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Presented By: Jarred O’Dell, ASP Safety Director Syracuse Utilities

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1 Presented By: Jarred O’Dell, ASP Safety Director Syracuse Utilities
The Engineering Economic Equations Every Safety Professional Should Know Presented By: Jarred O’Dell, ASP Safety Director Syracuse Utilities

2 House Keeping Please Silence Cell Phones

3 House Keeping Please Silence Cell Phones Sign-in Sheet

4 House Keeping Please Silence Cell Phones Sign-in Sheet
Sleeping during the presentation

5 Why Engineering Economics
For those of you pursuing your CSP: 7 – 15 questions on the ASP Exam 7 – 15 questions on the CSP Exam For everyone else: Ever buy a car or a house? Who has a credit card?

6 Engineering Economics
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11 Engineering Economics
You do not need an app for that!!!

12 Engineering Economics

13 Uff Da! Uff da is an expression of Norwegian origin adopted by Scandinavian-Americans. This exclamation is an announcement that, that person is going into a state of sensory overload.

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15 Uff Da! If you find yourself going into sensory overload and need to ask a question say: Uff Da!

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17 Uff Da!

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19 Question: 1

20 Engineering Economics: Q1
A wealthy relative died and left you her estate. You can choose to either accept $6,000,000 today or wait and receive $10,000,000 in five years. Assume the annual interest rate over this period is 10%. You decide to…

21 B) Take the money and run
A wealthy relative died and left you her estate. You can choose to either accept $6,000,000 today or wait and receive $10,000,000 in five years. Assume the annual interest rate over this period is 10%. You decide to: Wait the 5 years B) Take the money and run Do nothing, its probably the same people that brought you the Nigerian Lottery Question 1

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23 Engineering Economics: Q1
You decide to: Wait the 5 years B) Take the money and run Do nothing, its probably the same people that brought you the Nigerian Lottery

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25 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1 Page 1

26 Engineering Economics
F = P = A = n = i = Future Present Amount of periodic receipt/payment Number of years* Annual Interest* expressed in decimal form (e.g. 10% = .10) Page 1

27 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1 Page 1

28 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

29 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

30 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

31 Engineering Economics: Q1
𝐹= 𝑃(1+𝑖) 𝑛 F = P = A = i = n = 10,000,000 6,000,000 N/A 10% or 0.10 5 years 𝑃= 𝐹(1+𝑖) −𝑛 Page 2

32 Engineering Economics: Q1
𝐹= 𝑃(1+𝑖) 𝑛 F = P = 6,000,000 ( )5 10,000,000(1+0.10)-5 𝑃= 𝐹(1+𝑖) −𝑛 Page 2

33 Question: 2

34 Engineering Economics: Q2
You decided to go back to school and eared your Masters Degree in mathematics. Having heard this, your supervisor throws this scenario at you:

35 Engineering Economics: Q2
The chief financial officer of Widget Inc. expects a 10% annual return on investments for all capital projects. What is the maximum cost that will be approved from a project that is expected to save $8,000 per year over 10 years? Assume the project will be fully depreciated in the 10 years. Question 2

36 Uff Da!

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38 Engineering Economics: Q2
F = P = A = i = n = N/A ??? $8,000 10% or 0.10 10 years Page 3

39 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

40 Engineering Economics: Q3
𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= (1+.10) 10 −1 .10(1+.10) 10 Page 3

41 Question: 3

42 Engineering Economics: Q3
You recently obtained 6σ Black Belt status. Congratulations! Understandably, you are very anxious to test out your new skills. Soon you face this problem:

43 Engineering Economics: Q3
The financial policy of Acme requires that capital investments must have an annual return of 12%. An engineering solution to a safety problem will cost $250,000 for the initial installation, and it will cost $12,000 annually to maintain for 15 years. What is the required annual savings from this project in order for it to be approved? Question 3

44 Engineering Economics: Q3
F = P = A = i = n = N/A $250,000 ??? 12% or 0.12 10 years Page 4

45 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

46 Engineering Economics: Q3
𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1 A = (1+.12) 15 (1+.12) 15 −1 4-34 Page 4

47 Engineering Economics: Q3
𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1 P = i = n = 4-34

48 Question: 4

49 Engineering Economics: Q4
Having recently been conferred as a Doctor in Safety and Engineering Science, your are now in a position to poses this scenario to your employer:

50 Engineering Economics: Q4
Your company decided to hire an EHS/6σ, executive. If a balloon payment of $10,000,000 is due in 10 years, what amount would management have to deposit monthly into a savings account (paying interest of 6% per year) to accumulate adequate funds to pay the note? Question 4

51 Engineering Economics: Q4
F = P = A = i = n = $10,000,000 N/A ??? 6% or 0.06 10 years Page 5

52 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

53 Engineering Economics
𝐴=𝐹 𝑖 1+𝑖 𝑛 −1 A = 10,000, −1 𝐴=10,000, −1

54 Engineering Economics
F = P = A = n = i = Future Present Amount of periodic receipt/payment Number of years* Annual Interest* expressed in decimal form (e.g. 10% = .10) Page 1

55 Engineering Economics
𝐴=𝐹 𝑖 1+𝑖 𝑛 −1 A = 10,000, −1 𝐴=10,000, −1

56 Question: 5

57 Engineering Economics: Q4
One of your faceless pawns is having trouble figuring out the following scenario. He humbly/fearfully asks for your help:

58 Engineering Economics: Q5
Calculate the monetary value after ten years of a behavior based safety program that costs $40,000 per year at the start. Assume an inflation rate of 4.3% Question 5

59 Engineering Economics: Q4
F = P = A = i = n = ??? N/A $40,000 4.3% or 0.043 10 years Page 5

60 Engineering Economics
𝐹= 𝑃(1+𝑖) 𝑛 𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 𝑃=𝐴 (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 𝑃= 𝐹(1+𝑖) −𝑛 𝐴=𝐹 𝑖 (1+𝑖) 𝑛 −1 𝐴=𝑃 𝑖(1+𝑖) 𝑛 (1+𝑖) 𝑛 −1

61 Engineering Economics: Q5
𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 A = i = n =

62 Engineering Economics: Q5
𝐹=𝐴 (1+𝑖) 𝑛 −1 𝑖 F = 40, ( ) 10 −

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65 Uff Da!

66 Engineering Economics

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