Calculate the initial cost.

Question

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Answer 1

Question

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

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Answer 2

Question

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

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Answer 3

Question

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

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Answer 4

Question

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

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Answer 5

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

Answer 6

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

Answer 7

Chapter 13, Problem 25P

(a):

To determine

Calculate the initial cost.

Answer 8

(a):

Expert Solution

Explanation of Solution

Initial cost of Proposal A (FCA) is $250,000. Time period (n) is 4. Annual maintenance cost (MO) is $3,000. Repainting cost for every 2 year for Proposal B (RP) is $5,000. Time period (n1) is 2 and (n2) is 12. Interest rate (i) is 12%.

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−FCB(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−250,000(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−FCB(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−250,000(0.32923)−3,000=−FCB(0.16144)−5,000(0.4717)+5,000(0.04144)−82,307.5−3,000=−FCB(0.16144)−2,358.5+207.2FCB(0.16144)=85,307.5−2,358.5+207.2FCB=83,156.20.16144=515,090.44

Initial cost of Proposal B is $515,090.14.

Answer 9

(a):

Expert Solution

Answer 10

(a):

Expert Solution

Answer 11

Expert Solution

Answer 12

(b):

To determine

Calculate the initial cost of Proposal A.

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57

Initial cost of Proposal B is $340,671.57.

Answer 13

(b):

To determine

Calculate the initial cost of Proposal A.

Answer 14

(b):

Expert Solution

Explanation of Solution

Initial cost of Proposal B (FCB) can be calculated as follows:

−FCA(i(1+i)n(1+i)n−1)−MO=−700,000(i(1+i)n(1+i)n−1)−RP(i(1+i)n1−1)+RP(i(1+i)n2−1)−FCA(0.12(1+0.12)4(1+0.12)4−1)−3,000=(−700,000(0.12(1+0.12)4(1+0.12)4−1)−5,000(0.12(1+0.12)4−1)+5,000(0.12(1+0.12)4−1))−FCA(0.32923)−3,000=−700,000(0.16144)−5,000(0.4717)+5,000(0.04144)−FCA(0.32923)−3,000=−700,000(0.16144)−2,358.5+207.2−FCA(0.32923)=−113,008−2,358.5+207.2+3,000FCB=−112,159.3−0.32923=340,671.57