Ranking Task in a uniform manner

only looking for more information of blocks 1-3 and stopping at the reference1. Everything else is just extra information in case people do not understand what I am asking, which has been the case. This is the best I can explain, and really hoping someone can lead me even a nudge in the right direction to move forward.

the first block is the gist:
The last block of info is my attempt and logic : the full scope of how I can explain the scenario.
The blocks above that are my attempt to shorthand the quesstions:

determined by current job ( min wage @ 8.00 units/hr): paid rate for 30 minutes : 400 pennies / 1800 seconds == .22 pennies/sec.

determined by a task's rate of outcome/time : Self paid rate for 30 minutes : 2000 pennies / 1800 seconds == 1.11 pennies / sec. in theory,

I am trying to both RANK a task, and also decide how different tools can essentially alter that rank. By "normalizing" the task, the rank should apply to tasks which take different amounts of time and outcomes, but nevertheless boil down to a x cents/second rate.

Am I correct in saying that the ratio is not 1 / 5 , but rather, I should 1.11 c/s - .22 c/s, for an ACTUAL ratio of .89c/s / .22c/s.
How do i express the rate of the task that I did, as compared to the ‘baseline’ minimum wage task?
Is it fair to say the task is (.9/.2), or 4.5 greater than the baseline? if thats so, imagine the below scenario.
How do i use that number (4.5) to compare it to guy 1/guy 2 in the reference1 block. ;================================================================ below solution in reference1 is that guy 1 needs a rate of 12.5 cents/sec to catch guy2.
how does the task I did rank compared? since my task rate is 1.10 c/s and i needed 12.5 c/s AVERAGE to reach my yearly goal, ranking a task is as simple as 1.1/12.5?

in this context, this task could be given a single number score of .08 , and this number is effective at communicating the relative worth of task as I do more and more experiments and gain more and more data on different task’s, etc?

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whatever my task ranks as compared to the max or minimum, I want to use that number to quantify what is saved by upgrading from the standard method to a tool that makes the task faster.

for ex. if originally the rate is 1.10 cents/s , and a drill accomplishes the task in 80 seconds, compared to 120 seconds via screwdriver manually, a drill applies a multiplier of ⅔, or .66 as compared to the 'baseline'….

1.10 * .66 + 1.10 = 1.826 c/s

is it fair to say that a drill will result in 1.826 c/s /1.10 c/s? if i translate this its 1.66. therefore, if the baseline "rank" is .08, then the addition of a drill that increases efficiency by a factor of .33, or .66 % LESS TIME, or 1/3 (not entirely sure of the format to use)

then: .08 x 1.166 + .08, or Y = (Y x 1.66) + Y == 0.1328 + .08 ----->.2128 << the new "score" ;================================================================
reference 1:
Guy1: $18,000 annually from employment (8 hours per day, seven days a week) Spends 6 hours of free time every day on hobbies, earning X dollars per hour

Guy4: $1,000,000 annually from employment (14.4 hours per day, seven days a week) Spends 0 hours of free time every day on hobbies, earning nothing.

Determine rate X such that Guy1's total income matches Guy4's total income.

18000 + (6)(365)(X) = 1000000 18000 + 2190X = 1000000 Solve for X

(1) Subtract 18000 from both sides
(2) Divide both sides by 2190 X = (1000000 - 18000) / 2190 X = $448.401826 per hour (rounded)

In other words, Guy1 would need to earn about 12.45561 pennies per second from working on hobbies 6 hours every day, in order for his total annual income to match Guy4's $1,000,000.

CHECK: Hobby income for Guy1 12.45561 cents/second × 60 second/minute × 60 minute/hour × 6 hour/day × 365 day = 98,200,029 cents per year

98200029 cents = 982000.29 dollars $982,000.29 + $18,000 = $1,000,000.29 The extra 29 cents comes from rounding X to only six decimal places. If we were to retain additional decimal digits, that error would disappear ;================================================================ ;================================================================  ;================================================================ ;================================================================
purpose / setup of problem & my best attempt:

If a task is effective, we sell our time for the upper max, buy it back low as possible as lower max (introduced through the gain of value of doing a particular task), and now we have a data point to justify increasing our worth per second rate.

in the example that I tried, the total time took 30 minutes, for $20 return. thats easy enough to calculate.
However, I purposefully took out the outer 15 screws, 3 different times, using 3 different ‘level of tools.’ i.e. a manual screwdriver, an electric screwdriver, and a power drill/driver

- to glean information as it would relate to someone doing it from a beginner level / an experiment that helps put the task into a “global” perspective.

From the experiment:
for me : total time 1800 seconds,
$20 gain
15 screws by screwdriver took 120 seconds
15 screws by electric driver took 90 seconds.
15 screws by powerdrill took 80 seconds

Although various factors are the cause of the eventual time requirements of the task, I would like to use my experimental value of $20/1800seconds, and for the sake of the argument (math) use those figures as if it were done by the power drill alone, which is a 2 / 3 “multiplier” as compared to the screwdriver method, or .66 times faster.

and then use the $20/1800 figure as if it were done by the screwdriver method.

Because I have decided to get 2 different data points on 2 different tools/methods, I can now form a framework of how the tools, and ultimately the efficiency increase they bring, compare to one another, and also the actual value that the efficiency brings. i.e.

I am reasonably-and-justifyably able to attribute a “stat” to the tools, as they relate to one another, in the scope of a specific task. consider the below example :

If you were to do the same task, using a screwdriver,

you can know: $20/1800 will earn rate of 1.1 pennies/second same task with a power drill,
you can know: if you have a drill : $20/ 1188 (1800 x .66) = 1.7 pennies/second

therefore, I can conclude : Every second that I am using a power drill, it is netting me .6 pennies/second — and this applies to all tasks outside the scope of where I ran my initial experiment (taking apart the toaster oven)

This allows me to repeat the experiment again and again, and by using the same math, develop an average of the results, making the accuracy of the comparison greater over time. and by doing this, I reasonably-and-justifiably give “stats” to specific tools based on how they are comparing to their alternatives.

Now, the Second Time i do the task, lets say I have a drill, and previously didnt: i am using .22 pennies/sec in the value of my time, as corroborated by my employers rate of pay, THEN yielding a result of 1.1 pennies/second for 1800 seconds, (the same rate as before, only using a drill this time, and something else occurred that made it take JUST AS LONG, but nevertheless , I will apply the “multiplier” of the drill ) I need to apply the drill “multiplier” in the form of .6 pennies/second , im assuming, as opposed to simply taking the first time (1800) and multiplying by ( .66 ) as i showed in the first time when deriving the figure. because ULTIMATELY i DID take actual 1800 seconds to do it. and the reason that is super significant is because that 1800 of actual time needs to be used in the accountance, because again ULTIMATELY, it serves as a concrete and static fraction of my available FREE-TIME. 1800 / 10,512,000 == 1 / 5840 5840 instances of gaining a rate of .9 pennies / second for 1800 seconds (accounting for input time? .22 ₵/sec) If I did the task 5840 physical times in a year using the first method to calculate: screwdriver : “add” amount of time since 1.1 rate was rate from using a drill… “adding” the multiplier to the number 1800 (this is a simulated result using a lesser tool than originally used) 1800 - 1188 = 612 seconds to add that which would have been reduced if used drill. 5840 instances of gaining a rate of 1.1 pennies / second for 1800 1188 $76,317 drill:eek:riginal results 5840 instances of gaining a rate of 1.1 pennies / second for 1800 seconds $115,632 if I did the task 5840 times in a year using the second method to calculate: screwdriver : taking away .6 ₵/sec as opposed to adding to the “time” number 5840 instances of gaining a rate of 1.1 pennies / second for 1800 seconds $115,632 drill: effecting the rate of gain using the drills “multiplier” as .6 ₵/sec as opposed to .66 applied to only the time number 5840 instances of gaining a rate of 1.7 pennies / second for 1800 seconds 178,704 My Question is in 3 parts: 1. am I right that I should use the 2nd method of calculation, as the first was only necessary to serve as a starting point, and 2. Since with each completion of a task that is over my current baseline rate of .22 pennies/second, that number that is “worth-per-second” should hypothetically rise, because why would i STILL use my time at a calculated worth of .22 pennies / second after having just establishing a tremendous pattern of using my time to gain 1.1 pennies/second?? I should be averaging each result of every task as to effect the input parameter of “worth-per-second” of the subsequent task. HOW do I do it in this context. How does that change for someone who has a lesser number of FREE TIME than the 10,512,000 that I have, since all of the above is contingent on the idea that the task i did was 1/5840 of available time. e.i. How does it change the math if someone wanted to repeat the process for another set of tools, using a 1800 second task, but having half of that available time, because generally, as time becomes lesser and lesser, in theory the rate should go up as a result. at face value, this is relative and different between person to person, how much there final seconds are worth. EXCEPT this question is posed from the standpoint that only so many slots are left to be filled in their schedule, and there goal to reach necessarily being a set rate (way higher than such task will provide if chosen to engage in), should dictate the persons justification/ set an environmental “factor” that raises the buy AND sell price of their time. Now, by having done all this, I should be able to rank different tasks based on their average time/output ratio since the tasks have been “standardized” or “normalized” so to speak.. not sure if that is the correct definition.