Calculating Average Fees and Transaction Volumes Word Problem

uptickk

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If I have 415 transactions and know there are only two types of transactions (H&L) with a total average fee of 1.4 and I know transaction H has an average fee of 1.20 can I calculate the average fee for L and the individual transaction volumes for both H and L? I should also note that the total fees for both transactions (H&L) is 581.

total transactions*total average fee=total fees (415*1.4 = 581)
transH*1.2 = feeH
transL*avgL = feeL


It is possible I am not given enough information to come up with the missing values.

Any help would be greatly appreciated.

Thank you!
 
If I have 415 transactions and know there are only two types of transactions (H&L) with a total average fee of 1.4 and I know transaction H has an average fee of 1.20 can I calculate the average fee for L and the individual transaction volumes for both H and L? I should also note that the total fees for both transactions (H&L) is 581.

total transactions*total average fee=total fees (415*1.4 = 581)
transH*1.2 = feeH
transL*avgL = feeL


It is possible I am not given enough information to come up with the missing values.

Any help would be greatly appreciated.

Thank you!
Let h be the number of transactions of type H, let l be the number of transactions of type L and let x be the avg fee for a type L transaction
Then h+l = 451 and 1.2h + xl= 581
From what I am seeing there are two equations and three unknowns (h, l and x). This can not be solved uniquely.
 
Let h be the number of transactions of type H, let l be the number of transactions of type L and let x be the avg fee for a type L transaction
Then h+l = 415 and 1.2h + xl= 581
From what I am seeing there are two equations and three unknowns (h, l and x). This can not be solved uniquely.

Jomo, Thank you for your help! Of course I just realized I have one additional piece of data though I do not think it helps in solving the problem but please correct me if I am wrong. The additional data is that the number of transactions (H&L) occurs in 150 lockboxes. Not sure if it helps but this exercise deals with transactions in bank lockboxes.
 
Thank you both Jomo and Denis! I really do appreciate the help.

Note, however, that you can put some bounds on the numbers. Using Denis' notation we must have
0 < L < 415
1.41 \(\displaystyle \le\) a \(\displaystyle \le\) 84.20
assuming that the fee a must be expressible as whole pennies (a can't be 1.400484, for example)
 
Note, however, that you can put some bounds on the numbers. Using Denis' notation we must have
0 < L < 415
1.41 \(\displaystyle \le\) a \(\displaystyle \le\) 84.20
assuming that the fee a must be expressible as whole pennies (a can't be 1.400484, for example)

Thanks Ishuda for chiming in with the bounds. How is the 84.20 upper bound derived (feel like I am not thinking of something very apparent on this one)?

Also I should note I haven't done this type of math in quite some time. What type of math is this considered so I can do some sideline reading so I don't have so many questions? I am trying to do these calculations in excel and am not having much luck.

Denis, thank you for one of the outcomes! How did you derive it? Also how do I determine the number of possible outcomes using Ishuda bounds?

I do apologize as I am not relearning this as quickly as I hoped but am grateful for all the help!
 
Thanks Ishuda for chiming in with the bounds. How is the 84.20 upper bound derived (feel like I am not thinking of something very apparent on this one)? ...

Taking Denis's equation and re-writing it we have
L (a - 1.2) = 83
Since a value for the average fee for H exists, H is at least 1. Also, since the sum of H and L is 415, H is at least 1 and L can be no more than 414, i.e. less that 415. If L were zero, the above equation couldn't be satisfied so L is at least 1, i.e. greater than zero.

Now re-write the equation as
a = 83/L + 1.2
a is at a minimum when L is at a maximum and vice versa. Since the minimum of L is 1, then maximum of a is 84.20. Allowing all values for a, since the maximum of L is 414, the minimum of a is about 1.40048309178744 or, if it has to be a whole penny it is $1.41, since a=$1.40 would make L=415 which is greater than its max.

As far as further reading, you might look up word problems and (linear) simultaneous equations.
 
Taking Denis's equation and re-writing it we have
L (a - 1.2) = 83
Since a value for the average fee for H exists, H is at least 1. Also, since the sum of H and L is 415, H is at least 1 and L can be no more than 414, i.e. less that 415. If L were zero, the above equation couldn't be satisfied so L is at least 1, i.e. greater than zero.

Now re-write the equation as
a = 83/L + 1.2
a is at a minimum when L is at a maximum and vice versa. Since the minimum of L is 1, then maximum of a is 84.20. Allowing all values for a, since the maximum of L is 414, the minimum of a is about 1.40048309178744 or, if it has to be a whole penny it is $1.41, since a=$1.40 would make L=415 which is greater than its max.

As far as further reading, you might look up word problems and (linear) simultaneous equations.

Great explanation! Thanks again and thanks for the reading topics too!

Anyway to determine the number of possible outcomes given the bounds?
 
Last edited:
Great explanation! Thanks again and thanks for the reading topics too!

Anyway to determine the number of possible outcomes given the bounds?
It depends on your criterion. Since one would expect you would want L to be an integer, there would be at most 414 solutions. If L were not required to be an integer but could be any real number between zero and 415 exclusive, there are an infinite number of solutions.

If, in addition to requiring L to be an integer, you required a to be an exact amount to the penny, that is no partial cents rounded up or down was allowed, then one could determine the number of solutions. Since 8300/L would have to be an integer, L would also have to be a combination of the product of up to two 2's, up to two 5's and up to one 83. That is
L \(\displaystyle \epsilon\){1, 2, 4, 5, 10, 20, 25, 50, 83, 100, 166, and 332}
 
Thank you so much again Denis and Ishuda this has been so helpful!

Denis - What did you use to perform your computations? I am assuming you didn't run every instance manually. Just trying to figure out the most efficient way to perform similar calculations in the future. Example instead of H+L = 415, what if it was 415,000?

Again, thank you so much for your time.
 
These are the ones where transactions are integers:

249 @ 1.20 = 298.8
166 @ 1.70 = 282.2
=============
415 @ 1.40 = 581.0

332 @ 1.20 = 398.4
083 @ 2.20 = 182.6
=============
415 @ 1.40 = 581.0

405 @ 1.20 = 486.0
010 @ 9.50 = 095.0
=============
415 @ 1.40 = 581.0

410 @ 1.20 = 492.0
005 @17.80 = 089.0
=============
415 @ 1.40 = 581.0

413 @ 1.20 = 495.6
002 @42.70 = 085.4
=============
415 @ 1.40 = 581.0

414 @ 1.20 = 496.8
001 @84.20 = 084.2
=============
415 @ 1.40 = 581.0

Was also wondering why a few of these were not included, just want to make sure I am fully understanding the exercise. I just ended up making an excel spreadsheet but figure there is probably a more efficient way.


390 @ 1.20 = 468
025 @ 4.52 = 113
============
415 @ 1.40 = 581.0

395 @ 1.20 = 474
020 @ 5.35 = 107
============
415 @ 1.40 = 581.0
 
These are the ones where transactions are integers: ...
Why didn't you allow the rest?
L
a
H
b
total
allow
1
84.20
414
1.2
581
1
2
42.70
413
1.2
581
1
4
21.95
411
1.2
581
1
5
17.80
410
1.2
581
1
10
9.50
405
1.2
581
1
20
5.35
395
1.2
581
1
25
4.52
390
1.2
581
1
50
2.86
365
1.2
581
1
83
2.20
332
1.2
581
1
100
2.03
315
1.2
581
1
166
1.70
249
1.2
581
1
332
1.45
83
1.2
581
1

As far as generating the numbers you can either let a spread sheet do it for you [as I did here] or take the route of prime factors. Since
8300 = 22 * 55 * 83
and 100*a is an integer we have
100*a = 8300 / L + 120
and L must be a divisor of 8300

BTW: The allow column just a check to see whether 100*a is an integer. All of the rest of the L's had a zero in that column. So, set up the spread sheet to compute the numbers, do a Paste Special for Values and then do a sort.
 
Why didn't you allow the rest?
LaHbtotalallow
184.204141.25811
242.704131.25811
421.954111.25811
517.804101.25811
109.504051.25811
205.353951.25811
254.523901.25811
502.863651.25811
832.203321.25811
1002.033151.25811
1661.702491.25811
3321.45831.25811

As far as generating the numbers you can either let a spread sheet do it for you [as I did here] or take the route of prime factors. Since
8300 = 22 * 55 * 83
and 100*a is an integer we have
100*a = 8300 / L + 120
and L must be a divisor of 8300

BTW: The allow column just a check to see whether 100*a is an integer. All of the rest of the L's had a zero in that column. So, set up the spread sheet to compute the numbers, do a Paste Special for Values and then do a sort.


Ishuda, glad you had the same question regarding the others as I did.

Your spreadsheet looks almost identical to mine but your allow column was more efficient.

I want to thank you guys so much. You have been a huge help and have given me a much better understand of the process!
 
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