Practice Exam 1 Solutions PDF

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam

IE 343 Exam-1 Practice Exam Time: 50 minutes Write your printed name in the spaces provided above on every page. Show all of your work in the spaces provided. Interest rate tables are provided for you to use in questions that require numerical answers. For problems requiring expressions as answers, carry your solution to the point where the equation for each problem is specified. For example, 1,000 (P/A, 4%, 5) + 2,500 (P/F, 4%, 5) – 4,000 . If the question asks you to decompose the cash flow into Basic Components, you can only decompose the cash flow into the Five Basic Components: 1) Single Cash Flow, 2) Annuity, 3) Deferred Annuity, 4) Uniform (Arithmetic) Gradient Series, 5) Deferred Uniform (Arithmetic) Gradient Series. Exam 1 has totally 7 Problems with 105 Points: • Q: 1-6 make up 100 points • Q-7 is bonus problem for 5 points.

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam 1) Please answer the following T/F questions: (T/F) (2 points each): a) For engineering economic problems, the viewpoint should be changed while solving the problem. (T/F): FALSE b) Performance monitoring and post evaluation of results is part of the feedback loop in engineering economic analysis procedure. (T/F) TRUE c) A linear cost relationship with size (i.e., X = 1) will usually result less costly for a manufacturer than increasing economy of scale (T/F) TRUE d) A learning rate of 40% will result in more reduction in labor hours than a learning rate of 60%. (T/F). TRUE e) For i = 10%

P (at time 0) = 10,000 (P/A, 10%, 5) + 2,000 (P/G,10%, 4)

T/F. FALSE

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam 2) The purchase price of a natural gas-fired commercial boiler (capacity X) WAS $181,000 eight years ago. Another boiler of the same basic design, except with capacity 1.42X, is currently being considered for purchase. If it is purchased, some optional features presently costing $28,000 would be added for your application. If the cost index was 162 for this type of equipment when the capacity X boiler was purchased and is 221 now, and the applicable cost capacity factor is 0.8, what is your estimate of the purchase price of the new boiler? (20 points) ANSWER: CA = cost of new boiler, CB = cost of old boiler today, SA = X, SB = 1.42X, x (cost capacity factor) = .8, In = 221, Ik = 162 Using Cost Indexing technique, we calculate the current cost of the boiler (with capacity X). Cn = (221/162)*$181,000 = $246, 920 (8 points) Now, using power-sizing technique, we calculate the current cost of the boiler (with capacity 1.42X) CA = $246,920(1.42X/X)0.8 = $326,879 (8points) Final cost of the boiler = CA + cost of optional features: New boiler + optional features = $326,879 + $28,000 = $354,879

(4 points)

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam

3) You are analyzing a project with 5-year life. The project requires a capital investment of $50,000 now, and it will generate uniform annual revenue of $6,000 at the end of each year. Further, the project will have a salvage value of $4,500 at the end of the fifth year and it will require $3,000 each year for the operation. (20 points total) a) Develop the cash-flow diagram for this project from the investor’s viewpoint. (5 points) $6,000

$6,000

$6,000

$4500

$6,000 $6,000

1

2

$3,000

4

3

$3,000

$3,000

5

$3,000

$3,000

$50,000 $7,500

0

$3,000

$3,000

$3,000

$3,000

1

2

3

4

5

$50,000

b) Assuming an interest rate of 10% per year, what is the future equivalent value (at time 7) of the Cash Flow you created in part (a). Just write down the expression like “e.g. ฀฀0 = 2000(P/F, 4%, 10) + 3000 (P/A, 4%, 5) – 4,000 ”. You don’t need to calculate the final numerical answer. (10 points) ANSWER: From the above cash flow diagram in part (a), the future equivalent value at year 5 is:

Now, we calculate the future equivalent value at year 7.

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam

c) Solve the equation in part (b) and calculate the final numerical answer using tables. (5 points)

The future equivalent at year 7 is -$69,828.73

4) The following two cash flows Cash Flow (A) and Cash Flow (B) are economically equivalent. The effective interest rate is 10% per period. Please follow the questions to find out the unknown X. (15 points total)

Cash Flow (A) A = $2,000

0

1

2

3

4

5

6

7

8

$5,000

Cash Flow (B)

0

X/3

X/3

X/3

1

2

3

9

X

X

X

4

5

6

7

8

9

(a) What is the present equivalent value (at time 0) ฀฀฀฀ of Cash Flow (A). Please refer to the interest rate tables for numerical answers. (5 Points) ANSWER: ฀฀฀ ฀ = $฀฀, ฀฀฀฀฀฀(฀฀|฀฀, ฀฀฀฀%, ฀฀) = $฀฀, ฀฀฀฀฀฀(฀฀. ฀฀฀฀฀฀฀฀) = $฀฀฀฀, ฀฀฀฀฀฀

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam (b) Write an expression in terms of the unknown X: What is the present equivalent value ฀฀฀฀ of Cash Flow (B). You are free to decompose Cash Flow (B) further if you want to. Just write down the expression like “e.g. ฀฀฀ ฀ = X (P/F, 4%, 10) + X/2 (P/A, 4%, 5) – 4,000 ”. You don’t need to calculate the final numerical answer. (10 Points) ANSWER: Cash Flow (B) can be decomposed into 3 Basic Components as follows: Standard Annuity

0

X/3

X/3

X/3

1

2

3

4

5

6

7

8

9

7

8

9

Deferred Annuity

0

1

2

3

X

X

X

4

5

6

$5,000

Single Cash Flow

0

1

2

3

4

5

6

7

8

9

฀฀ ฀฀฀ ฀ = (฀฀|฀฀, ฀฀฀฀%, ฀฀) + ฀฀(฀฀|฀฀, ฀฀฀฀%, ฀฀)(฀฀|฀฀, ฀฀฀฀%, ฀฀) + ฀฀, ฀฀฀฀฀฀(฀฀|฀฀, ฀฀฀฀%, ฀฀) ฀฀

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam 5) Consider the following cash flow diagram. Assume the interest rate is 10% per period. (20 points total) $5,000 $4,000 $3,000 $2,000 $1,000

0

1

2

3

4

5

6

7

8

9

10

-$1,000 -$2,000 -$3,000 -$4,000 -$5,000

(a) Decompose the Cash Flow into several Basic Components. (10 Points) ANSWER: The cash flow diagram can be decomposed into the following three Basic Components

Single Cash Flow 0

1

2

3

4

5

6

7

8

9

10

-$5,000

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam Standard Annuity 0

1

2

3

4

5

6

7

8

9

10

A = -$4,000

Standard Uniform Gradient Series $9,000 $8,000

$3,000

$4,000

$5,000

$6,000

$7,000

$2,000 $1,000

0

1

2

3

4

5

6

7

8

9

10

(b) Write an expression: Based on your cash flow decompositions in part (a), what is the annual equivalent value ฀฀ over the 10-year period given an interest rate of 10% per period. Just write down the expression like “e.g. A = 1,000 (A/P, 10%, 10) + 2,500 (A/F, 10%, 10) – 4,000”. You don’t need to calculate the final numerical answer. (10 Points) ANSWER: ฀ ฀ = −฀฀, ฀฀฀฀฀฀(฀฀|฀฀, ฀฀฀฀%, ฀฀฀฀) − ฀฀, ฀฀฀฀฀฀ + ฀฀฀฀฀฀฀฀(฀฀|฀฀, ฀฀฀฀%, ฀฀฀฀).

6) Suppose you took out a bank loan of $25,000 at an annual interest rate of 12% compounded monthly. The loan is repayable over a period of 5 years. (15 points total) 8

NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam a) What is your monthly payment? Just write down the expression like “e.g. A = 1,000 (A/P, 10%, 10) + 2,500 (A/F, 10%, 10) – 4,000”. You don’t need to calculate the final numerical answer. (8 points) ANSWER: Monthly payment should match with monthly effective interest rate. The effective monthly interest rate = 12% / 12 = 1% per interest period. Here the interest period is monthly. For 5 years of repayable period, we need to make 5*12 = 60 total payments, so A=$25,000(A/P,1%,60)

b) Monthly payments are made at the end of every month. After making 40 such payments, you could pay a lump sum now (right after the 40th payment) to close out the loan. How much do you need to pay? Just write down the expression like “e.g. F = 1,000 (A/F, 10%, 10) + 2,500 (A/F, 10%, 10) – 4,000”. You don’t need to calculate the final numerical answer. (7 points) ANSWER: We need to calculate the present equivalent value of the last 20 payments at time end of period 40, which is P(at End of Period 40) = (Annuity from part (a))(P/A,1%,20) = 25,000(A/P,1%,60)(P/A,1%,20) Hence the required lump sum amount after 40th payment is $25,000(A/P,1%,60)(P/A,1%,20)

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam 1 Bonus Problem, 5 Points 7) What is the accumulated value after 2 years of $500 monthly payments with nominal interest rate of 12% compounded semi-annually? Just write down the expression like “e.g. F = 1,000 (A/F, 10%, 10) + 2,500 (A/F, 10%, 10) – 4,000”. You don’t need to calculate the final numerical answer (5 points) ANSWER: Since the compounding frequency is less than the payment frequency, we can add up the payments till we reach a compounding period. 6 month duration 0

6 month duration

6 month duration

1

6 month duration

2

A = $500 Compounding Period

In the above figure, the crosses represent the compounding, which happens after every 6 months. The arrows represent monthly payment of $500. Since, this is a case where compounding frequency is less than the payment frequency, we can define a new compounding period and a corresponding interest rate for the same. New compounding period= 6 months = twice in 1 year. (Shown by the dashed rectangle in the figure) Interest rate per compounding period = 12% / 2 = 6% (It makes sense, because we have an interest rate of 12% yearly, which would become 6% for every 6-months). New Annuity: We add up the payments till we reach the first compounding (or cross in the above figure) = $500 *6 = $3000 N= Number of payments of new annuity = 4.

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NAME (Printed): _________________,________________ (Last) (First) Fall 2019 – IE 343 Exam-1- Practice Exam

Hence, we have the new cash flow diagram as shown below:

6 month duration

6 month duration

6 month duration

6 month duration

0

1

2

3

4

Annuity = $500*6= $3,000

Hence the accumulated value after 2 years is $3000(F/A, 6%,4)

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