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Marxian economics: Differentiating the Rate of Profit Equation

  1. Apr 2, 2014 #1
    Marxian economics: Differentiating the "Rate of Profit Equation"

    While watching a recent Youtube video on Marxian crisis theory and the Tendency Of The Rate Of Profit To Fall, I decided to apply calculus to the equations to see what I got.


    The rate of profit is defined as ∏ = S / (C + V)
    where ∏ = rate of profit, S = total surplus value appropriated by the capitalist employer, C = total capital expenditure and V = total labor expenditure (wages and payrolls).

    For those not familiar with Marxian economic theory, the argument is that as production becomes more "capital intensive" (machines taking the place of workers), then it causes the rate of profit to fall over time. Capital intensity is also called the "Organic composition of capital"


    Organic composition of capital:

    O = C/V

    Algebraic mutations:
    C = OV
    V = C/O

    Substituting to express ∏ in terms of O:

    ∏= S / (OV + C/O)

    Now we must take the partial derivative of ∏ with respect to O. I used an online derivative calculator and got:

    ∂∏/∂O = = -S(V – C/O^2) / (OV + C/O)^2

    Next we can substitute the variable O back out of the formula:

    ∂∏/∂(C/V) = -S[V - C/(C^2/V^2)] / [ V(C/V) + C/(C/V)]^2

    And simplify down to:

    ∂∏/∂(C/V) = -S(V – V^2/C) / (C + V)^2

    A few things can be said about this derivative formula:

    1. It is a rational function with the numerator –S(V-V^2/C) and the denominator (C + V)^2.

    2. The sign of the derivative is key: if it’s positive, than a higher organic composition of capital (i.e. more capital-intensive production) leads to a higher profit rate. But if it’s negative, then a higher organic composition leads to a lower profit rate.

    3. Since the denominator is a perfect square regardless of the value of C and V, we can conclude it will always be positive. Hence, the sign of the derivative depends entirely on the sign of the numerator –S(V - V^2/C).

    4. The numerator is a product of the quantities –S and (V – V^2/C). Since we can safely assume that S is almost always positive (unless most companies across the economy are running red ink!), then –S will almost always be a negative value.

    5. This leaves the quantity (V – V^2/C). Its sign must therefore determine the sign of the profit-rate derivative. If it is positive, the derivative is negative. If it is negative, though, it cancels out the negativity of –S and the derivative is positive. We know that V is positive, so the quantity must therefore be positive if V > V^2/C.

    6. Since the falling rate of profit is contingent upon a negative derivative, which is in turn contingent on the positivity of (V – V^2/C), a falling rate of profit must also be contingent upon the inequality V > V^2/C.

    7. Divide both sides by V:
    1 > V/C

    8. Convert back to O:
    1 > 1/O

    9. Multiply both sides by O:
    O > 1.

    10. So we can finally conclude that a falling rate of profit is contingent on O > 1, which essentially means that C > V. So the capitalist production becomes more relatively profitable up to the point where C = V, then becomes less relatively profitable once C > V.

    =============

    Again, this all depends on whether the online differentiator (derivative-calculator.net) actually gave me the correct partial derivative with respect to O, and whether my steps involving O = C/V substitution are actually mathematically kosher.
     
  2. jcsd
  3. Apr 3, 2014 #2

    Mark44

    Staff: Mentor

    This seems flaky to me right from the get-go. What does "rate of profit" mean? A rate of some quantity is always measured with respect to some other quantity. For example, velocity is the rate of change of position, with respect to time.

    In economics, they talk about marginal quantities, which are akin to derivatives, with, say, marginal profit being the change in profit for a dollar (or Euro or whatever) increase in revenue.

    It doesn't help that the explanation of the variables is couched in Marxian jargon; e.g. "total surplus value appropriated by the capitalist employer" and "organic composition of capital.
    I don't see any problems with the mathematics, but I have questions about the initial assumptions (how rate of profit is defined) and the conclusion. In your final point above, you conclude that "a falling rate of profit is contingent upon C > V." In other words, a higher capital expenditure than labor expenditure.

    So what? It seems possible to me to have an increasing rate of profit, with very little actual profit, as well as to have a decreasing rate of profit, with profits pouring in like crazy.

    This analysis seems to me to confuse the ideas of rates of change of quantities with the quantities themselves. Here's a concrete example. Suppose I'm driving my car, and am stopped at a traffic light. When the light turns green, I step on the gas. At that moment my speed (the rate of change of position with respect to time) is very low, but my acceleration (the rate of change of velocity with respect to time) is very high.

    After some time, I am on the freeway, moving along at 70 miles/hour. My acceleration is zero, but I'm whizzing along, covering lots of ground.

    A few minutes later, I see a car ahead of me that has stopped. I press down on the brake pedal. My acceleration is very negative, but my velocity is still close to 70 miles/hour.

    My point is this: a discussion about rates in a vacuum, with no discussion about the ratios that these rates measure seems to me to be an exercise in sophistry.
     
  4. Apr 3, 2014 #3
    You start out with a process with a profit S and capital and labour costs C and V.
    Then the process is changed to get a new C' = OC and V' = V/O. This will keep the product of C' and V' constant. I see no reasonwhy the costs must change this way.
     
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