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gladu [14]
3 years ago
5

Now suppose firm x undertakes a process innovation that reduces its marginal cost of production from $20 to $15. the fixed cost

of undertaking this process innovation is f > 0. is this a tough commitment or a soft commitment? why? for what values of f would it makes sense for firm x to make this commitment?
Business
1 answer:
baherus [9]3 years ago
7 0

Answer:

(A). It is a tough commitment.

(B). when the value of f is less or equal to 198.

Explanation:

So, we are given from the question above that;

=> there is reduction in the marginal cost of production from $20 to $15.

=> Also, the fixed cost of undertaking this process innovation is f > 0.

Recall that the demand functions is given as;

Qx = 80 – 2Px + Py. ------------------(**).

Qy = 80 – 2Py + Px a).---------------------(***).

Hence, if we Differentiate πx with respect to Ax we will have that;

(Px - 15) × (-2) + ( 80 - 2Px + Py) = 0. ----------------------------------------------------------(1).

=> 110 + Py = 4 × Px.

Solving for Py and Px using the Bertrand equilibrium gives;

Px = 37.3 and Py = 39.3.

If we slot in the values of Py and Px above into the demand function in equation (**) and (***) we will have;

Qx = 44.7 and Qy = 38.7.

Before innovation = (40 - 20) × 40. = 800

Then, (39.3 - 20) × 38.7 = 747.3 is the after innovation.

(37.3 - 15) × 44.7 - f.

=> 997.5 - f (after innovation).

The values of f would it makes sense for firm x to make this commitment when;

after innovation > before innovation.

997.5 - f > 800.

=>f is less or equal to 198.

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You are planning to save for retirement over the next 25 years. To do this, you will invest $700 per month in a stock account an
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withdraw each month is $6,902.37

Explanation:

given data

time = 25 year

invest = $700 per month

stock amount = $300 per month

expected rate = 9% = \frac{0.09}{12}

bond account = 5%

return =  6%

to find out

withdraw each month from account for 20 year withdrawal period

solution

we will apply here future value formula that is

FV = P \frac{(1+r)^t -1}{r}      ...............1

here P is principal amount i.e $700 given and r is are and t is time

so

The value of the stock account at retirement will be

value of the stock account =  700 \frac{(1+\frac{0.09}{12})^{25*12} -1}{\frac{0.09}{12}}  

value of the stock account = $784,785.36

and

value of the bond account at retirement will be

value of the bond account =  300 \frac{(1+\frac{0.05}{12})^{25*12} -1}{\frac{0.05}{12}}  

value of the bond account = $178,652.91

and

so  value of the two accounts combined is here

= $178,652.91+$784,785.36    = $963,438.27

so

monthly withdrawal from combined account is

amount = \frac{Pv}{\frac{1- \frac{1}{(1+r)^t}}{r} }      ...............2

amount = \frac{963438.27}{\frac{1- \frac{1}{(1+\frac{0.06}{12})^{20*12}}}{\frac{0.06}{12}} }  

amount =  $6,902.37

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