10 Uniform Gradient Present Worth Problem Set
UNIFORM GRADIENT PRESENT WORTH | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: A proprietary piece of mechanical equipment has an expected maintenance cost of $500 after the first year, and is expected to increase $100 every year thereafter for the 8 year expected lifespan. Assuming a 6% interest, the sum of money the company should put aside now to maintain it for the entire 8 year period is most close to: A. $5,089 B. $9,921 C. $3,105 D. $2,065
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SOLUTION 1: 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 8 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. Uniform Gradient problems begin to get a little bit more complicated, but with a few extra steps, they can be simplified to be no different than those we are already familiar with.
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The first step is to graphically illustrate the scenario using a cash flow diagram:
If the gradient maintenance costs increased uniformly starting at year 0 (i.e. Year 0 $0, Year 1 - $100, Year 2 - $200, etc.) then we could simply use the Uniform Gradient Present Worth Formula found in the NCEES Supplied Reference Handbook to convert the Gradient costs (G) to a Present sum (P). However, since the first cost is $500 and then the uniform gradient increase begins, we need to approach the problem in parts.
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We can take the Cash Flow Diagram and dissect it in to two segments such that:
The FORMULA for UNIFORM SERIES 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 lower segment should look familiar; itβs simply a Uniform Cost scenario. To convert this in to a Present sum we can use the Uniform Series Present Worth Formula, as we have before, which is:
ππ = π΄π΄
1 + ππ ' β 1 ππ 1 + ππ '
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The lower segment represents the cash flow that will be CONSTANT throughout the lifetime of the investment. The upper segment shows 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 provided in the NCEES Supplied Reference Handbook. The triangle represents the cash flow DECREASING throughout the lifetime of the investment. You can imagine the gradient as the SLOPE of the triangle. 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. 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:
ππ = πΊπΊ
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Where: β’ 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 β’ G is a uniform gradient sum of money
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β’ 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. β’ n is the number of interest periods. It can be any length of time, but most commonly will be one year. If we look at this problem, we are essentially trying to calculate the AREA of the trapezoid shown, using two different formulas specific to Engineering Economics that represent the TOTAL AREA of the trapezoid. To achieve the total sum of money that the company should invest now at 6% interest to maintain this piece of equipment over the 8 year period, we must combine the two values such that: ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ) + π΄π΄(ππ/π΄π΄, ππ, ππ)
We can then re-write the total sum of the projects total investment in terms of the formulas provided in the NCEES Supplied Reference Handbook as:
ππ = π΄π΄
1 + ππ ' β 1 + πΊπΊ ππ 1 + ππ '
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
In this problem, we are given: β’ P = Is the unknown value we are solving for. β’ A = $500 β’ i = .06 (6%) β’ n = 8 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.06 5 β 1 1 + 0.06 5 β 1 8 ππ = $500 + $100 β 5 * 5 (0.06) 1 + 0.06 (0.06) 1 + 0.06 (0.06) 1 + 0.06
5
ππ = $5,089
The company should set aside $5,089 NOW to ensure they are able to pay the maintenance costs for this machine over its 8 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 second method to solving this problem is to use both the Uniform Series Compound Amount Formula and the Uniform Gradient Present Worth Formula written in functional notation for a Present Worth, which is ππ = π΄π΄(ππ/π΄π΄, ππ, ππ) and ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ) where the term (ππ/π΄π΄, ππ, ππ) and (ππ/πΊπΊ, ππ, ππ) can be defined using the given
values (ππ, ππ) and the Compound Interest Tables provided in the NCEES Supplied
Reference Handbook.
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The Uniform Gradient Present Worth Formula written in functional notation for a Present Worth is: ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ)
Where the term (ππ/πΊπΊ, ππ, ππ) can be defined using the given values (i, n) and the
Compound Interest Table provided in the NCEES Supplied Reference Handbook. In the problem, we are given: β’ i = .06 (6%) β’ n = 8 years β’ The FACTOR TABLE for an INTEREST RATE of 6% can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 135 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Referencing the Compound Interest Table for i=6% in the NCEES Supplied Reference Handbook, we locate n=8 (far left column) and work our way horizontally to the factor P/A and find that:
For the Uniform Series Present Worth Formula written in functional notation as Present Worth is ππ = π΄π΄(ππ/π΄π΄, ππ, ππ), the economic factor is determined to be: (ππ/π΄π΄, 6%, 8) = 6.2098
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Referencing the Compound Interest Table for i=6% in the NCEES Supplied Reference Handbook, we locate n=8 (far left column) and work our way horizontally to the factor P/G and find that:
For the Uniform Gradient Present Worth Formula written in functional notation as Present Worth is ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ), the economic factor is determined to be: (ππ/πΊπΊ, 6%, 8) = 19.8416
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Recall, to achieve the total sum of money that the company should invest now at 6% interest to maintain this piece of equipment over the 8 year period, we must combine the two values such that:
ππ = π΄π΄
1 + ππ ' β 1 + πΊπΊ ππ 1 + ππ '
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Plugging these values into the equation, we get: ππ = $500 6.2098 + $100 19.8416 = $5,089.06
The company should set aside $5,089 NOW to ensure they are able to pay the maintenance costs for this machine over its 8 year lifespan.
Therefore, the correct answer choice is A. $5,089
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PROBLEM 2: A concrete company wants to purchase 2 new pump trucks at $200,000 apiece. It is estimated that each year following the purchase, it will cost them $1,000 to maintain each of the trucks, incrementally increasing by $1,000 over the first 5 years. Assuming an interest rate of 6%, the sum of money should the company set aside now to ensure that they are able to cover the fees as they arise over the 5 year period is most close to: A. $7,935 B. $11,968 C. $2,330 D. $15,869
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SOLUTION 2: 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 the total sum of money that the company should invest now at 6% interest to cover projected maintenance costs for the first 5 years after purchasing the trucks. The maintenance costs are projected to be uniform, incrementally increasing by $1,000 starting the year after the purchase. We are looking to convert these GRADIENT costs in to a PRESENT sum. 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.
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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. 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:
ππ = πΊπΊ
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Where: β’ 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 β’ G is a uniform gradient sum of money β’ 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. β’ 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: β’ P = Is the unknown value we are solving for. β’ G = $1,000 β’ i = .06 (6%) β’ n = 5 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.06 < β 1 5 ππ = $1,000 β (0.06)* 1 + 0.06 < (0.06) 1 + 0.06
PROBLEM 1: A proprietary piece of mechanical equipment has an expected maintenance cost of $500 after the first year, and is expected to increase $100 every year thereafter for the 8 year expected lifespan. Assuming a 6% interest, the sum of money the company should put aside now to maintain it for the entire 8 year period is most close to: A. $5,089 B. $9,921 C. $3,105 D. $2,065
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SOLUTION 1: 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 8 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. Uniform Gradient problems begin to get a little bit more complicated, but with a few extra steps, they can be simplified to be no different than those we are already familiar with.
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The first step is to graphically illustrate the scenario using a cash flow diagram:
If the gradient maintenance costs increased uniformly starting at year 0 (i.e. Year 0 $0, Year 1 - $100, Year 2 - $200, etc.) then we could simply use the Uniform Gradient Present Worth Formula found in the NCEES Supplied Reference Handbook to convert the Gradient costs (G) to a Present sum (P). However, since the first cost is $500 and then the uniform gradient increase begins, we need to approach the problem in parts.
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We can take the Cash Flow Diagram and dissect it in to two segments such that:
The FORMULA for UNIFORM SERIES 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 lower segment should look familiar; itβs simply a Uniform Cost scenario. To convert this in to a Present sum we can use the Uniform Series Present Worth Formula, as we have before, which is:
ππ = π΄π΄
1 + ππ ' β 1 ππ 1 + ππ '
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by Prepineer | Prepineer.com
The lower segment represents the cash flow that will be CONSTANT throughout the lifetime of the investment. The upper segment shows 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 provided in the NCEES Supplied Reference Handbook. The triangle represents the cash flow DECREASING throughout the lifetime of the investment. You can imagine the gradient as the SLOPE of the triangle. 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. 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:
ππ = πΊπΊ
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Where: β’ 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 β’ G is a uniform gradient sum of money
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β’ 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. β’ n is the number of interest periods. It can be any length of time, but most commonly will be one year. If we look at this problem, we are essentially trying to calculate the AREA of the trapezoid shown, using two different formulas specific to Engineering Economics that represent the TOTAL AREA of the trapezoid. To achieve the total sum of money that the company should invest now at 6% interest to maintain this piece of equipment over the 8 year period, we must combine the two values such that: ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ) + π΄π΄(ππ/π΄π΄, ππ, ππ)
We can then re-write the total sum of the projects total investment in terms of the formulas provided in the NCEES Supplied Reference Handbook as:
ππ = π΄π΄
1 + ππ ' β 1 + πΊπΊ ππ 1 + ππ '
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
In this problem, we are given: β’ P = Is the unknown value we are solving for. β’ A = $500 β’ i = .06 (6%) β’ n = 8 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.06 5 β 1 1 + 0.06 5 β 1 8 ππ = $500 + $100 β 5 * 5 (0.06) 1 + 0.06 (0.06) 1 + 0.06 (0.06) 1 + 0.06
5
ππ = $5,089
The company should set aside $5,089 NOW to ensure they are able to pay the maintenance costs for this machine over its 8 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 second method to solving this problem is to use both the Uniform Series Compound Amount Formula and the Uniform Gradient Present Worth Formula written in functional notation for a Present Worth, which is ππ = π΄π΄(ππ/π΄π΄, ππ, ππ) and ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ) where the term (ππ/π΄π΄, ππ, ππ) and (ππ/πΊπΊ, ππ, ππ) can be defined using the given
values (ππ, ππ) and the Compound Interest Tables provided in the NCEES Supplied
Reference Handbook.
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The Uniform Gradient Present Worth Formula written in functional notation for a Present Worth is: ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ)
Where the term (ππ/πΊπΊ, ππ, ππ) can be defined using the given values (i, n) and the
Compound Interest Table provided in the NCEES Supplied Reference Handbook. In the problem, we are given: β’ i = .06 (6%) β’ n = 8 years β’ The FACTOR TABLE for an INTEREST RATE of 6% can be referenced under the SUBJECT of ENGINEERING ECONOMICS on page 135 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Referencing the Compound Interest Table for i=6% in the NCEES Supplied Reference Handbook, we locate n=8 (far left column) and work our way horizontally to the factor P/A and find that:
For the Uniform Series Present Worth Formula written in functional notation as Present Worth is ππ = π΄π΄(ππ/π΄π΄, ππ, ππ), the economic factor is determined to be: (ππ/π΄π΄, 6%, 8) = 6.2098
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Referencing the Compound Interest Table for i=6% in the NCEES Supplied Reference Handbook, we locate n=8 (far left column) and work our way horizontally to the factor P/G and find that:
For the Uniform Gradient Present Worth Formula written in functional notation as Present Worth is ππ = πΊπΊ(ππ/πΊπΊ, ππ, ππ), the economic factor is determined to be: (ππ/πΊπΊ, 6%, 8) = 19.8416
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Recall, to achieve the total sum of money that the company should invest now at 6% interest to maintain this piece of equipment over the 8 year period, we must combine the two values such that:
ππ = π΄π΄
1 + ππ ' β 1 + πΊπΊ ππ 1 + ππ '
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Plugging these values into the equation, we get: ππ = $500 6.2098 + $100 19.8416 = $5,089.06
The company should set aside $5,089 NOW to ensure they are able to pay the maintenance costs for this machine over its 8 year lifespan.
Therefore, the correct answer choice is A. $5,089
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PROBLEM 2: A concrete company wants to purchase 2 new pump trucks at $200,000 apiece. It is estimated that each year following the purchase, it will cost them $1,000 to maintain each of the trucks, incrementally increasing by $1,000 over the first 5 years. Assuming an interest rate of 6%, the sum of money should the company set aside now to ensure that they are able to cover the fees as they arise over the 5 year period is most close to: A. $7,935 B. $11,968 C. $2,330 D. $15,869
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SOLUTION 2: 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 the total sum of money that the company should invest now at 6% interest to cover projected maintenance costs for the first 5 years after purchasing the trucks. The maintenance costs are projected to be uniform, incrementally increasing by $1,000 starting the year after the purchase. We are looking to convert these GRADIENT costs in to a PRESENT sum. 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.
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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. 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:
ππ = πΊπΊ
1 + ππ ' β 1 ππ β ππ * 1 + ππ ' ππ 1 + ππ
'
Where: β’ 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 β’ G is a uniform gradient sum of money β’ 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. β’ 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: β’ P = Is the unknown value we are solving for. β’ G = $1,000 β’ i = .06 (6%) β’ n = 5 years
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Plugging these values in to the Uniform Series Compound Amount Formula we get: 1 + 0.06 < β 1 5 ππ = $1,000 β (0.06)* 1 + 0.06 < (0.06) 1 + 0.06
