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Daniel [21]
3 years ago
15

Most businesses replace their computers every two to three years. Assume that a computer costs $2,000 and that it fully deprecia

tes in 3 years, at which point it has no resale value and is thrown away.
If the interest rate for financing the equipment is equal to I, show how to compute the minimum annual cash flow that a computer must generate to be worth the purchase.
Business
1 answer:
sineoko [7]3 years ago
3 0

Answer:

$2000=Z/(1+i)^1+Z/(1+i)^2+Z/(1+i)^3

Explanation:

let Z be the annual minimum cash flow

The internal rate of approach can be used here, in other words, the rate of return at which capital outlay of $2000 is equal present values of future cash flows

In year 1, present value of cash =X/discount factor

year 1 PV=Z/(1+i)^1

year 2 PV=Z/(1+i)^2

year 3=Z/(1+i)^3

Hence,

$2000=Z/(1+i)^1+Z/(1+i)^2+Z/(1+i)^3

Solving for Z above would give the minimum annual cash flow that must be generated for the computer to worth the purchase

Assuming i, interest rate on financing is 12%=0.12

Z can be computed thus:

$2000=Z(1/(1+0.12)^1+(1/(1+0.12)^2+(1+0.12)^3)

$2000=Z*3.09497902

Z=$2000/3.09497902

Z=$646.21

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Burger Prince buys top-grade ground beef for $1.00 per pound. A large sign over the entrance guarantees that the meat is fresh d
Gnoma [55]

Answer:

The optimal order quantity is 316 pounds

Explanation:

In order to calculate What daily order quantity is optimal, we have to calculate first The cost of underestimating the demand Cu and cost of overestimating demand Co

Cu = ($0.60 - $0.50)*4 = $0.40

Co = $1 - $0.80 = $0.20

Next we have to calculate the Service Level = Cu / (Cu + Co)

= 0.40 / (0.40 + 0.20)

= 0.40/0.60

= 0.6667

So, Z Value at above service level = 0.430727

Therefore, in order to calculate the Optimal Order quantity, we would have to use the following formula

Optimal Order quantity= Mean + Z Value × Std Deviation

= 301 + 37 * 0.430727

= 301 + 15.36899

= 316 pounds

7 0
2 years ago
Wilson Co. has three segments -- Tennis, Golf, and Fishing. The Tennis segment is currently producing 2,000 units annually. The
frutty [35]

Answer:

$400,000

Explanation:

Calculation to determine the differential revenue if Wilson Co. were to eliminate the Tennis segment

Differential revenue= $200x2,000 units

Differential revenue= $400,000

Therefore the differential revenue if Wilson Co. were to eliminate the Tennis segment will be $400,000

6 0
2 years ago
During the "mass production" era, operations management focused primarily on:
fiasKO [112]
The answer to your question is: "<span>Internal production"

hope this helped :)</span>
5 0
3 years ago
g Liabilities of the commercial banking system include Question 69 options: A) deposits. B) loans and deposits. C) reserves and
netineya [11]

Answer:

A) deposits

Explanation:

In the case of the commercial banking system, the  liabilities is deposits as the deposit is the amount of the depositors

So as per the given situation, the option A is correct as the deposits represents the commercial banking liabilities

hence, all the other options are incorrect

Therefore, the same is to be considered

8 0
2 years ago
Agan Interior Design provides home and office decorating assistance to its customers. In normal operation, an average of 2.5 cus
Phantasy [73]

Answer:

A) Single-server single-phase model (M/M/1).

\lambda=2.5 \,customers/hour\\\\\mu=6\,customers/hour

B) The goal is not met, as the average time waiting for service is 5.56 minutes.

C) The new mean service rate is 7.5 customers/hour.

In this case, the average time waiting for service is 4 minutes, so the goal is met.

Explanation:

A) This situation can be modeled as a single-server single-phase model (M/M/1).

The mean arrival rate is 2.5 customers per hour.

\lambda=2.5 \,customer/h

The mean service rate is 6 customers per hour, calculated as:

\mu=\frac{60\, min/h}{10 \,min/customer}=6\, customer/h

B) The average waiting time for a customer can be expressed as:

W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda}  =\frac{2.5}{6}\frac{1}{6-2.5} =0.417*0.222=0.093\,hours\\\\W_q=0.093\,hours*(60min/h)=5.56 \,min

The average waiting time is 5.56 minutes, so it is more than the goal of 5 minutes.

C) If the average time spent per customer to 8 minutes, the mean service rate becomes

\mu=\frac{60\, min/h}{8 \,min/customer}=7.5\, customer/h

An the average waiting time for the service now becomes:

W_q=\frac{\lambda}{\mu}\frac{1}{\mu-\lambda}  =\frac{2.5}{7.5}\frac{1}{7.5-2.5} =0.333*0.2=0.067\,hours\\\\W_q=0.067\,hours*(60min/h)=4 \,min

The average time is now 4 minutes, so the goal is achieved.

6 0
2 years ago
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