A little bit of Engineering Econ

[adbutcher]

Since the summer is dragging on I figured I would spice things up with a little bit of engineering economics. What better way to kill the time then doing boring problems? These are problems that I am taking from my text in preparation for my exam and I will post the solutions as I solve them. Enjoy.

A company has a manufacturing division with a normal manufacturing capacity of 500,000 units. Each unit sells for $60. The unit price consists of variable labor and material costs at $35, fixed cost at $15, and profit at $10. During a recession, sales decrease so much that the company on average has only been able to charge $52 per unit. As a result of reduced output, the total fixed cost can be reduced 20% below normal if the plant remains open and 30% below normal if it closes. The variable cost is directly proportional to output. What is the breakeven level of production necessary during a recession to make the company indifferent to remaining open or closing (assuming that money is the only consideration)?

[trickblue]

Originally Posted by adbutcher

Since the summer is dragging on I figured I would spice things up with a little bit of engineering economics. What better way to kill the time then doing boring problems? These are problems that I am taking from my text in preparation for my exam and I will post the solutions as I solve them. Enjoy.

Crap... I hated those break-even problems in college... maybe I'll drag out the old college textbooks... then again... maybe not...

[Yeagermeister]

I'll just look in the back of the book for the answer

[adbutcher]

Originally Posted by trickblue

Crap... I hated those break-even problems in college... maybe I'll drag out the old college textbooks... then again... maybe not...

It is no walk in the park for me either. Business like thingies doesn't mesh with the way engineers are wired. Whoever put this nonsense in the curriculum had a very sick sense of humor. When I propose a project or resolution to a problem, I could care less how much it cost as long as it works. They should just leave the fit, form, and function to the engineers and the cost to the bean counters, lol. Now they screwed up by making us cost justify everything which takes the fun out of engineering.

[adbutcher]

Originally Posted by Yeagermeister

I'll just look in the back of the book for the answer

The back of the book only have the odd problem answers, this is an even number problem.

[Yeagermeister]

Originally Posted by adbutcher

The back of the book only have the odd problem answers, this is an even number problem.

DANG IT!!!!!!!!!!!!!! I hated when they did that

[davidyee]

Originally Posted by adbutcher

Since the summer is dragging on I figured I would spice things up with a little bit of engineering economics. What better way to kill the time then doing boring problems? These are problems that I am taking from my text in preparation for my exam and I will post the solutions as I solve them. Enjoy.

... 625,000 units.

[adbutcher]

No takers? Here is the solution:

If the plant remains open and 'X' is the number of units sold in a recession
Total Cost = Fixed cost + Variable cost
= [(1- 0.20) (15) (500,000)] + [(35) (X)] = $6,000,000 + 35X
Revenue = (52) (X)
Profit (Open) = Revenue - Total Cost = 52(X) - [6,000,000 + 35(X)] = 17(X) - 6,000,000

If the plant closes
Total Cost = Fixed cost + Variable cost
= [(1- 0.30) (15) (500,000) + 0 = $5,250,000
Revenue = 0
Profit (Closes) = Revenue - Total Cost = 0 - 5,250,000 = -$5,250,000

Profit (Open) = Profit (Closes)
17(X) - 6,000,000 = -5,250,000
17(X) = 750,000
X = 44,117.65

At $52 per unit, the company has to s o sell at least 44,118 units to consider remaining open.

[adbutcher]

Originally Posted by davidyee

... 625,000 units.

Close, how did you arive at your solution?

Edit: I forgot the smiley

[adbutcher]

Here is another one:

A company currently has an annual taxable income of $80,000 from its regular business operations. The company is considering expanding its business by adding an extra manufacturing line. The extra line would bring in an extra $100,000 in annual revenues, but would also add an additional $60,000 in annual expenses. A.) What is the incremental tax rate due to the addition of the extra manufacturing line? B.) What will be the company’s marginal tax rate if they add the extra manufacturing line?

[adbutcher]

And another one:

Two alternative machines will produce the same product. The following are relevant data:

Machine A:
First Cost (Capital Investment) = $50,000
Life = 15 years
Terminal Book (Market) Value = $5,000
Annual Receipts = $190,000
Annual Expenses = $150,000

Machine B:
First Cost (Capital Investment) = $35,000
Life = 10 years
Terminal Book (Market) Value = $4,000
Annual Receipts = $180,000
Annual Expenses = $140,000

Perform an after-tax analysis to determine which is the better investment alternative assuming “repeatability” and based on using straight line depreciation, an income tax rate of 30%, and an after-tax minimum attractive rate of return of 10%.

I am about to eat and then take the little one to the aquarium so you all have a little more time with these two. Also for extra credit try this one out.

Develop (derive) the MACRS recovery allowance percentages (depreciation schedule) for a 6 year class with a half-year convention. Assume a double declining balance rate switching to straight line when advantageous. For the percentages, carry the decimals out to four places (example, 0.2593 or you would get some serious truncation errors ).

[Zaxor]

Originally Posted by adbutcher

Since the summer is dragging on I figured I would spice things up with a little bit of engineering economics. What better way to kill the time then doing boring problems? These are problems that I am taking from my text in preparation for my exam and I will post the solutions as I solve them. Enjoy.

44,118 units

[Zaxor]

Originally Posted by adbutcher

And another one:



I am about to eat and then take the little one to the aquarium so you all have a little more time with these two. Also for extra credit try this one out.

Dang man I think I pulled something on the first one...if I did another they may have to call the paddy wagon

[adbutcher]

Problem:
A company currently has an annual taxable income of $80,000 from its regular business operations. The company is considering expanding its business by adding an extra manufacturing line. The extra line would bring in an extra $100,000 in annual revenues, but would also add an additional $60,000 in annual expenses. A.) What is the incremental tax rate due to the addition of the extra manufacturing line? B.) What will be the company’s marginal tax rate if they add the extra manufacturing line?

Solution:
Before Project = $13,750 + 0.39 (X – $75,000) = $13,750 + 0.34 ($80,000 – $75,000) = $15,450.00

After Project Taxable Income = Revenue – Expenses = $180,000 - $60,000 = $120,000.00
After Project = $22,250 + (0.39(X - $100,000) = $22,250 + (0.39($120,000 - $100,000) = $30,050.00

Project = ($120,000 - $80,000) (.15) = $6000.00

Marginal: before ->0.34, after ->0.39, and project ->.15

Average
Before = $15,450 / $80,000 = 0.1931 = 19.31%
After = $30,050 / $120,000 = 0.2504 = 25.04%
Project = $6000 / $40,000 = 0.15 = 15%

Incremental
Extra Revenue = $40,000
Extra Tax = $30,050 - $15,450 = $14,600
Incremental Tax Rate = $14,600 / $40,000 = 0.365 = 36.5%

Note: The effective corporate income tax rate is based on 2001 Corporate Federal Income Tax Rates

Now I am off to the Aquarium.

[adbutcher]

This one was a doozy.

Bonus Problem:

Develop (derive) the MACRS recovery allowance percentages (depreciation schedule) for a 6 year class with a half-year convention. Assume a double declining balance rate switching to straight line when advantageous. For the percentages, carry the decimals out to four places (example, 0.2593 or you would get some serious truncation errors ).

Solution:

Assuming B = 1, N = 6, R = 2/N = 2/6

d1 = (2/6) (B) (1/2)
d1 = (2/6) (1) (1/2) = 0.1667
BV1 = 1 – 0.1667 = 0.8333

d2 = (2/6) (0.8333) = 0.2778
dSL2 = 0.8333 / 5.5 = 0.1515
BV2 = 0.8333 – 0.2778 = 0.5555

d3 = (2/6) (0.5555) = 0.1852
dSL3 = 0.5555 / 4.5 = 0.1234
BV2 = 0.5555 – 0.1852 = 0.3703

d4 = (2/6) (0.3703) = 0.1234
dSL4 = 0.3703 / 3.5 = 0.1058
BV4 = 0.3703 – 0.1234 = 0.2469

d5 = (2/6) (0.2469) = 0.0823
dSL5 = 0.2469 / 2.5 = 0.0988
BV5 = 0.2469 – 0.0988 = 0.1481

BV6 = 0.1481 – 0.0988 = 0.0493

BV7 = 0.0493 – (.5 x 0.0988) = 0.0000