## Is the US at Full Employment?

This is a common question. It is important to know if an economy is at full employment. However, the question is wrong. or at least the way it is answered is wrong…

Full employment is based on employment of labor. However, full employment must consider both labor and capital since they are both resources for production. You might see a situation where capital reaches full employment before labor. Actually that is what I always see. Then labor becomes more fully employed as capital stops being employed. That is a normal process for a business cycle.

So is the US at full employment? As I see it, the economy reached full employment in the second half of 2014, when capital utilization was optimized.

The bottom line in the graph is the optimization of capital in the economy. When the line reaches zero, a resource is optimized for profits.

The top line is the optimization of labor. The graph shows that capital is always optimized but labor never is. So in essence, capital always reaches full employment but labor does not.

Still, labor and capital need to be taken together to see full employment. The economy has reached full employment considering labor and capital, but employment of labor can still increase even after full employment of capital. The economy is on the other side of full employment going through a subtle cascading process into a contraction.

The fact that in your model labor is never optimized makes me wonder about the model.

You are taking the derivative of z-vertical = (m + k – am2k2)/L.

Why should the coefficients for m and k both be 1/L? (or for that matter constants?)

Why should your cobra equation be symmetrical in m and k when the costs associated with capital and labor are different in character?

EL – Why is your formula dependent on 0.76 and 1.46?

Did you end up with these modifiers so that the results suited you?

The only difference in the two is that, in the second equation, you are squaring (Capacity Utilization: Total Industry/100).

Why should squaring that one term turn Optimization of Capital into Optimization of Labor?

The idea that we are at full employment in belied by the fact that the Labor Force Participation Rate (ages 25-54) for 2015 was 2% below it’s level in 2008.

Capital might be at full utilization, since our manufacturing output is almost back to its pre-recession peak, but with businesses sitting on $1.9T in cash, I don’t think we can say that we are at full capital employment.

We may however, be at OPTIMAL (not full) capital employment, by which I mean that putting their cash in T-bills earns those businesses the best risk-reward ratio that they think they can get.

EL – I see that your chart has the last data series from July 1, 2016. So you are looking in the rear view mirror by at least 6 months.

A model that looks backward is not of much use. Come up with something that uses real time data and looks forward and you will have accomplished something.

Warren,

Edward is taking partial derivatives of his function for profits in terms of labor and capital utilization. When the derivative wrt capital is zero, then no change to capital will produce higher profits. You can find the “Cobra Equation” he is differentiating here: http://effectivedemand.typepad.com/ed/synopsis-of-the-effective-demand-research.html or in many AB posts.

Having tried to describe it, I see the error of my question. The surface of the cobra has a saddle, but anywhere else the partial derivatives are not zero at the same time. Since the equation maximizes on the boundary, there is actually no reason why any profit increasing trajectory need have a zero wrt capital (or labor) at any time during the cycle. If you accept that the equation is meaningful (which I do not follow) then by geometric inspection, labor and capital cannot both be ‘optimized’.

Arne,

Thanks for answering the question. I have not had access to the internet for a couple days. Your answer is perfect.

Ah, I see the problem. The legend is cut off, so I cannot see the squaring in the first equation.