如何使用我的使用量和量对成本的括号动态计算水费成本？

Question

I'm trying to predict my water bill using a flowmeter connected to my mainline.

I have the price rates used for my bill which are: 27.54kL @ $2.392 91.80kL @ $3.413 300.00kL @ $3.699

So for example, if I used 250kL, my bill would be broken down as:

1st rate: There are more than 27.54kL in 250kL, so cost of this rate is: 27.54*$2.392 = $65.88 (250 - 27.54 = 222.46kL remaining for the next rate)
2nd rate: There are more than 91.80kL in 222.46kL, so cost of this rate is = 91.80*$3.413 = $313.31 (222.46 - 91.80 = 130.66kL remaining for next rate)
3rd rate: There are less than 300kL in 130.66kL, but more than 0, so cost of this rate is: 130.66*$3.699 = $483.31 (130.66 - 130.66 = 0L remaining for next rate (if there was a 4th))

Total = sum of all rate costs = $862.50

How can I calculate this in a dynamic way, so that if the rates or even the number of rates change, it still works? I can easily reproduce the above, but I'd like to use lists and list comprehension or other loops to produce the final cost.

This will calculate what I need, but it's not dynamic:

volumeCostDict = [[27.54,2.392], [91.80,3.413], [350.00,3.699]]
volume = 271 # kilolitres

rate1Volume = volumeCostDict[0][0] if volume > volumeCostDict[0][0] else volume
rate1Cost = rate1Volume * volumeCostDict[0][1]

rate2Volume = volumeCostDict[1][0] if volume > (volumeCostDict[0][0] + volumeCostDict[1][0]) else volume - volumeCostDict[0][0]
rate2Cost = rate2Volume * volumeCostDict[1][1]

rate3Volume = volumeCostDict[2][0] if volume > (volumeCostDict[0][0] + volumeCostDict[1][0] + volumeCostDict[2][0]) else volume - volumeCostDict[0][0] - volumeCostDict[1][0]
rate3Cost = rate3Volume * volumeCostDict[2][1]

print(rate1Cost)
print(rate2Cost)
print(rate3Cost)

totalCost = rate1Cost + rate2Cost + rate3Cost

Thanks!

Answer 1

I got it, just needed to code it statically first and then work out the pattern to apply it dynamically. There's still probably a simpler way, but this works EDIT: to work for all cases

volumeCostDict = [[27.54,2.392], [91.80,3.413], [350.00,3.699]]
volume = 271 # kilolitres

rateVolumes = [volume_kL \
                 if volume_kL < volumeCostDict[i][0] and i == 0 \
                 else volumeCostDict[i][0] \
                 if volume_kL > sum([volumeCostDict[j][0] for j in range(i+1)]) \
                 else volume_kL - sum([volumeCostDict[k][0] for k in range(i)]) \
                 if (volume_kL - sum([volumeCostDict[k][0] for k in range(i)])) > 0 \
                 else 0 \
                 for i in range(len(volumeCostDict))]
totalCost = sum([rateVolumes[i]*volumeCostDict[i][1] for i in range(len(rateVolumes))])
print(rateVolumes)
print(totalCost)

Answer 2

I have found a way out, if you say that, the range would be same, it is only data change thing. So what we can do, we can make use of while loop and follow this algo

1. While volume != 0
2. Two extreme cases and two normal cases are there:
   a) {If the volume becomes less the least value `volumeCostDict[0][0]`} and {If volume > `volumeCostDict[0][0]` and volume > `volumeCostDict[0][0]`}
   b) {If volume >= `volumeCostDict[0][0]`} and {If volume >= `volumeCostDict[1][0]`}
3. Deal with the both the cases, in case one do the computation for both the extreme cases. 
4. For normal cases, computation is there, which you have shown already
5. Print after coming from the while loop 

FINAL SOLUTION

volumeCostDict = [[27.54,2.392], [91.80,3.413], [350.00,3.699]]
volume = 250 # kilolitres
totalCost = 0

# this will run until volume becomes 0
while volume != 0 :
    # extreme case, what if the value becomes less than the least
    # in our case 27.54, so we calculate the same for 27.54
    # and make the value go to 0 
    # this is required, since for Volume = 200, you won't get any result
    if volume < volumeCostDict[0][0]:
        totalCost += (volume * volumeCostDict[0][1])
        volume = 0
      
    # this is normal case, which will check if the volume is
    # greater than equals least value  
    if volume >= volumeCostDict[0][0]:
        volume -= volumeCostDict[0][0]
        totalCost += (volumeCostDict[0][0] * volumeCostDict[0][1])
        
    # this is normal case, which will check if the volume is
    # greater than equals second rate value 
    if volume >= volumeCostDict[1][0]:
        volume -= volumeCostDict[1][0]
        totalCost += (volumeCostDict[1][0] * volumeCostDict[1][1])
      
    # extreme case again, if the volume after deduction  is still 
    # greater than the normal cases, that is 27.54 and 91.80
    if volume > volumeCostDict[0][0] and volume > volumeCostDict[1][0]:
        totalCost += (volume * volumeCostDict[2][1])
        volume = 0

print("${:.2f}".format(totalCost))

# OUTPUT for volume = 250 will be 
>>> $862.50

You can check this with other volumes as well. I know this solution will sound layman to you. But tell me how much useful it was for you:) For any suggestions, I am open to learn.