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ICE Princess25 [194]
2 years ago
7

4. Builtrite D is considering the purchase of a new machine for $500,000 which would require a $5000 installation charge. Employ

ees would need to go through a brief training session to operate the machine properly which would cost $25,000. In addition, an increase in net working capital of $30,000 would be required. The machine has an expected life of 10 years and due to efficiencies, labor costs are expected to decrease $150,000 annually (before depreciation and taxes). Assume straight-line depreciation, a 34% marginal tax rate and a cost of capital of 15%. Should Builtrite D purchase the machine?
Mathematics
1 answer:
Paraphin [41]2 years ago
3 0

Answer:

Builtrite D should purchase the machine

Step-by-step explanation:

Cash outflow in year zero = $ 500,000 + $ 25,000 ( training cost ) + $ 30,000 ( Net working capital)

Cash outflow in year zero = $ 555,000

Terminal cash flow in year 10  = $ 150,000 + $ 30,000 ( NWC)

Terminal cash flow in year 10  = $ 180,000

Operating cash flow per year = [ Savings - expenses - depreciation ] X ( 1 - tax rate) + depreciation

Net present value = -500,000 + \frac{116,000}{1.15^1} + \frac{116,000}{1.15^2} +...+ \frac{180,000}{1.15^{10}}

The Net present value of purchasing the machine = $32,071.42

Builtrite D should purchase the machine

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On a snow day, Mason created two snowmen in his backyard. Snowman A was built to a height of 51 inches and Snowman B was built t
mr_godi [17]

Answer:

A ( t ) = -4t + 51

B ( t ) = -2t + 29

t < 11 hours ... [ 0 , 11 ]

Step-by-step explanation:

Given:-

- The height of snowman A, Ao = 51 in

- The height of snowman B, Bo = 29 in

Solution:-

- The day Mason made two snowmans ( A and B ) with their respective heights ( A(t) and B(t) ) will be considered as the initial value of the following ordinary differential equation.

- To construct two first order Linear ODEs we will consider the rate of change in heights of each snowman from the following day.

- The rate of change of snowman A's height  ( A ) is:

                           \frac{d h_a}{dt} = -4

- The rate of change of snowman B's height ( B ) is:

                           \frac{d h_b}{dt} = -2

Where,

                   t: The time in hours from the start of melting process.

- We will separate the variables and integrate both of the ODEs as follows:

                            \int {} \, dA=  -4 * \int {} \, dt + c\\\\A ( t ) = -4t + c

                            \int {} \, dB=  -2 * \int {} \, dt + c\\\\B ( t ) = -2t + c

- Evaluate the constant of integration ( c ) for each solution to ODE using the initial values given: A ( 0 ) = Ao = 51 in and B ( 0 ) = Bo = 29 in:

                            A ( 0 ) = -4(0) + c = 51\\\\c = 51

                           B ( 0 ) = -2(0) + c = 29\\\\c = 29

- The solution to the differential equations are as follows:

                          A ( t ) = -4t + 51

                          B ( t ) = -2t + 29

- To determine the time domain over which the snowman A height A ( t ) is greater than snowman B height B ( t ). We will set up an inequality as follows:

 

                              A ( t ) > B ( t )

                          -4t + 51 > -2t + 29

                                  2t < 22

                               t < 11 hours

- The time domain over which snowman A' height is greater than snowman B' height is given by the following notation:

Answer:                     [ 0 , 11 ]

   

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