Best choice between option A and B,by calculating the present worth.

Question

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Answer 1

Question

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

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Answer 2

Question

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

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Answer 3

Question

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

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Answer 4

Question

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

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

Answer 5

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

Answer 6

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M

The above formula gives the Present Worth for choice A of −$45.26M

Present Worth for choice B

PWB =−22−3( ( 1+.18 ) 16 −1 .18 ( 1+.18 ) 16 )+7.5(1 ( 1+.18 ) 16 )=−14−3(5.16)+7.5(0.07)=−$36.96M

The above formula gives the Present Worth for choice Bof −$36.96

Therefore, the best choice between A & B is B as it saves $8.30M

Answer 7

Chapter 5, Problem 38P

To determine

To find:Best choice between option A and B,by calculating the present worth.

Answer 8

Expert Solution & Answer

Answer to Problem 38P

The better option is B which would save $8.30M

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

First CostO & M Cost/yearSalvageLifeA$14M$3.5M$2M8 yearsB22M3M7.5M16 years

Formula used:

PWA=−14−A( ( 1+i )N−1i ( 1+i )N)+F(1 ( 1+i )N)− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PWB=−22− A( ( 1+i )N−1i ( 1+i )N)+ F(1 ( 1+i )N)

PW represents Present Worth

A is end of period cash receipt or disbursement in a uniform series continuing for “n” periods

F is Future sum of money at the end of the n^t^h interest period, which is equivalent to “P” with interest rate “i”.

P isPresent sum of money

i is Effective interest rate per interest period

N is Number of interest period

Calculation:

By using the above-mentionedformulas, Present Worth is calculated.

Present Worth for choice A

PWA =( −14−3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+( 2−14 )( 1 ( 1+.18 ) 8 ) − 3.5( ( 1+.18 ) 8 −1 .18 ( 1+.18 ) 8 )+2( 1 ( 1+.18 ) 8 ))=−14−3.5(4.08)−12(0.27)−3.5(4.08)+2(0.27)=−$45.26M