Calculate the initial cost.

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Answer

M” method Given the following model, solve by the method of “M”. (see image)

As indicated in the attached image, U.S. earnings for high- and low-skill workers as measured by educational attainment began diverging in the 1980s. The remaining questions in this problem set use the model for the labor market developed in class to walk through potential explanations for this trend. 1. Assume that there are just two types of workers, low- and high-skill. As a result, there are two labor markets: supply and demand for low-skill workers and supply and demand for high-skill workers. Using two carefully drawn labor-market figures, show that an increase in the demand for high skill workers can explain an increase in the relative wage of high-skill workers. 2. Using the same assumptions as in the previous question, use two carefully drawn labor-market figures to show that an increase in the supply of low-skill workers can explain an increase in the relative wage of high-skill workers.

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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