Brentwood is part of the Nashville-Davidson–Murfreesboro–Franklin consolidated metropolitan area which has an unemployment rate of 9% and a population of 1,666,566 people. Even if you wrongfully assume that the entire population is looking for work it only puts unemployment at roughly 150,000 people. A simple search using a single resource populates a combined 66,349 job positings within 30 days and 25 miles. And thats just 1 resource, within just 25 miles, and within only 30 days. If she’s still unemployed then she is clearly doing something wrong.
How do you know that each of those job postings are not redundant, and each actually represents a real job opening? Multiple resources aren’t likely to help us, since one cannot just add the jobs from different resources, because employers can post the same job at multiple resources.
I do not think using the U-3 of 9% is a good assumption; using the Shadow Stats alternative is the real measure of unemployment which is more like 20%. But if we assume that the labor force participation rate is 50%, and and an unemployment rate of 9%, the unemployed rate is 75,000 people at a given time. So, at a given time, the number of unemployed is slightly more than the number of available jobs. These numbers do not tell us much about difficulty of getting a job.
I’ve been playing around with these equations trying to construct a toy model of unemployment and job openings during the commercial breaks of the Yankee/Red Sox game.
One can approximate the change in job openings:
(1) dA/dt = B - C ; A = job openings, B = jobs created per unit of time (PUT), C = people being hired PUT .
The equation is rather since the number of job openings increase with the number of jobs created and decreases when people are hired to fill those openings.
The change in unemployed:
(2a) dU/dt = F - C
U = number of unemployed; F = people getting fired/laid off PUT
This is rather intuitive, but the real equation is:
(2b) dU/dt = F - C - D
the D parameter represents the number of discouraged workers that leave the work force PUT so the term would have a negative sign. It could have a positive sign if discouraged workers enter the workforce. For simplification, we will ignore people entering the labor force such as high school and college graduates.
Of course, your survey of the labor market only tells us the value of A and U not, dA/dt or dU/dt. “A” could be maintained at a high number if parameter “C” (people hired) is small. Of course, if parameter “A” increases (dA/dt > 0) because B, the job creation parameter, is high, rather than “C” being low, it would be a good thing. If “C” is a low number, then one can expect the period of unemployment to be high (unless they could easily be discouraged and remove themselves from the pool of unemployed as expressed by the “D” parameter in 2b).
Your survey assumes that A is proportional to the number of people being hired which is reasonable and intuitive. This can be expressed as:
(3) kA = C
where k is a proportionality constant of the proportion of job openings by the unemployed filled PUT
Also, we’ll assume that the rate of discouragement is proportional to the number of unemployed:
(4) mU = D
where m is a proportionality constant of the unemployed becoming discouraged PUT
We’ll plug in (3) and (4) for the differential equations (1) and (2b):
(5) dA/dt = B - kA
(6) dU/dt = F - kA – mU
The equilibrium of (5) (dA/dt = 0) is:
(7) B = kA ; B/k = A ; k = B/A
In other words, if the number of job openings is at an equilibrium and the rate of hiring per job opening “k” is low, then the number of created job openings (B) would also be low. In differential equation (6), a low (k) means that the term, -kA, would be small, so, in order to produce a substantial drop, then mU, the discouragement term, would have to act as a sink for the unemployed, but this would be an undesirable policy goal.
Plugging in (7) into (6) gets us:
(8) dU/dt = F – B – mU
Equations (6) and (8) tells us that knowing “A” (or its equilibrium value) alone cannot inform us of the value of dU/dt. A high equilibrium value “A” could be merely an “illusion” created by a very low “k” and consequently a low “B”, reflecting a lack of hiring. But one can derive “B” is one knows what is the value of “k” and “A”. To put it another way, a high “A” does not necessarily mean that the unemployed can easily find jobs. You also implicitly assume that “k” is a high, but where is the evidence for this?
If F and B are constants, then we can solve (8) (the solution is too long to be shown here):
(9) U = (F-B)/m + C[sub]o[/sub]e[sup]-mt[/sup]
where e = euler number; C[sub]o[/sub] is the
constant of integration derived by solving the differential equation.
Over the long run (t= infinity), then U = (F-B)/m because the “C[sub]o[/sub]e[sup]-mt[/sup]” term approaches zero as time goes on. So unemployment is proportional to the rate of job destruction and inversely proportional to discouragement.
Get any extra debt paid off and hunkier down and what out the storm.