09 Uniform Gradient Present Worth Concept Overview
UNIFORM GRADIENT PRESENT WORTH | CONCEPT OVERVIEW The TOPIC of UNIFORM GRADIENT PRESENT WORTH can be referenced on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: In this review, we will define the topic of Compound Interest and the use of Uniform Gradient Payment Formulas, walk through the general work flow of solving such problems, and jump in to working an example of something we may see on the exam. As explained in previous reviews, money does not have the same value at different points in time. For this reason, we need tools, tables, formulas, and various economic factors to reference when it is necessary to compare two complex alternatives. Uniform Gradient Series of transactions are always increasing or decreasing by a fixed amount and denoted with a “G”. Recall that we are concerned with the effects of interest when using these formulas, and more specifically, the compound interest. Compound interest is the interest for a period calculated off the principle and interest from a previous period.
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All engineering economic analysis is based off compound interest, and for that reason, special tables with various pre-calculated conversion factors have been developed for our use. Uniform Gradient Payment Formulas take a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time. Uniform Gradient Series of transactions are always increasing or decreasing by a fixed amount and denoted with a “G”. We are concerned with the effects of compound interest when using Uniform Gradient Series formulas, and more specifically, the compound interest. Compound interest is the interest for a period calculated off the principle and interest from a previous period. The goal of any Uniform Gradient Series Payment problem is to determine what single monetary value or uniform annual value would be equivalent based off specific economic factors. The first step to solving a Uniform Gradient Series Payment problem is to determine the various factors of importance.
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These factors include: 1. The trend and magnitude of the Uniform Gradient Series, given as “G” 2. The equivalent value to be determined, FUTURE, PRESENT, or ANNUAL. 3. The interest rate given as “i” 4. The number periods given as “n” Uniform Gradient Payment Formula problems can be solved either by using the Uniform Gradient Present Worth Formula or using the functional notation version of the equation and referencing the Compound Interest Tables. The FORMULA for UNIFORM GRADIENT PRESENT WORTH PRESENT WORTH can be referenced in the TABLE of ENGINEERING ECONOMICS FACTORS under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The formula for UNIFORM GRADIENT PRESENT WORTH takes a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time, and is given as:
𝑃𝑃 = 𝐺𝐺
1 + 𝑖𝑖 ' − 1 𝑛𝑛 − 𝑖𝑖 ) 1 + 𝑖𝑖 ' 𝑖𝑖 1 + 𝑖𝑖
'
Where: • P is a present sum of money taking in to account some uniform gradient, G, costs at a certain interest rate, i, over some number of periods, n • G is a uniform gradient sum of money • i is the effective interest rate for a specific interest period. It is important to note that the problem will give this interest rate as a percentage, but when we are working it in to these formulas, it must be represented as a decimal value. • n is the number of interest periods. It can be any length of time, but most commonly will be one year.
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CONCEPT EXAMPLE: A company purchases a machine that is projected to have a maintenance cost of $250 after the first year, and is expected to increase $250 every year thereafter for the 15 year expected lifespan. Assuming a 10% interest, the amount money the company should put aside now (present time) to maintain this machine for the duration of its lifespan is most close to: A. $9,937 B. $10,038 C. $10,382 D. $10,862
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SOLUTION: The TOPIC of UNIFORM GRADIENT PRESENT WORTH can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The goal is to determine how much money the company should put aside NOW to cover the 15 years of increasing maintenance costs for this piece of equipment. Uniform Gradient Payment Formula problems can be solved either by using the Uniform Gradient Present Worth Formula or using the functional notation version of the equation and referencing the Compound Interest Tables. We will first solve the problem using the Uniform Gradient Present Worth Formula, and then solve it again using the functional notation version of the equation. This problem deals with GRADIENT (G) costs over a period of time. To convert these costs in to a Present sum we can use the Uniform Gradient Present Worth Formula. The FORMULA for UNIFORM GRADIENT PRESENT WORTH can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The formula for UNIFORM GRADIENT PRESENT WORTH takes a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time, and is given as:
𝑃𝑃 = 𝐺𝐺
1 + 𝑖𝑖 ' − 1 𝑛𝑛 − 𝑖𝑖 ) 1 + 𝑖𝑖 ' 𝑖𝑖 1 + 𝑖𝑖
'
Where: • P is a present sum of money taking in to account some uniform gradient, G, costs at a certain interest rate, i, over some number of periods, n • G is a uniform gradient sum of money • i is the effective interest rate for a specific interest period. It is important to note that the problem will give this interest rate as a percentage, but when we are working it in to these formulas, it must be represented as a decimal value. • n is the number of interest periods. It can be any length of time, but most commonly will be one year. In this problem, we are given: • P = Is the unknown value we are solving for. • G = $250 • i = .10 (10%) • n = 15 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.1 01 − 1 15 𝑃𝑃 = $250 − (0.1)) 1 + 0.1 01 (0.1) 1 + 0.1
01
= $10,038
The company should set aside $10,038 NOW to ensure they are able to pay the maintenance costs for this machine over its 15 year lifespan. Next, we will go ahead and solve the problem again using the functional notation version of the problem and referencing the Compound Interest Table provided in the NCEES Supplied Reference Handbook. As mentioned earlier, as Engineering Economic problems get more complicated, it is best to get comfortable using the functional notation version of the equations and referencing the Compound Interest Tables as it will lead to a much more efficient use of your time. The Uniform Gradient Present Worth Formula written in functional notation for a Present Worth is: 𝑃𝑃 = 𝐺𝐺(𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛)
Where the term (𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛) can be defined using the given values (i, n) and the
Compound Interest Table provided in the NCEES Supplied Reference Handbook.
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In the problem, we are given: • i = .10 (10%) • n = 15 years The FACTOR TABLE for an INTEREST RATE of 10% can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 136 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Referencing the Compound Interest Table for i=10% in the NCEES Supplied Reference Handbook, we locate n=15 (far left column and work our way horizontally to the fact P/G and find that:
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For the Uniform Gradient Present Worth Formula written in functional notation as Present Worth is 𝑃𝑃 = 𝐺𝐺(𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛), the economic factor is determined to be: (𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛) = 40.1520
Plugging this value into the equation, we get: 𝑃𝑃 = $250 40.1520 = $10,038
The company should set aside $10,038 NOW to ensure they are able to pay the maintenance costs for this machine over its 15 year lifespan.
Therefore, the correct answer choice is B. $𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎
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CONCEPT INTRO: In this review, we will define the topic of Compound Interest and the use of Uniform Gradient Payment Formulas, walk through the general work flow of solving such problems, and jump in to working an example of something we may see on the exam. As explained in previous reviews, money does not have the same value at different points in time. For this reason, we need tools, tables, formulas, and various economic factors to reference when it is necessary to compare two complex alternatives. Uniform Gradient Series of transactions are always increasing or decreasing by a fixed amount and denoted with a “G”. Recall that we are concerned with the effects of interest when using these formulas, and more specifically, the compound interest. Compound interest is the interest for a period calculated off the principle and interest from a previous period.
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All engineering economic analysis is based off compound interest, and for that reason, special tables with various pre-calculated conversion factors have been developed for our use. Uniform Gradient Payment Formulas take a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time. Uniform Gradient Series of transactions are always increasing or decreasing by a fixed amount and denoted with a “G”. We are concerned with the effects of compound interest when using Uniform Gradient Series formulas, and more specifically, the compound interest. Compound interest is the interest for a period calculated off the principle and interest from a previous period. The goal of any Uniform Gradient Series Payment problem is to determine what single monetary value or uniform annual value would be equivalent based off specific economic factors. The first step to solving a Uniform Gradient Series Payment problem is to determine the various factors of importance.
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These factors include: 1. The trend and magnitude of the Uniform Gradient Series, given as “G” 2. The equivalent value to be determined, FUTURE, PRESENT, or ANNUAL. 3. The interest rate given as “i” 4. The number periods given as “n” Uniform Gradient Payment Formula problems can be solved either by using the Uniform Gradient Present Worth Formula or using the functional notation version of the equation and referencing the Compound Interest Tables. The FORMULA for UNIFORM GRADIENT PRESENT WORTH PRESENT WORTH can be referenced in the TABLE of ENGINEERING ECONOMICS FACTORS under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The formula for UNIFORM GRADIENT PRESENT WORTH takes a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time, and is given as:
𝑃𝑃 = 𝐺𝐺
1 + 𝑖𝑖 ' − 1 𝑛𝑛 − 𝑖𝑖 ) 1 + 𝑖𝑖 ' 𝑖𝑖 1 + 𝑖𝑖
'
Where: • P is a present sum of money taking in to account some uniform gradient, G, costs at a certain interest rate, i, over some number of periods, n • G is a uniform gradient sum of money • i is the effective interest rate for a specific interest period. It is important to note that the problem will give this interest rate as a percentage, but when we are working it in to these formulas, it must be represented as a decimal value. • n is the number of interest periods. It can be any length of time, but most commonly will be one year.
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CONCEPT EXAMPLE: A company purchases a machine that is projected to have a maintenance cost of $250 after the first year, and is expected to increase $250 every year thereafter for the 15 year expected lifespan. Assuming a 10% interest, the amount money the company should put aside now (present time) to maintain this machine for the duration of its lifespan is most close to: A. $9,937 B. $10,038 C. $10,382 D. $10,862
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SOLUTION: The TOPIC of UNIFORM GRADIENT PRESENT WORTH can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The goal is to determine how much money the company should put aside NOW to cover the 15 years of increasing maintenance costs for this piece of equipment. Uniform Gradient Payment Formula problems can be solved either by using the Uniform Gradient Present Worth Formula or using the functional notation version of the equation and referencing the Compound Interest Tables. We will first solve the problem using the Uniform Gradient Present Worth Formula, and then solve it again using the functional notation version of the equation. This problem deals with GRADIENT (G) costs over a period of time. To convert these costs in to a Present sum we can use the Uniform Gradient Present Worth Formula. The FORMULA for UNIFORM GRADIENT PRESENT WORTH can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 131 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The formula for UNIFORM GRADIENT PRESENT WORTH takes a Uniform Gradient Series, which is cash flow either increasing or decreasing by a fixed amount over a period of time, and converts it in to either a uniform annual value or a single equivalent value at some other point in time, and is given as:
𝑃𝑃 = 𝐺𝐺
1 + 𝑖𝑖 ' − 1 𝑛𝑛 − 𝑖𝑖 ) 1 + 𝑖𝑖 ' 𝑖𝑖 1 + 𝑖𝑖
'
Where: • P is a present sum of money taking in to account some uniform gradient, G, costs at a certain interest rate, i, over some number of periods, n • G is a uniform gradient sum of money • i is the effective interest rate for a specific interest period. It is important to note that the problem will give this interest rate as a percentage, but when we are working it in to these formulas, it must be represented as a decimal value. • n is the number of interest periods. It can be any length of time, but most commonly will be one year. In this problem, we are given: • P = Is the unknown value we are solving for. • G = $250 • i = .10 (10%) • n = 15 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.1 01 − 1 15 𝑃𝑃 = $250 − (0.1)) 1 + 0.1 01 (0.1) 1 + 0.1
01
= $10,038
The company should set aside $10,038 NOW to ensure they are able to pay the maintenance costs for this machine over its 15 year lifespan. Next, we will go ahead and solve the problem again using the functional notation version of the problem and referencing the Compound Interest Table provided in the NCEES Supplied Reference Handbook. As mentioned earlier, as Engineering Economic problems get more complicated, it is best to get comfortable using the functional notation version of the equations and referencing the Compound Interest Tables as it will lead to a much more efficient use of your time. The Uniform Gradient Present Worth Formula written in functional notation for a Present Worth is: 𝑃𝑃 = 𝐺𝐺(𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛)
Where the term (𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛) can be defined using the given values (i, n) and the
Compound Interest Table provided in the NCEES Supplied Reference Handbook.
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In the problem, we are given: • i = .10 (10%) • n = 15 years The FACTOR TABLE for an INTEREST RATE of 10% can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 136 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Referencing the Compound Interest Table for i=10% in the NCEES Supplied Reference Handbook, we locate n=15 (far left column and work our way horizontally to the fact P/G and find that:
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For the Uniform Gradient Present Worth Formula written in functional notation as Present Worth is 𝑃𝑃 = 𝐺𝐺(𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛), the economic factor is determined to be: (𝑃𝑃/𝐺𝐺, 𝑖𝑖, 𝑛𝑛) = 40.1520
Plugging this value into the equation, we get: 𝑃𝑃 = $250 40.1520 = $10,038
The company should set aside $10,038 NOW to ensure they are able to pay the maintenance costs for this machine over its 15 year lifespan.
Therefore, the correct answer choice is B. $𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎
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