salary will reduce the cost of wages, motivate them to work hard
and increase production.
Ameliorating in quality and services is an engineering economic decision made daily.
The firm has to improve on its services and quality of products to gain a competitive advantage.
For example, many automobile companies sell the same makes of cars. For a firm to attain some
uniqueness it has to source high-quality cheap materials, add some accessories and sell at the
same price as the competitors.
Equipment and process selection are engineering economic decisions that ensure that a
firm chooses high-quality material at a reasonable price putting into mind that the material will

ENGINEERING ECONOMIC DECISIONS

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determine the manufacturing process. For example, the production process for an automobile’s
body panels is affected by the material, the equipment, and the machinery. The engineers should
consider factors such as the cost of labor, material, machinery, and equipment.
Equipment replacement is decisions engineers make to enhance production and reduce
costs. For example, a firm can buy 20 large presses to manufacture stamped metal parts for 15
years. They then disc...


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