Calculate the equivalent annual worth.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

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 2

Question

Chapter 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

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 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

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 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

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 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

Answer 6

Chapter 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

Answer 7

Chapter 6, Problem 17P

(a)

To determine

Calculate the equivalent annual worth.

Answer 8

(a)

Expert Solution

Explanation of Solution

Model G: First cost (C) is $36,000. Operating cost (A) is $4,000. Salvage value (SV) is $15,000. Time period (n) is 3 years. Interest rate (i) is 15%.

Model F: First cost (C) is $32,000. Operating cost (A) is $3,100. Salvage value (SV) is $15,000. Time period (n) is 4 years. Interest rate (i) is 15%.

The equivalent annual worth of Model G (AWG) can be calculated as follows:

AWG=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−36,000(0.15(1+0.15)3(1+0.15)3−1)−4,000+15,000(0.15(1+0.15)3−1)=−36,000(0.15(1.520875)1.520875−1)−4,000+15,000(0.151.520875−1)=−36,000(0.2281310.520875)−4,000+15,000(0.150.520875)=−36,000(0.437976)−4,000+15,000(0.287977)=−15,767.136−4,000+4,319.655=−15,447.48

The annual worth of Model G is -$15,447.48.

The equivalent annual worth of Model F (AWF) can be calculated as follows:

AWF=−C(i(1+i)n(1+i)n−1)−A+SV(i(1+i)n−1)=−32,000(0.15(1+0.15)4(1+0.15)4−1)−3,100+15,000(0.15(1+0.15)4−1)=−32,000(0.15(1.749006)1.749006−1)−3,100+15,000(0.151.749006−1)=−32,000(0.2623510.749006)−3,100+15,000(0.150.749006)=−32,000(0.350266)−3,100+15,000(0.200265)=−11,208.512−3,100+3,003.975=−11,304.54

The annual worth of Model F is -$11,304.54. Since the annual worth of Model F is greater than Model G, select Model F.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Calculate the present worth.

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41

The annual worth of Model F is -$61,277.41.

Answer 13

(b)

To determine

Calculate the present worth.

Answer 14

(b)

Expert Solution

Explanation of Solution

The time period to calculate the present worth is 12 years (4×3). The present worth of Model G (PG) can be calculated as follows:

PG=AWG((1+i)n−1i(1+i)n)=−15,447.48((1+0.15)12−10.15(1+0.15)12)=−15,447.48(5.35025−10.15(5.35025))=−15,447.48(4.350250.80254)=−15,447.48(5.420602)=−83,734.64

The present worth of Model G is -$83,734.64.

The present worth of Model F (PF) can be calculated as follows:

PF=AWF((1+i)n−1i(1+i)n)=−11,304.54((1+0.15)12−10.15(1+0.15)12)=−11,304.54(5.35025−10.15(5.35025))=−11,304.54(4.350250.80254)=−11,304.54(5.420602)=−61,277.41