Cost comparison calculation

the cost comparison calculation is a static procedure of the investment calculation and serves for the comparison of several investment alternatives. Here the total costs of the alternatives are determined and the most economical is selected.

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Calculation

the total costs result from the fixed and the variable costs. Since the cost comparison calculation regards the average costs of one period, the acquisition disbursement must be considered accordingly within the fixed costs. These capital costs result from the calculatory writings-off and the calculatory interest. Those Total costs are thus defined as follows:

[itex] K_ {entirely} = K_ {f} + k_ {v} \ cdot x + \ begin {matrix} \ underbrace {\ frac {I_ {0} - L_ {T}} {n}} \ \ {} ^ {\ rm lime. Writing-off} \ \ [- 4.0ex] \ end {to matrix} + \ begin {matrix} \ underbrace {\ frac {I_ {0} + L_ {T}} {2} \ cdot i} \ \ {} ^ {\ rm lime. Interest} \ \ [- 4.0ex] \ end {to matrix}< /math>

Are:

• [itex] K_ {f}< /math> -The fixed total costs
• [itex] k_ {v}< /math> - The average variable unit cost prices
• [itex] x< /math> - On the average set off/produced quantity
• [itex] I_ {0}< /math> - Initial costs/amount of investment at the time 0
• [itex] L_ {T}< /math> - The liquidation proceeds/residual value at the end of the service life
• [itex] n< /math> - Prospective service life; this can be taken appropriate depr. - tables
• [itex] from i< /math> - That Example
within an enterprise

the following data for two excluding investment alternatives are given calculation interest rate [work on]. It is to be selected with the help of the cost comparison calculation one of the two alternatives (the paragraph is secured for the next 8 years - for this reason).

Plant I Plant II
data of the plants
acquisition value (€) 80,000, - 120,000, -
service life 8 years 8 years
capacity (year) 15,000 LE 15,000 LE
extent of utilization (year) 10,000 LE 10,000 LE
calculation interest rate (year) 10% 10%
fixes to costs (€ /Jahr) 1,000, - 1,700, -
variable costs (€ /Jahr)
wages and ancillary wages 16,000, - 8,000, -
Tools, fuels among other things3.800, - 4,000, -
energy and other variable costs 1,900, - 2,700, -

(LE = unit of power)

calculation for plant I

< math> \ begin {matrix} \ mbox {writings-off} &=& \ frac {80,000, 00 EUR - 0.00 EUR} {8 years} = 10,000, 00 \ frac {EUR} {year} \ \ && \ \

\ mbox {calculatory interest} &=& \ frac {80,000, 00 EUR + 0.00 EUR} {2}\ 10 \ frac {\ %} {year cdot} = 4,000, 00 \ frac {EUR} {year} \ \ && \ \

k_ {v} &=& \ frac {16,000, 00 \ frac {EUR} {year} + 3,800, 00 \ frac {EUR} {year} + 1,900, 00 \ frac {EUR} {year}} {10,000, 00 \ frac {LE} {year}} = 2.17 \ frac {EUR} {LE} \ end {to matrix} [/itex]

[itex] \ begin {matrix} K_ {entirely} &=& 1,000, 00 \ frac {EUR} {year} + 2.17 \ frac {EUR} {LE} \ cdot 10,000, 00 LE + 10,000, 00 \ frac {EUR} {year} + 4,000, 00 \ frac {EUR} {year} \ \ && \ \

```          &=& 36,700, 00 \ frac {EUR} {year}
```

\ end {to matrix} [/itex]

calculation for plant II

< math> \ begin {matrix} \ mbox {writings-off} &=& \ frac {120,000, 00 EUR - 0.00 EUR} {8 years} = 15,000, 00 \ frac {EUR} {year} \ \ && \ \

\ mbox {calculatory interest} &=& \ frac {120,000, 00 EUR + 0.00 EUR} {2} \ \ frac {\ %} {year cdot} = 6,000, 00 \ frac {EUR} {year} \ \ && \ \

k_ {v} &=& \ frac {8,000, 00 \ frac {EUR} {year} + 4,000, 00 \ frac {EUR} {year} +2.700, 00 \ frac {EUR} {year}} {10,000, 00 \ frac {LE} {year}} = 1.47 \ frac {EUR} {LE} \ \ \ end {to matrix} [/itex]

[itex] \ begin {matrix} K_ {entirely} &=& 1,700, 00 \ frac {EUR} {year} + 1.47 \ frac {EUR} {LE} \ cdot 10,000, 00 LE + 15,000, 00 \ frac {EUR} {year} + 6,000, 00 \ frac {EUR} {year} \ \ && \ \ &=& 37,400, 00 \ frac {EUR} {year} \ end {to matrix} [/itex]

For the given input data the plant I is to be preferred. One sees however that thosePlant II substantially lower variable unit cost prices possesses. Here it is interesting to know starting from which number of items plant II is prefer worthy. This critical number of items can be determined by simply equating of the functions.

[itex] K_ {f1} + k_ {v1} \ cdot n = K_ {f2} + k_ {v2} \ cdot n< /math>

[itex] n= \ frac {K_ {f2} - K_ {f1}} {k_ {v1} - k_ {v2}}< /math>

[itex] n = \ frac {22,700, 00 \ frac {EUR} {year} - 15,000, 00 \ frac {EUR} {year}} {2.17 \ frac {EUR} {LE} - 1.47 \ frac {EUR} {LE}}< /math>

[itex] n = 11,000 \ frac {LE} {year}< /math>

Starting from a number of items of 11.000 units of power per year plant II of the plant I is to be preferred purely cost-related. Whether higher production however at the marketto be set off and also a higher profit does not bring can the cost comparison calculation knows not answer. Here however the profit comparative calculation helps .

with it

the alternatives to be economically meaningfully compared, must they can meet criticism the following acceptance:

• There here only the costs, the yields are compared must be alike with all alternatives. i.e.:
• the yields per period must be alike.
• the service life of the different alternatives must be alike.
• For the same reason no general Vorteilhaftigkeitsaussage is possible.
• By the consideration of only one period the statement is generalto see critically
• this investment counting method presupposes safe expectations .

During achievement-economical investments the application of the cost comparison calculation is meaningful during the spare replacements and rationalization decisions.

As static procedure of the investment calculation the cost comparison calculation counts their application for the “auxiliary methods in such a way specified of practice”,

• only in few cases really meaningfullyand
• with complex decisions with varying deposit surplus is

unsuitable, because they, contrary to the dynamic procedures, do not consider the Zeitverlauf.