Calculate the rate of return .

Question

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Answer 1

Question

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

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not use ai please

Not use ai please

Exercise 6 Imagine that you head production of a multinational food processing company. The ongoing uncer- tainty about costs means that you are unsure of the future cost of one of your inputs, x2. Your firm's production function is y = f(x1, x2) = x²x²² The output price p is 1000, x1 = 27, and wx₁ = 60. 1. Suppose the current input price is Wx2 = 50. Solve for the optimal choice of x2. 2. Now suppose that the probability the input price remains 50 is 0.65 and the probability that Wx2 60 is 0.35. Solve for the optimal choice of x2. Round down to the nearest integer. = 3. Finally, suppose the costs do actually rise, i.e., Wx2 = 60. Calculate the difference in profit from the uncertainty in (2) vs. the certainty in (1).

Answer 2

Question

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

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Answer 3

Question

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

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Answer 4

Question

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

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Answer 5

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

Answer 6

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

Answer 7

Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

Answer 8

(a)

Expert Solution

Explanation of Solution

Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=IN1,300,000((1+i)10−1i(1+i)10)=4,970,000((1+i)10−1i(1+i)10)=4,970,0001,300,000((1+i)10−1i(1+i)10)=3.8231$

Substitute the rate of return as 22% by trial and error method in the above Equation.

$((1+0.22)10−10.22(1+0.22)10)=3.8231(7.304631−10.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231$

When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.

$((1+0.228)10−10.228(1+0.228)10)=3.8231(7.798008−10.228(7.798008))=3.8231(6.7980081.777946)=3.82313.8235≈3.8231$

The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.

The interest rate can be calculated through spreadsheet function that given below:

=RATE(10,1300000,-4970000)

The above functions gives the value of 22.8%.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Calculate the rate of return.

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

Answer 13

(b)

To determine

Calculate the rate of return.

Answer 14

(b)

Expert Solution

Explanation of Solution

Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.

Rate of return (i) can be calculated as follows:

$S((1+i)n−1i(1+i)n)=G×PG1,100,000((1+i)10−1i(1+i)10)=113×72,000((1+i)10−1i(1+i)10)=8,136,0001,100,000((1+i)10−1i(1+i)10)=7.3964$

Substitute the rate of return as 6% by trial and error method in the above Equation.

$((1+0.06)10−10.06(1+0.06)10)=7.3964(1.790848−10.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964$

When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.

$((1+0.05896)10−10.05896(1+0.05896)10)=7.3964(1.773354−10.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964$

The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.

The interest rate can be calculated through spreadsheet function that given below:

= RATE(10,1100000,-72000*113)

The above functions gives the value of 5.896%.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 7 - Prob. 1P Ch. 7 - Prob. 2P Ch. 7 - Prob. 3P Ch. 7 - Prob. 4P Ch. 7 - Prob. 5P Ch. 7 - Prob. 6P Ch. 7 - Prob. 7P Ch. 7 - Prob. 8P Ch. 7 - Prob. 9P Ch. 7 - Prob. 10P

Calculate the rate of return .

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Calculate the rate of return.

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Calculate the rate of return.

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Chapter 7 Solutions