-
1. What is the capitalized value of an infinite series of $1,000 biennial (every two years) payments? MARR = 10%
$4,950
$4,025
. $4,350
$4,762
-
2. Frank West Enterpises would like to choose one of the following plans for expansion. Using present worth analysis, find the best alternative among the three alternatives and the Do-Nothing.
Alt. A
Alt. B
Alt. C
Initial Cost
$10,000
$21,000
$8,000
EUAB
$6,000
$7,000
$4,000
EUAC
$1,500
$1,000
$2,000
Salvage Value
$2,000
$8,000
$3,000
Life
2 Years
3 Years
2 Years
Use a MARR of 10%.
. Do-Nothing
. Alt. C
. Alt. B
. Alt. A
3. -
U. S. Engineering is investigating the possibility of acquiring new automated packaging equipment at a cost of $12,000. It is expected that the equipment will have a salvage value of $1,000 at the end of its useful life of 10 years. It is determined by the plant engineering department at the company that the operation and maintenance cost will be $500 in the first year and will gradually increase every year starting year 2 at the rate of $50 until the equipment is retired. Determine the equivalent uniform annual cost (EUAC) if MARR for the company is 10%.
$2,575.95
$2,150.50
. $2,176.60
. $2,184.40
4. -
The cash flows given in table below are for two different alternatives. MARR =10%
Data
C
D
Initial Cost
$20,000
$80,000
Uniform Annual Benefits
$6,000
$10,000
Salvage Value
$5,000
$20,000
Useful Life in years
5
infinity
The equivalent uniform annual worth (EUAW) of alternative N is ______________.
Hint: Assume the Salvage Value is never realized since N is infinite.
$2,000
$1,920
$1,762
$2,476
5.
-
Compare the following plans using a MARR of 6%.
Data
Plan X
Plan Y
Equipment First Cost
$50,000
$75,000
Annual Operation & Maintenance Cost
$3,000
$2,500
Salvage Value
$10,000
$0
Service Life, Years)
25
50
$6,728; $7,255
$7,255; $9,000
. $6,728; $7,728
. $6,728; $8,200
6.
-
LAEP, Inc. wants to evaluate two methods of shipping their products. The following cash flows are associated with each alternative:
Data
Method
A
B
Life (Years)
10
10
First Cost
$700,000
$1,512,000
M&O Cost
$18,000
$9,000
M&O Cost Gradient
$900
$775
Annual Benefit
$154,000
$303,000
Salvage Value
$142,000
$210,000
Annual profits are based on amount of products which can ordinarily be shipped each year as a function of the amount of vehicles or service purchased with the first cost and the M&O costs.
Using a MARR of 15%, calculate the equivalent uniform annual cash flow (EUAB - EUAC) for each alternative.
Determine the most desirable alternative based on the results.
$389.57, $445.90; Method A
$445.90, $389.57; Method A
. $445.90, $389.57; Method B
. $389.57, $445.90; Method B
7. -
For the cash flow diagram shown, which of the following equations properly calculates the uniform equivalent (A)?
n
0
3
6
9
12
15
Dollars
$100
$100
$100
$100
$100
$100
A=100(A/P,i,15)+ 100(A/F,i,3)
A=100(A/P,i,15)
A=100(A/F,i,3) + 100(A/F,i,15)
A=100(A/P,i,3) + 100(A/F,i,3)
8.
-
Three different alternatives shown in table below are being considered by U. S. Engineering systems.
Assume that alternatives X and Z are replaced at the end of their lives.
Data
Alternative X
Alternative Y
Alternative Z
Initial Cost
$6,000
$1,000
$1,500
Uniform Annual Benefits
$810
$125
$ 230
Useful Life in Years
20
infinite
10
MARR
12%
The NPW for alternative
-
1. What is the capitalized value of an infinite series of $1,000 biennial (every two years) payments? MARR = 10%
$4,950
$4,025
. $4,350
$4,762
-
2. Frank West Enterpises would like to choose one of the following plans for expansion. Using present worth analysis, find the best alternative among the three alternatives and the Do-Nothing.
Alt. A
Alt. B
Alt. C
Initial Cost
$10,000
$21,000
$8,000
EUAB
$6,000
$7,000
$4,000
EUAC
$1,500
$1,000
$2,000
Salvage Value
$2,000
$8,000
$3,000
Life
2 Years
3 Years
2 Years
Use a MARR of 10%.
. Do-Nothing
. Alt. C
. Alt. B
. Alt. A
3.-
U. S. Engineering is investigating the possibility of acquiring new automated packaging equipment at a cost of $12,000. It is expected that the equipment will have a salvage value of $1,000 at the end of its useful life of 10 years. It is determined by the plant engineering department at the company that the operation and maintenance cost will be $500 in the first year and will gradually increase every year starting year 2 at the rate of $50 until the equipment is retired. Determine the equivalent uniform annual cost (EUAC) if MARR for the company is 10%.
$2,575.95
$2,150.50
. $2,176.60
. $2,184.40
4.-
The cash flows given in table below are for two different alternatives. MARR =10%
Data
C
D
Initial Cost
$20,000
$80,000
Uniform Annual Benefits
$6,000
$10,000
Salvage Value
$5,000
$20,000
Useful Life in years
5
infinity
The equivalent uniform annual worth (EUAW) of alternative N is ______________.
Hint: Assume the Salvage Value is never realized since N is infinite.
$2,000
$1,920
$1,762
$2,476
5.
-
Compare the following plans using a MARR of 6%.
Data
Plan X
Plan Y
Equipment First Cost
$50,000
$75,000
Annual Operation & Maintenance Cost
$3,000
$2,500
Salvage Value
$10,000
$0
Service Life, Years)
25
50
$6,728; $7,255
$7,255; $9,000
. $6,728; $7,728
. $6,728; $8,200
6.
-
LAEP, Inc. wants to evaluate two methods of shipping their products. The following cash flows are associated with each alternative:
Data
Method
A
B
Life (Years)
10
10
First Cost
$700,000
$1,512,000
M&O Cost
$18,000
$9,000
M&O Cost Gradient
$900
$775
Annual Benefit
$154,000
$303,000
Salvage Value
$142,000
$210,000
Annual profits are based on amount of products which can ordinarily be shipped each year as a function of the amount of vehicles or service purchased with the first cost and the M&O costs.
Using a MARR of 15%, calculate the equivalent uniform annual cash flow (EUAB - EUAC) for each alternative.
Determine the most desirable alternative based on the results.
$389.57, $445.90; Method A
$445.90, $389.57; Method A
. $445.90, $389.57; Method B
. $389.57, $445.90; Method B
-
For the cash flow diagram shown, which of the following equations properly calculates the uniform equivalent (A)?
n
0
3
6
9
12
15
Dollars
$100
$100
$100
$100
$100
$100
8.A=100(A/P,i,15)+ 100(A/F,i,3)
A=100(A/P,i,15)
A=100(A/F,i,3) + 100(A/F,i,15)
A=100(A/P,i,3) + 100(A/F,i,3)
-
Three different alternatives shown in table below are being considered by U. S. Engineering systems.
Assume that alternatives X and Z are replaced at the end of their lives.
Data
Alternative X
Alternative Y
Alternative Z
Initial Cost
$6,000
$1,000
$1,500
Uniform Annual Benefits
$810
$125
$ 230
Useful Life in Years
20
infinite
10
MARR
12%
The NPW for alternative
-
-
-
-
-
-