Not sure where to start in a large logistics problem

[Tybras]

Homework Statement

You are trying to clean up a mess. This mess extends along a 28 mile path between points A and B. You have 61 workers (including yourself) that can lift upto 150 lbs.

You have up to 4 wagons that can carry a total of 12000 lbs each, and weigh 400 lbs each, pulled buy upto 7 horses. Each wagon needs 1 driver if there are 4 or less horses on it, and 2 drivers if there are 5 to 7 horse. While driving a wagon, a worker cannot carry any weight. Each worker weighs 200 lbs.

Every 2 feet, there is a pile of debris with 722 - 50 lbs pieces. Each worker eats 2 lbs of food a day. Each horse eats 10 lbs of feed a day. Each person has a container that weigh 5 lbs when full, 1 lb when empty) that holds enough water for 2 days. Each horse needs to drink 7 gallons of water a day. A 300 gallon trough weighs 560 lbs when empty and 2810 lbs when full.

You have access to a river to refill all the containers for water once every 5 miles. If the weight per horse pulled on a wagon is less than 540 lbs, the wagon can travel 15 miles per day on any area of path not cleared by debris, and 30 miles per day on any area cleared of debris.If the weight per horse pulled on a wagon is between 540 lbs and 1080 lbs, the wagon can travel 12 miles per day on any area of path not cleared of debris and 15 miles per day on any area of path that has been cleared of debris.

If the weight per horse pulled on a wagon is between 1080 lbs and 2160 lbs, the wagon can travel at 9 miles per day on any area of path that has not been cleared of debris, and 12 miles per day on any area of path that has been cleared of debris. If the weight per horse pulled on a wagon is more than 2160 lbs, the wagon cannot move.

Any worker carrying less than 100 lbs can move at 15 miles per day on any area of path not cleared of debris, and 30 miles per day on any area of path that has been cleared of debris. If a worker is carrying between 100 lbs and 150 lbs, they move at 9 miles per hour regardless of if the path has or has not been cleared of debris.

Assuming they start from point A, and can only obtain food for the workers and feed for the horses at point A, and must carry enough food and feed to preform each trip, How many days would it take them to move all the debris to Point A? How many days would it take them to move all the debris to point B, then return to point A?

Homework Equations

Time = distance/ Rate of travel
Weight per horse= (weight of wagon+ weight on wagon)/Number of Horses per wagon
Weight on wagon = weight of drivers + weight of cargo; must be less than or equal to 12000 lbs

unsure of other equations

The Attempt at a Solution

not sure of where to go

 

[berkeman]

Holy smokes!

I'm not sure where to start either. What kind of class is this from? Maybe a final exam in an algebra class?

Anyway, I'd start with drawing some sketches of the geotraphy, and little figures for the workers, horses, carts, food, water, etc. Then look at how you can start to write equations for how much work can be done with different mixes of resources. Just start sketching to start to see if any clear trends show up.

 

[Tybras]

well, believe it or not, this is not from any class. This was a question posed to me by my dungeons and dragons group because i was the best at math.

best if can figure for transporting everything to Point B would be to take the awnser for point A, then add however much time it would take to transport all the food and feed you would need to cover that to point b before you start clearing any of the debris using just the wagons. possibly cut a little time by having workers start clearing as the wagons go back and forth with food.

 

[Ray Vickson]

Holy smokes!

I'm not sure where to start either. What kind of class is this from? Maybe a final exam in an algebra class?

Anyway, I'd start with drawing some sketches of the geotraphy, and little figures for the workers, horses, carts, food, water, etc. Then look at how you can start to write equations for how much work can be done with different mixes of resources. Just start sketching to start to see if any clear trends show up.

This is not the type of problem seen in a typical algebra class. It is similar to the types of problems dealt with by Operations Research (specifically, Logistic models)

Be warned, however, that such problems can be very difficult (NP-hard in the jargon) and often must be attacked by heuristic methods rather than exact algorithms. On the other hand, real-world companies really do deal with models of such problems (whether exact or heuristic) and can quite often save $millions in annual "logistics costs"----not an exaggeration.

 

[mfb]

That problem sounds overly complicated just to make it complicated.

28 miles are 147,840 feet.
Every 2 feet, there are 722 piles of 50 lbs each? Then we have 36100 lbs (3 wagons) of debris every 2 feet, or 2.7 billion lbs of debris, about 220,000 wagons.

How fast can workers load a wagon right at the point of debris? For a real life situation, this looks like the largest effort. If it happens instantaneous, then your workers are not very useful: Move the wagons to where the debris is, starting at A. Outbound from A, let them deliver water and food as necessary, then unload that and load them with debris, go back to A. You just need 2 drivers per wagon (they can also do the magic instant-loading), which means all other workers can run around and clear debris on their own, with the same approach.

Once the debris is cleared up so much that wagons or workers need more than a day to go back, make depots on the way.

 

[Tybras]

Well, in the system that the question was posed, it is given that a person can go at maximum 30 miles a day if carrying less than 15 lbs, 24 if carrying upto 100 lbs and 18 if carrying upto 150 lbs.. It is assumed they can reach anything within 5 feet without using any movement and wagons can get within probably 5 feet of anyone of the piles of debris. It is also assumed that it takes them 6 seconds to move 1 50 lb object upto 30 feet.

 

[mfb]

With so much debris every 2 feet (a bit over 3 full wagons), there is no need to walk to the wagons with debris.

I'm surprised the wagons don't have a finite length.

 

[Tybras]

With so much debris every 2 feet (a bit over 3 full wagons), there is no need to walk to the wagons with debris.

I'm surprised the wagons don't have a finite length.

with the way the question was phrased to me, i was assuming that the debris would all be stackable to get the maximum weight in, reguardless of the size of the wagon, to simplify the problem some.

 

[Tybras]

so had some time to work some more on this. looked up lengths and what not. The length of the wagon, with a team of 7 horses is roughly 34 feet. and it takes them about 11 feet to be able to turn around, so the wagons wouldn't be useful for the first 45 feet of clearing.
Based on the rules of where the problem was posed, a person can move 158400 feet (30 miles) in a day. to pick up and set down an object of upto 50 lbs takes 5 feet of movement. If a person is carrying more than 100 lbs, it takes them 10/3 the movement to go any where, and twice the movement if they move through debris that hasn't been cleared.

so, number of workers times distance covered by workers = number of feet covered in a day by the workforce 60 times 158400 feet per day= 9504000 feet covered per day

every 2.5 piles moved requires all workers to move 5 feet forward; so every 2.5 piles requires a total of 300 extra feet moved. upto the 45 foot mark (so 22.5 piles must be moved before wagons can actively matter)

722 objects per pile each object takes 5 feet of movement to pick up and set down (combined), so 722*2.5*5 to clear a before movement is added (9025 feet movement used to clear each 5 feet of space

so need to clear 45 feet of space there are 9025 feet of space used every 5 feet cleared, then 300 feet of space used to move forward, then it for every 2 pieces it takes them 20 feet of movement to clear (5*2 to pick up and set down + 5 back to point a then 5 back to the pile) a space or they can carry 3 but it then costs them 15 feet + 10/3*5+5 which would be 36 2/3 feet to clear 3, making it more efficient to move 2 at a time.

so it looks like the first 45 feet would be figured by: 9025+300+9025/2*20+300+9025/2*30+300+9025/2*40+300+9025/2*50+300+9025/2*60+300+9025/2*70+300+9025/2*80+300+9025/2*90+300+9025/2*100+300=2448775 feet used

after that, the workers would be loading the wagons. Once the wagons come into play, you lose 8 workers though to be drivers. which means just 9025 feet used to clear a 5 foot area then 260 feet to move up 5 feet, so 9285 feet per 5 feet moved of the remaining 147795 feet
so the first day would cover the distance used for clearing 45 feet + 9285x/5 = 9504000

2448775+(9285x/5)=9504000
x=(9504000-2448775)(5/9285)
x=3799.2595584276
so day one would have cleared about 3800 feet.
then you take total distance (147840) - day one distance cleared (3800) then multiplied that by distance used per 5 feet move forward (9285/5). Giving you a remaining distance to cover. (267482280)
You divide that by the total distance the work group less the drivers can cover in a day (158400*(60-8) =8236800)), to give you remaining days of work. 32.4740530303
From there you take the fraction of day left, subtract it from one them multiply it by the distance 1 person can travel, and subtract it from the total distance from point b to point A. If the number is positive it will take more than a day to get back, if negative, they make it back that day.
so 147840-(1-.4740530303)(158400)=64530

so then you take total days worked and add 1 for travel and round up you get a total working day of 35 for the way to be cleared assuming the wagons can keep up with their weight limits.please correct me if that doesn't sound right

 

[Tybras]

now for the wagons

a wagon fully loaded weighs 12400 lbs (max weight of the wagon +weight of the wagon), divided by the number of horses (upto 7) would need a minimum of 6 horses to move at 12 miles a day along a cleared path, and a 7th wouldn't increase it's speed.
(7 horses make it 1771.4 lbs per horse, 6 is 2066.7 lb per horse, 5 can't pull the wagon due to the weight, speed increase would have to be less than 1080 lbs per horse).

each day a horse needs 7 gallons of water, and 10 lbs of feed, each driver need 2 lbs of food and half a water skin of water (5 lbs when full), they also need to carry the food requirement for the workers for the time it take the wagons to be loaded, return to town, be unloaded, resupplied and return to them)

a trough of water weighs 2810 lbs when full and holds 300 gallons. It holds almost 2 day worth of water for all the horses, so you only need one trough. It can be refilled at point a or b or every 5 miles in between, so it should be considered always full for simplicity.

Counting the weight of the trough, the wagons will only need to 4410 worth of supplies to feed let everyone have food and water for 1 day (the further they go from point a the more they will have to carry)
They will not have to carry food or water for the first day.

so for the first day each wagon can carry upto 11600 lbs of debris or 232 pieces, with 4 wagons, so they can moce 928 pieces in each load. on the first day

They can cover a total of 30 miles (158400 feet) a day, movement when loaded costs 10/3 times as much.
They must end at point A, unless they have enough food and water on them for another day (either by staying at one of the 5 mile marks and having brought food or by bringing the trough and food with them on the last part of the trip.

since there are 9025 pieces to load per 5 feet, and they can only hold 928 pieces per trip the first day, it takes (9025/928) 9.7525215517241379 trips per 5 foot space.

each round trip (from A back to A) is distance away times 13/3 (1 for trip too + 10/3 for trip back).
need the summation of (9025/928)*(13/3)*(5(X!+8)) being less than or equal to 158400.

they would be able to clear from 45 feet to 195 feet, plus take 9 full load (8352) back from the pile at 195 feet, leaving 840 pieces. While they do not have enough total movement to make another load, they do have enough movement to bring food for the next work day and water for the horses out past the point where the work crew could get by themselves, so after clearing the 195 feet + 9 loads, the work crew (still having a lot of distance left) could carry back debris and the wagons could met them out as far as they could get that way with food for the next day and reduce the wagons trip out by 1 for the next work day.

so the work crew working out to 195 feet and moving all the debris to leave 840 pieces from the pile there would have used the distance for 45 feet (2448775) multiplied by current workers divided by origional workers (52/60) + (feet used to move 5 feet)*(feet moved/5) (9285*190/5) + Number of pieces moved times 5 ((9025-840)*5)
equals 2516025

we already determined that it is more efficeint to move 2 blocks than it is to move 3, so they can move 104 blocks per trip. it will take 7.5 trips for the remainder of this pile

7.5*195*2=2925

then you take the total number of feet the workgroup can cover in a day (8236800) subtract the total amount of feet covered so far (2516025+2925), then divided that number by number of feet needed to cover 5 (9285/5) feet to find out where the work group ends up. 3079.0791599354

you round down to the nearest 5, subtract that, then divide it by 5 and multiply by number of pieces in a 5 foot area to see how many pieces are left in the next area round up. leaves 1663 piecessince the wagons are there with them at the start of day 2, the workers load the wagon (using 928*5 feet of movement, assuming they unloaded the trough and the rations and feed for the next day the night before, which they would have had the movement left to do). The wagons make the trip back from 3075 feet, and then we progress with their previous calculation to figure ourt how much they can move for day 2. so their starting movement is 10/3 *3075 = 10250 feet used, then we used Starting + Remander/928*13/3 + (9025/928)*(13/3)*(5(X!+D)) being less than or equal to 158400.
Where D is (the distance to the closes pile -5)/5; D in this trip starts at 614.

so: 10250+ ((1663-928)/928)*(13/3) + (9025/928)*(13/3)*(5(X!+614))

This means they can go back to a, use the 10 trips to clear the next 5' space, and take one load from the next and come back with food to met the worker group however far forward they can make it clearing further on themselves. They would also leave 8097 blocks at 3090 ft. 8097/104=78 trips

78*3090*2=482040 to remove the rest of that pile
Add in the feet used to load the wagons (1663 + 928 + 9285 + 260)
to figure out how much movement they have used currently (484176)
subtract that from total move for the day to figure out remaining move (8236800-484176=7742624)

then use their movement formula (9025/2)*(2(X!+5))+ 260X! being less than or equal to distance remaining, where X starts at current distance (of 3080 ft) simplified formula 9285x!+45125 less than or equal to 7742624; it comes out to a fraction, where they are only able to preform 23 trips, getting rid of 2392 of the remaining 8097 pieces, and ending up back the the pile.

carrying the process on like this, I'm getting approximately 11500 days to move all the debris to point A, unless i screwed up somewhere, and reversing it so that everything is transported to B and all the crew returns to A after the job is done adds about an additional 120 days to move all the supplies needed for the job to point B by wagon while the crew clears from point B to A first. Does this sound about right to anyone else, or am i way off?