Calculate the present value.

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Answer 1

Question

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

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Answer 2

Question

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

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Answer 3

Question

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

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Answer 4

Question

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

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Answer 5

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

Answer 6

Chapter 3, Problem 53P

To determine

Calculate the present value.

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21

The present value is $9,921,758.21.

Answer 7

Chapter 3, Problem 53P

To determine

Calculate the present value.

Answer 8

Expert Solution & Answer

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

P=C((1+i)n1−1i(1+i)n1)+(C((1+i)n−n1−1i(1+i)n−n1)−d×1i((1+i)n−n1−1i(1+i)n−n1−n−n1(1+i)n−n1))(1(1+i)n1)=1,800,000((1+0.12)3−10.12(1+0.12)3)+(1,800,000((1+0.12)10−3−10.12(1+0.12)10−3)−30,000×10.12((1+0.12)10−3−10.12(1+0.12)10−3−10−3(1+0.12)10−3))(1(1+0.12)3)=1,800,000(1.404928−10.12(1.404928))+(1,800,000(2.210681−10.12(2.210681))−30,000×8.3333(2.210681−10.12(2.210681)−72.210681))(11.404928)=1,800,000(0.4049280.168591)+(1,800,000(1.2106810.265282)−30,000×8.3333(1.2106810.265282−3.1664451))(0.71178)=1,800,000(2.401836)(+1,800,000(4.563751)−30,000×8.3333(4.563751−3.1664451))(0.71178)=4,323,304.8+(8,214,751.8−30,000×8.3333(1.397306))(0.71178)=4,323,304.8+(8,214,751.8−349,325.1027)(0.71178)=4,323,304.8+5,598,453.4146=9,921,758.21