Writing tiered pricing as a linear program - linear-programming

Question

Is there a way to write a cost function that uses tiered pricing as a linear program?
supplier volume A price A volume B price B
1 0 $1 300 $2
2 0 $3 500 $1.50
By tiered pricing I mean for supplier 1, the first 300kg costs $1 and the second 200kg costs $2. (Not if I reach over 300kg, then all units cost $2). This is the cost function that I am currently looking at.
cost = P1a * V1 + P1b * V1 + P2a * V2 + P2b * V2
s.t V1+V2 = 1000
V2 < 8000, V1 < 800
where
P1a = first price bracket for supplier 1
P1b = second price bracket for supplier 1
V1 = volume for supplier 1
If it was the case where all volume is becomes price B when volume B is reached, then a I can add a binary variable so the function looks like this
cost = u * P1a * V1 + v * P1b * V1 + x * P2a * V2 + y * P2b * V2
s.t u+v = 1, x+y = 1
u,v,x,y = 1 if price bracket is used, 0 if price bracket isnt used
Is there a way I could do something similar?
Edit:
V1 and V2 are both integers and have a set upper cap, lets say 800kg. There is also a demand constraint, lets say V1 + V2 = 1000.

Answer 1

It seems to be feasable, provided there is a constant M>=0 which is a common upper bound for the volume variables V1 and V2 (e.g if there is a resriction V1+V2 <=M).
Provided that there is such an M, let us introduce two binary variables b_1 and b_2 which indicate (meaning having value 1) if V1>= 300 and V2>=500 respectively:
300 * b_1 <= V1 <= 300 * (1 - b_1) + M* b_1
500 * b_2 <= V2 <= 500 * (1 - b_2) + M* b_2
Now we introduce non-negative variables a_1, h_1 which will decompose V1 into a part less than 300 (denoted by a_1)and a part holding the part exceeding (denoted by variable h_1):
V1= a_1 + h_1
h_1 <= M b_1
300 b_1 <= a_1 <= 300
The last two constraints with the binary variable enforce that h1 will only be positive if V1 >= 300 and a_1 will always be 300 if V1 >= 300.
Let us do the same for V2 respectively
V2= a_2 + h_2
h_2 <= M b_2
500 b_2 <= a_2 <= 500
Then the cost function will be
$1 * a_1 + $2 * h_1 + $3 * a_2 + $1.5 * h_2
`
Let me remark that this solution might not be correct on boundaries e.g. when V1=300, because then the binary variable b_1 might attain value 0 and thus a_1=0, h_1>0.
However this problem could be solved if the volumes V1 and V2 would be always integer values, e.g. see
here
*Addendum to J.Doe's comment:
if V1 is an integer then the boundary problem would be treated by 301 b_1 <=V_1 <=301 (1-b_1)+ M b_1 instead of the defining condition for b_1 above.
No, h_1 < = M b_1 - a_1 is not valid because this would mean V1 = a_1 + h_1 <= M b_1 which would imply V1 = 0 if V1 >=300 - which is not true.
Yes, h_1 is the excess volume above the first bracket right.
If we have more than one price bracket we would need 2 further variables: One representing the amount which represents the volume V1's share between the current price bracket and the next pricebracket.
The second variable should be a binary indicator variable which is 1 if the volume V1 is between the current price bracket and the next pricebracket.