School bus: A scenario.

[Dakini]

Start some extracurricular activities. One group can be picked up earlier in the morning arriving before school starts and have some time doing band (or some sports) and the other group can be dropped off later in the afternoon and spend their time doing band (or whatever) while they wait at the end of the day.

[New Korongo]

Here is the solution:
Make them get off there fat asses and walk to school

[IL Ruffino]

Before making a decision, I need to know the ethnicity of the children

I like you.

[New Korongo]

Give each child a shotgun and make them fight to get on the bus

[Dazchan]

I'd be working on domesticating the pterodactyls and getting the kids to ride those to and from school.

In the meantime, I'd take the 60 kids from the other side of town on trip #1. Trip #2 would be the 30 kids who live on this side of town, plus the kids from the outskirts.

[Dazchan]

If you don't want to play, no-one's forcing you to post in the thread.

As I said earlier, I messed up the scenario and would have been happy to see the thread die. Towards that end, I won't reply any further to you unless you're going to answer the question I asked. Have a nice day.

I answered your question. It's not my fault you won't accept people thinking outside the tiny box you built. You construct this scenario, thinking that there's only a few select answers. Then, when someone comes along and breaks your scenario with a bit of lateral thought, you throw a hissy fit and leave the thread? No offense, but that seems a little childish to me.

The question was "What is the fairest way to use your school bus?"
"Buy another bus" does not answer that question.

[Geniasis]

I simply give priority to whichever group can ante up the most cash. I let the Free Market Fairy sort out the details.

Also, I take the kid who didn't pay and use him as bait to lure a pterodactyl so I may tame him as my steed.

[Dinner Theatre Actors]

Give them all guns. Make sure they learn gun safety so they don't 'accidentally' hurt each other.

[The Floridian Coast]

what we used to do when i was a kid was have one area picked up first in the morning and taken home first in the afternoon and one group picked up last in the morning and taken home last in the afternoon.

that way no kid has to spend way more time than the others AT school.

That's nice and fair but hardly efficient by any means.

[Zeth Rekia]

The government installs flak cannons along the roads most traveled upon by buses. As flying dinosaurs are a hazard to children's health. Even though ptero's are endangered, they still must be dealt with accordingly.

[Ryadn]

what we used to do when i was a kid was have one area picked up first in the morning and taken home first in the afternoon and one group picked up last in the morning and taken home last in the afternoon.

that way no kid has to spend way more time than the others AT school.

Or just stagger the day.

[Vortiaganica]

Actually, look at the following:

60 kids at four hours. = 15 units of efficiency

30 kids at one hour. = 30 units of efficiency, that is kids delivered per hour.

10 kids at....say, five hours? = 2 units of efficiency.

Yes, unrealistic (10 hours to finish the whole run at the end of the day AND the start), but the hours bit isn't really part of this, more just a secondary element. For all we know, it could be a unit of 10 minutes, but for now I'll just say its hours and we have a 144 hour day. It has nothing to do with the scenario.

HOWEVER, if we add all these units together, we get 47. 47 total units of efficiency. Therefore, we should do the following.

2/47 chance that the 10 kids get delivered first, 30/47 that the 30 kids get delivered first, 15/47 chance that the 60 get delivered first.

Then, you subtract the required number, and redraw.

For example, by a stroke of luck, this is one of the times the 10 kids go home first.

It's a 30/45 (Or 2/3 chance) that the 30 kids go home second, and a 1/3 chance that the 60 go home second.

ALTERNATIVELY, we could say that since the lowest unit is 2 and the highest is 30, we should decide that there is to be a 1/15 chance that the 10 kids go home first, and since the middle is 15, a half chance that THEY should go home first, take the 1/15 that is already included. Therefore 15/30, 13/30, 2/30. I prefer the top solution because this one was only lightly based on the attempt at making a logical answer in my brain that got corrupted on the way here.

[Nobel Hobos]

what we used to do when i was a kid was have one area picked up first in the morning and taken home first in the afternoon and one group picked up last in the morning and taken home last in the afternoon.

that way no kid has to spend way more time than the others AT school.

Or just stagger the day.

If you mean to alternate routes (every second day) I wonder what you'd think of this. So that parents don't have to keep track of a 10-day timetable, use whichever schedule seems marginally fairer on monday wednesday and friday. Use the other schedule on tuesday and thursday.

I think we an agree that knowing when their kid is supposed to get home is important to parents, and some would be too busy to or too stupid to keep track of alternating weekdays?

(And don't you dare say "issue a school calendar". That would cost money, money which is already budgeted for the repaving of the Principal's pterodactyl pad.)

[Nobel Hobos]

Actually, look at the following:

60 kids at four hours. = 15 units of efficiency

30 kids at one hour. = 30 units of efficiency, that is kids delivered per hour.

10 kids at....say, five hours? = 2 units of efficiency.

Yes, unrealistic (10 hours to finish the whole run at the end of the day AND the start), but the hours bit isn't really part of this, more just a secondary element. For all we know, it could be a unit of 10 minutes, but for now I'll just say its hours and we have a 144 hour day. It has nothing to do with the scenario.

HOWEVER, if we add all these units together, we get 47. 47 total units of efficiency. Therefore, we should do the following.

2/47 chance that the 10 kids get delivered first, 30/47 that the 30 kids get delivered first, 15/47 chance that the 60 get delivered first.

Then, you subtract the required number, and redraw.

For example, by a stroke of luck, this is one of the times the 10 kids go home first.

It's a 30/45 (Or 2/3 chance) that the 30 kids go home second, and a 1/3 chance that the 60 go home second.

ALTERNATIVELY, we could say that since the lowest unit is 2 and the highest is 30, we should decide that there is to be a 1/15 chance that the 10 kids go home first, and since the middle is 15, a half chance that THEY should go home first, take the 1/15 that is already included. Therefore 15/30, 13/30, 2/30. I prefer the top solution because this one was only lightly based on the attempt at making a logical answer in my brain that got corrupted on the way here.

I like this! You're addressing the question of fairness without any regard to practicality. Thinking outside the box is a vastly over-rated skill; thinking accurately within constraints is harder.

I must say though, that lotteries are only fair in principle. It is possible that a random selection produces the result "the ten kids get taken home first" in half the cases within a finite time despite how you've weighed the odds. Though the process is fair by design, the actual result might not be.

So dispense with the lottery. Plan a repeating timetable which has 2 days out of 47 in which the smallest group gets taken home first, etc. You agree?

[Vortiaganica]

dispense with the lottery. Plan a repeating timetable which has 2 days out of 47 in which the smallest group gets taken home first, etc. You agree?

That would work.

Although practicality is sort of a non-issue when there are spec-ops attack pterodactyls keeping kids from going home because you can't afford a new bus since all the money goes to the Principal's pterodactyl pad.

[Ryadn]

Or just stagger the day.

If you mean to alternate routes (every second day) I wonder what you'd think of this. So that parents don't have to keep track of a 10-day timetable, use whichever schedule seems marginally fairer on monday wednesday and friday. Use the other schedule on tuesday and thursday.

I think we an agree that knowing when their kid is supposed to get home is important to parents, and some would be too busy to or too stupid to keep track of alternating weekdays?

(And don't you dare say "issue a school calendar". That would cost money, money which is already budgeted for the repaving of the Principal's pterodactyl pad.)

No, just do the early bird/late bird thing where half come in an hour early, then they overlap, then the other half stay an hour late.

[Nobel Hobos]

dispense with the lottery. Plan a repeating timetable which has 2 days out of 47 in which the smallest group gets taken home first, etc. You agree?

That would work.

Although practicality is sort of a non-issue when there are spec-ops attack pterodactyls keeping kids from going home because you can't afford a new bus since all the money goes to the Principal's pterodactyl pad.

Yeah. You can sort of see why I don't start many threads. They've gone worse than this I can tell you.

I assumed that the 30 kids and 60 kids and faraway 10 were in completely different areas, meaning you'd have to return to school anyway to get to the faraway area.

I wish I hadn't fallen over drunk within minutes of the OP, because this is exactly what I meant. I would have edited the OP to make it plainer if I'd seen this post of yours at the time.

The three runs are mutually exclusive ... kids from one area CAN be loaded into the bus and delivered after kids from another area the bus visits first, but it involves driving past the school and there is no advantage gained which wouldn't also be got by leaving the kids at the school and picking them up later.

[Nobel Hobos]

If you mean to alternate routes (every second day) I wonder what you'd think of this. So that parents don't have to keep track of a 10-day timetable, use whichever schedule seems marginally fairer on monday wednesday and friday. Use the other schedule on tuesday and thursday.

I think we an agree that knowing when their kid is supposed to get home is important to parents, and some would be too busy to or too stupid to keep track of alternating weekdays?

(And don't you dare say "issue a school calendar". That would cost money, money which is already budgeted for the repaving of the Principal's pterodactyl pad.)

No, just do the early bird/late bird thing where half come in an hour early, then they overlap, then the other half stay an hour late.

Yes. It has a practical advantage that everyone knows when a kid from a particular area has to be ready to catch the bus, and when they're expected home. And that stays the same for every day.

There might still be some trace of unfairness in it? An hour in the morning is not equivalent to an hour in the afternoon. They're both valuable gaming time lost, but isn't having to get out of bed earlier worse?

[Blitzkrenia]

If one is to take efficiency as one's main concern,

The school is at point P.
The same side of town is region A.
The other side of town is region B.
"Far away" is region C.

The possible (and somewhat sensible) combinations of going home are
a)
1st trip. bring students living in B home. (B)
2nd trip. bring students living in A home and then bring students living in C home. (A-C)
b)
1. A-C
2. B
c)
1. A
2. B
3. C
d)
1. C
2. B
3. A

Term definitions:
NM is the amount of time required to travel from region N to M.
N_AvgDrop is the average waiting time for drop off for students living in N after arrival at region N (assuming return to starting point in region N).
N_AllDrop is the time required drop off all students living in region N after arrival at N (assuming return to starting point in region N).

---

Choice a) requires a gross of 60*PB(=bringing B students to point B) +60*B_AvgDrop(=average time of waiting to be dropped off after arrival at B) + (2*40*PB + 40*B_AllDrop)(=waiting time of A and C students for dropping off and return from first trip) + 40*PA(=bringing A and C students to point A) + 30*A_AvgDrop(=average time of waiting to be dropped off after arrival at A)+ 10*A_AllDrop(=waiting time of C students for dropping off A students) + 10*AC(=bringing C students to point C from A) + 10*C_AvgDrop(=average time of waiting to be dropped off after arrival at C) =

140*PB + 60*B_AvgDrop + 40*B_AllDrop + 40*PA + 30*A_AvgDrop + 10*A_AllDrop + 10*AC + 10*C_AvgDrop hours of waiting.

---

Choice b) requires a gross of 40*PA + 30*A_AvgDrop + 10*A_AllDrop + 10*AC + 10*C_AvgDrop + 60*PA + 60*A_AllDrop + 60*AC + 60*C_AllDrop + 60*PC + 60*PB + 60*B_AvgDrop =

100*PA + 30*A_AvgDrop + 70*A_AllDrop + 70*AC + 10*C_AvgDrop + 60*C_AllDrop + 60*PC + 60*PB + 60*B_AvgDrop hours of waiting.

---

Choice c) requires a gross of 30*PA + 30*A_AvgDrop + 70*PA + 70*A_AllDrop + 70*PA + 60*PB + 60*B_AvgDrop + 10*PB + 10*B_AllDrop + 10*PB + 10*PC + 10*C_AvgDrop=

170*PA + 30*A_AvgDrop + 70*A_AllDrop + 80*PB + 60*B_AvgDrop + 10*B_AllDrop + 10*PC + 10*C_AvgDrop hours of waiting.

---

Choice d) requires a gross of 10*PC + 10*C_AvgDrop + 90*PC + 90*C_AllDrop + 90*PC + 60*PB + 60*B_AvgDrop + 30*PB + 30*B_AllDrop + 30*PB + 30*PA + 30*A_AvgDrop=

190*PC + 10*C_AvgDrop + 90*C_AllDrop + 120*PB + 60*B_AvgDrop + 30*B_AllDrop + 30*PA + 30*A_AvgDrop hours of waiting.

---

Assuming that I did not mess up on the above calculations, which is quite likely, in the scenario, PB = 2PA, and B_AllDrop = 2PB, or PB:PA:B_AllDrop = 1:2:4.

plugging this into the formulas,
Choice a) requires a gross of 480*PA + 30*A_AvgDrop + 10*A_AllDrop + 10*AC + 60*B_AvgDrop + 10*C_AvgDrop hours of waiting.

Choice b) requires a gross of 220*PA + 30*A_AvgDrop + 70*A_AllDrop + 70*AC + 60*B_AvgDrop + 60*PC + 10*C_AvgDrop + 60*C_AllDrop hours of waiting.

Choice c) requires a gross of 370*PA + 30*A_AvgDrop + 70*A_AllDrop + 60*B_AvgDrop + 10*PC + 10*C_AvgDrop hours of waiting.

Choice d) requires a gross of 390*PA + 30*A_AvgDrop + 60*B_AvgDrop + 190*PC + 10*C_AvgDrop + 90*C_AllDrop hours of waiting.

Going to school is similar, except that the sequence is reversed.

The three runs are mutually exclusive ... kids from one area CAN be loaded into the bus and delivered after kids from another area the bus visits first, but it involves driving past the school and there is no advantage gained which wouldn't also be got by leaving the kids at the school and picking them up later.

In that case,
Choice a) requires a gross of 490*PA + 30*A_AvgDrop + 10*A_AllDrop + 60*B_AvgDrop + 10*PC + 10*C_AvgDrop hours of waiting.

Choice b) requires a gross of 290*PA + 30*A_AvgDrop + 70*A_AllDrop + 60*B_AvgDrop + 130*PC + 10*C_AvgDrop + 60*C_AllDrop hours of waiting.

Choice c) and d) stay the same.

In conclusion, it depends - more info is required.

^^;

[Sibirsky]

The ones that live closest to the school walk.
The farthest away get picked up first, middle kids second.
The middle kids get dropped off first, farthest kids last.

This is how my old school did it.

Your old school isn't in the OP. The kids cannot walk.

[Risottia]

I'd get the 60 at home first. So I have to keep only 40 of them in the school. It's easier to watch only 40 kids instead of 60 while they wait for the bus.

If the parents of the 40 don't like their kids waiting so much, the can move their arse and take their kids home by themselves. Considering that 30 of them live near the school anyway! No-good socialist slackers, always waiting for nanny School Administrator to take care of their problems.

[Sibirsky]

Take the 60 kids first. Then the 30+10