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KiRa [710]
3 years ago
13

Jane is the manager of a local bank branch in College Station where he consumes bundles of two commodities x and y. Prices in Co

llege Station are px=1 and py=5. He is offered a transfer to Dallas where prices are px=4 and py=5; Jane is guaranteed a salary in Dallas with which he would be able to buy exactly what he buys in College Station. Jane’s utility function is U(x,y)=xy2 and his income in College Station is $6000. What happens to Janes’s utility if she accepts the transfer (Jane’s utility maximization is always characterized by the tangency rule)?
Business
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

Remain the same

Explanation:

U(x,y) = xy^{2} ......................................................... (1)

ICS = Income in College Station = $6,000

CSpx = Price of x in College Station = 1

CSpy = Price of y in College Station = 5

ID = Income in Dallas = ?

Dpx = Price of x in Dallas = 4

Dpy = Price of y in Dallas = 5

Step 1

Assume that Jane always divides his income in College Station equally into two, i.e. $3,000 each, to buy x and y, the quantities of x and y he can buy in College Station can be calculated by dividing the $3,000 by the prices of x and y. This is calculated as follows:

CSqx = Quantity of x in College Station = $3,000 ÷ 1

         = 3,000 units

CSqy = Quantiy of y in College Station = 3,000 ÷ 5

         = 600 units

Jane's utility in College Station can be calculated by amending equation (1) and substituting 3,000 units for x and 600 units for y as follows:

CSU(CSqx,CSqy) = (CSqx.CSqy^{2})

 CSU(3000,600) = (3000*600^{2})

                           = 3,000 * 360,000  

CSU(3000, 600) = 1,080,000,000 utils .......................... (2)

Step 2

Since Jane is guaranteed a salary in Dallas with which he would be able to buy exactly what he buys in College Station, this implies that the salary in Dallas will make him to be able to buy 3,000 units of good x and 600 units of good which he currently buys in College Station.

Since

CSpx = 1, which is less than Dpx = 4

But

CSpy = 5, is equal to Dpy = 5

We need to calculate how much his Income will increase in Dallas to be able to buy 3,000 units of good x in Dallas given that its price is $4. Therefore, his income will increase by multiplying $4 by 3000 units and deduct $3,000 he was spending in College Station on x as follows:

IID = Increase in Income in Dallas = (3,000 * $4) - $3,000

    = $12,000 - $3,000

     = $9,000

Therefore, ID (Income in Dallas) is the addition of IDD and ICS (Income in College Station) calculated as:

ID = IID + ICS

    = $9,000 + $6,000

    = $15,000

Conclusion

With the ID of $15,000, Jane will be spending $12,000 to buy 3,000 units of good x in Dallas and continue to spend $3,000 to buy 600 units of good y in Dallas.

This will make Jan's utility in Dallas (DU) to be equal to 1,080,000,000 utils as obtained in equation (2) above.

Therefore, Jane's utility will remain the same based on the tangency rule which states that  a consumer will choose a combination of two goods at which an indifference curve is tangent to the budget line, i.e. his income.

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Answer:

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Answer:

Machine I

capitalized cost:  230,271.28

EAC: $ 27,047.58

Machine II

EAC:  $ 27,377.930  

As Machine I cost per year is lower it is better to purchase that one.

Annual deposits to purchase Machine I in 20 years: $ 1,396.770  

return of machine I with savings of 28,000 per year: 10.51%

Explanation:

WE calculate the present worth of each machine and then calculate the equivalent annual cost:

MACHINE 1

Operating cost:

C \times \frac{1-(1+r)^{-time} }{rate} = PV\\  

C 18,000

time 20

rate 0.1

18000 \times \frac{1-(1+0.1)^{-20} }{0.1} = PV\\  

PV $153,244.1470  

Salvage value:

\frac{Maturity}{(1 + rate)^{time} } = PV  

Maturity  $20,000.0000  

time   20.00  

rate  0.1

\frac{20000}{(1 + 0.1)^{20} } = PV  

PV   2,972.87  

Total: -80,000 cost - 153,244.15 annual cost + 2,972.87 salvage value:

Total: 230,271.28

PV \div \frac{1-(1+r)^{-time} }{rate} = C\\  

Present worth  $(230,271.28)

time 20

rate 0.1

-230271.28 \div \frac{1-(1+0.1)^{-20} }{0.1} = C\\  

C -$ 27,047.578  

Fund to purchase in 20 years:

FV \div \frac{(1+r)^{time} -1}{rate} = C\\  

FV  $80,000.00  

time 20

rate 0.1

80000 \div \frac{(1+0.1)^{20} -1}{0.1} = C\\  

C  $ 1,396.770  

IF produce a 28,000 savings:

we must solve using a financial calcualtor for the rate at which the capitalized cost equals 28,000

PV \div \frac{1-(1+r)^{-time} }{rate} = C\\  

PV  $230,271.28  

time 20

rate 0.105126197

230271.28 \div \frac{1-(1+0.105126197287798)^{-20} }{0.105126197287798} = C\\  

C  $ 28,000.000  

rate of 0.105126197 = 10.51%

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100,000 cost

25,000 useful life

15,000 operating cost during 10 years

20,000 for the next 15 years

Present value of the operating cost:

C \times \frac{1-(1+r)^{-time} }{rate} = PV\\  

C 15,000

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rate 0.1

15000 \times \frac{1-(1+0.1)^{-10} }{0.1} = PV\\  

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in the timeline this is at the end of the 10th year we must discount as lump sum for the other ten years:

\frac{Maturity}{(1 + rate)^{time} } = PV  

Maturity  $152,121.5901  

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rate  0.1

\frac{152121.590126167}{(1 + 0.1)^{10} } = PV  

PV   58,649.46  

salvage value

\frac{Maturity}{(1 + rate)^{time} } = PV  

Maturity  $25,000.0000  

time   25.00  

rate  0.1

\frac{25000}{(1 + 0.1)^{25} } = PV  

PV   2,307.40  

Total cost: 100,000 + 92,168.51 + 58,649.46 - 2,307.40 = $248,510.57

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PV  $248,510.57  

time 25

rate 0.1

248510.57 \div \frac{1-(1+0.1)^{-25} }{0.1} = C\\  

C  $ 27,377.930  

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