Being really good at your job is a good way to achieve impact in general, because your “impact above replacement” is what counts. If a replacement level employee who is barely worth hiring has productivity 100, and the average productivity is 150, the average employee will get 50 impact above replacement. If you do your job 1.67x better than average (250 productivity), you earn 150 impact above replacement, which is triple the average.
Thanks for the nice point, Thomas. Generalising, if the impact is 0 for productivity P_0, and P_av is the productivity of random employees, an employee N times as productive as random employees would be (N*P_av—P_0)/(P_av—P_0) as impactful as random employees. Assuming the cost of employing someone is proportional to their productivity, the cost-effectiveness as a fraction of that of random employees would be (P_av—P_0/N)/(P_av—P_0). So the cost-effectiveness of an infinitely productive employee as a fraction of that of random employees would be P_av/(P_av—P_0) = 1/(1 - P_0/P_av). In this model, super productive employees becoming more proctive would not increase their cost-effectiveness. It would just make them more impactful. For your parameters, employing an infinitely productive employee would be 3 (= 1/(1 − 100⁄150)) times as cost-effective as employing random employees.
Being really good at your job is a good way to achieve impact in general, because your “impact above replacement” is what counts. If a replacement level employee who is barely worth hiring has productivity 100, and the average productivity is 150, the average employee will get 50 impact above replacement. If you do your job 1.67x better than average (250 productivity), you earn 150 impact above replacement, which is triple the average.
Thanks for the nice point, Thomas. Generalising, if the impact is 0 for productivity P_0, and P_av is the productivity of random employees, an employee N times as productive as random employees would be (N*P_av—P_0)/(P_av—P_0) as impactful as random employees. Assuming the cost of employing someone is proportional to their productivity, the cost-effectiveness as a fraction of that of random employees would be (P_av—P_0/N)/(P_av—P_0). So the cost-effectiveness of an infinitely productive employee as a fraction of that of random employees would be P_av/(P_av—P_0) = 1/(1 - P_0/P_av). In this model, super productive employees becoming more proctive would not increase their cost-effectiveness. It would just make them more impactful. For your parameters, employing an infinitely productive employee would be 3 (= 1/(1 − 100⁄150)) times as cost-effective as employing random employees.