Inventory optimization problem.
$begingroup$
I have n products. $(y_1, y_2, ...y_n)$ is my inventory vector. This should last for a certain period.
Assume $m$ transactions will be made in this period.
Each transaction will have only one of these $n$ items (they are substitutional products). The probabilities of purchase of each product are known, $p_1,p_2,p_3,...,p_n.$ They add up to 1.
Revenue given by each product is known $r_1,r_2,...r_n$.
OBJECTIVE
Maximize :- $R_m(y_1, y_2, ...,y_n)$
which is the revenue made in this period, given $m$ purchases are made and given an inventory vector.
Decision variable is obviously the inventory vector.
CONSTRAINT
$y_1+y_2+...+y_n leq C$ [capacity constraint]
How do you approach this problem? It's more tricky than your usual lp problem which is what I'm familiar with. I have done a bit of searching but didn't across similar problems.
Expected revenue for one transaction is just sumproduct of probablities and profits. So we have $m$ transactions and just multiplying by $m$ is wrong since we don't know if the products still remain in inventory by the time we reach $m-th$ transaction (quantity is what we have to decide).
optimization dynamic-programming
edited Jan 1 at 7:48
amWhy
1
asked Jan 1 at 7:01
Goutham Goutham
12
$endgroup$
add a comment |
$begingroup$
I have n products. $(y_1, y_2, ...y_n)$ is my inventory vector. This should last for a certain period.
Assume $m$ transactions will be made in this period.
Each transaction will have only one of these $n$ items (they are substitutional products). The probabilities of purchase of each product are known, $p_1,p_2,p_3,...,p_n.$ They add up to 1.
Revenue given by each product is known $r_1,r_2,...r_n$.
OBJECTIVE
Maximize :- $R_m(y_1, y_2, ...,y_n)$
which is the revenue made in this period, given $m$ purchases are made and given an inventory vector.
Decision variable is obviously the inventory vector.
CONSTRAINT
$y_1+y_2+...+y_n leq C$ [capacity constraint]
How do you approach this problem? It's more tricky than your usual lp problem which is what I'm familiar with. I have done a bit of searching but didn't across similar problems.
Expected revenue for one transaction is just sumproduct of probablities and profits. So we have $m$ transactions and just multiplying by $m$ is wrong since we don't know if the products still remain in inventory by the time we reach $m-th$ transaction (quantity is what we have to decide).
optimization dynamic-programming
edited Jan 1 at 7:48
amWhy
1
asked Jan 1 at 7:01
Goutham Goutham
12
$endgroup$
1
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13
add a comment |
$begingroup$
I have n products. $(y_1, y_2, ...y_n)$ is my inventory vector. This should last for a certain period.
Assume $m$ transactions will be made in this period.
Each transaction will have only one of these $n$ items (they are substitutional products). The probabilities of purchase of each product are known, $p_1,p_2,p_3,...,p_n.$ They add up to 1.
Revenue given by each product is known $r_1,r_2,...r_n$.
OBJECTIVE
Maximize :- $R_m(y_1, y_2, ...,y_n)$
which is the revenue made in this period, given $m$ purchases are made and given an inventory vector.
Decision variable is obviously the inventory vector.
CONSTRAINT
$y_1+y_2+...+y_n leq C$ [capacity constraint]
How do you approach this problem? It's more tricky than your usual lp problem which is what I'm familiar with. I have done a bit of searching but didn't across similar problems.
Expected revenue for one transaction is just sumproduct of probablities and profits. So we have $m$ transactions and just multiplying by $m$ is wrong since we don't know if the products still remain in inventory by the time we reach $m-th$ transaction (quantity is what we have to decide).
optimization dynamic-programming
edited Jan 1 at 7:48
amWhy
1
asked Jan 1 at 7:01
Goutham Goutham
12
$endgroup$
I have n products. $(y_1, y_2, ...y_n)$ is my inventory vector. This should last for a certain period.
Assume $m$ transactions will be made in this period.
Each transaction will have only one of these $n$ items (they are substitutional products). The probabilities of purchase of each product are known, $p_1,p_2,p_3,...,p_n.$ They add up to 1.
Revenue given by each product is known $r_1,r_2,...r_n$.
OBJECTIVE
Maximize :- $R_m(y_1, y_2, ...,y_n)$
which is the revenue made in this period, given $m$ purchases are made and given an inventory vector.
Decision variable is obviously the inventory vector.
CONSTRAINT
$y_1+y_2+...+y_n leq C$ [capacity constraint]
How do you approach this problem? It's more tricky than your usual lp problem which is what I'm familiar with. I have done a bit of searching but didn't across similar problems.
Expected revenue for one transaction is just sumproduct of probablities and profits. So we have $m$ transactions and just multiplying by $m$ is wrong since we don't know if the products still remain in inventory by the time we reach $m-th$ transaction (quantity is what we have to decide).
optimization dynamic-programming
optimization dynamic-programming
edited Jan 1 at 7:48
amWhy
1
asked Jan 1 at 7:01
Goutham Goutham
12
edited Jan 1 at 7:48
amWhy
1
asked Jan 1 at 7:01
Goutham Goutham
12
edited Jan 1 at 7:48
amWhy
1
edited Jan 1 at 7:48
amWhy
1
edited Jan 1 at 7:48
amWhy
1
1
asked Jan 1 at 7:01
Goutham Goutham
12
asked Jan 1 at 7:01
Goutham Goutham
12
asked Jan 1 at 7:01
Goutham Goutham
12
12
1
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13
add a comment |
1
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13
1
1
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If in each transaction only one unit is sold, I guess your objective function is something like
$$ sum_j^m sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.
answered Jan 1 at 8:22
PatricioPatricio
3377
$endgroup$
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
add a comment |
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1 Answer
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$begingroup$
If in each transaction only one unit is sold, I guess your objective function is something like
$$ sum_j^m sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.
answered Jan 1 at 8:22
PatricioPatricio
3377
$endgroup$
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
add a comment |
$begingroup$
If in each transaction only one unit is sold, I guess your objective function is something like
$$ sum_j^m sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.
answered Jan 1 at 8:22
PatricioPatricio
3377
$endgroup$
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
add a comment |
$begingroup$
If in each transaction only one unit is sold, I guess your objective function is something like
$$ sum_j^m sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.
answered Jan 1 at 8:22
PatricioPatricio
3377
$endgroup$
If in each transaction only one unit is sold, I guess your objective function is something like
$$ sum_j^m sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.
answered Jan 1 at 8:22
PatricioPatricio
3377
answered Jan 1 at 8:22
PatricioPatricio
3377
answered Jan 1 at 8:22
PatricioPatricio
3377
answered Jan 1 at 8:22
PatricioPatricio
3377
3377
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
add a comment |
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
Thanks a lot. I think I've understood it.
$endgroup$
– Goutham
Jan 1 at 9:11
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
The indicator function's boolean expression would have (j-1) though not j right?
$endgroup$
– Goutham
Jan 3 at 11:14
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Yes, you're right
$endgroup$
– Patricio
Jan 4 at 10:49
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
$begingroup$
Can this function's maxima be actually found using a solver? I used scipy's minimze function in python and it failed. It doesn't seem like a convex function so it can't be done?
$endgroup$
– Goutham
Jan 4 at 11:42
add a comment |
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1
$begingroup$
what are you asking? what are your own thoughts about how to approach this problem?
$endgroup$
– gt6989b
Jan 1 at 7:13
$begingroup$
Oh i guess I should have written a bit more.. I'll edit
$endgroup$
– Goutham
Jan 1 at 7:24
$begingroup$
Do you mean that if before the initial inventory is $(y_1,y_2,...,y_i,...,y_n)$, it will be $(y_1,y_2,...,y_i-1,...,y_n)$ after the first transaction, or can it be $(y_1,y_2,...,y_i-k,...,y_n)$ and $k$ is also a random variable?
$endgroup$
– Patricio
Jan 1 at 7:50
$begingroup$
No just one would be sold
$endgroup$
– Goutham
Jan 1 at 9:13