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Solving an LP problem with an upper limit for the variables
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Im trying to solve the Above LP. Now, can we apply just simplex considering that we can treat the upper bounds of the variables as constraints.
Meaning, we need to add $6$ slack variables, and it would be a bit harder to do the algorithm by hand. Now, the dual will give six variables as well, so it doesnt help much.
In my notes, it says to consider "bounded variable simplex method", but I haven't seen how this method works. how can we apply this method to solve this problem? Is it possible to solve it using the standard primal simplex?
linear-programming
asked Nov 16 at 7:11
Jimmy Sabater
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reputation from Jimmy Sabater ending in 3 days.
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Im trying to solve the Above LP. Now, can we apply just simplex considering that we can treat the upper bounds of the variables as constraints.
Meaning, we need to add $6$ slack variables, and it would be a bit harder to do the algorithm by hand. Now, the dual will give six variables as well, so it doesnt help much.
In my notes, it says to consider "bounded variable simplex method", but I haven't seen how this method works. how can we apply this method to solve this problem? Is it possible to solve it using the standard primal simplex?
linear-programming
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
This question has an open bounty worth +150
reputation from Jimmy Sabater ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
1
Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im trying to solve the Above LP. Now, can we apply just simplex considering that we can treat the upper bounds of the variables as constraints.
Meaning, we need to add $6$ slack variables, and it would be a bit harder to do the algorithm by hand. Now, the dual will give six variables as well, so it doesnt help much.
In my notes, it says to consider "bounded variable simplex method", but I haven't seen how this method works. how can we apply this method to solve this problem? Is it possible to solve it using the standard primal simplex?
linear-programming
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
Im trying to solve the Above LP. Now, can we apply just simplex considering that we can treat the upper bounds of the variables as constraints.
Meaning, we need to add $6$ slack variables, and it would be a bit harder to do the algorithm by hand. Now, the dual will give six variables as well, so it doesnt help much.
In my notes, it says to consider "bounded variable simplex method", but I haven't seen how this method works. how can we apply this method to solve this problem? Is it possible to solve it using the standard primal simplex?
linear-programming
linear-programming
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
asked Nov 16 at 7:11
Jimmy Sabater
1,829218
1,829218
This question has an open bounty worth +150
reputation from Jimmy Sabater ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +150
reputation from Jimmy Sabater ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
1
Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago
add a comment |
1
Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago
1
1
Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago
add a comment |
1 Answer
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Let's write down our simplex table, remembering our upper bounds for each variables.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -2 & -4 & color{blue}{-7} & -1 & -5 & 0 & 1 & 0 & - & - \ hline
s & 3 & 5 & 11 & 2 & 7 & 1 & 0 & 10 & frac{10}{11} & 1 \ hline
end{array}
At the beginning, we are at $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$. The entering variable would be $x_3$. since the minimum ratio is less than the upper bound, we will do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & color{blue}{-frac9{11}} & 0 & frac3{11} & -frac6{11} & frac7{11} & 1 & frac{70}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{10}{11} & 2 & 1 \ hline
end{array}
Upon doing the simplex update, we reach $(0,0,frac{10}{11},0,0)$. Our next entering variable is $x_2$. The difference from the previous step this time is the ratio is more than the upper bound. We will increment the $x_2$ by the upperbound amount, which is $1$. We will also update the RHS and mark $x_2$ as $bar{x}_2$ to indicate that it is now non-basic.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & -frac9{11} & 0 & frac3{11} & color{blue}{-frac6{11}} & frac7{11} & 1 & frac{79}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{5}{11} & frac57 & 1 \ hline
end{array}
Now, we arrive at $(0,1,frac5{11},0,0)$. Our next entering variable is $x_5$, since the ratio is less than the upper bound, we do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & frac1{7} & -frac3{7} & frac67 & frac3{7} & 0 & frac5{7} & 1 & frac{53}{7} & - & - \ hline
x_5 & frac3{7} & frac5{7} & frac{11}7 & frac2{7} & 1 & frac1{7} & 0 & frac{5}{7} & - & - \ hline
end{array}
Now, we arrive at $(0,1,0,0,frac57)$. Upon checking the reduced cost, we can see that we are optimal with objective value $frac{53}{7}$.
R code to check the final solution:
> library(lpSolve)
> f.obj <- c(2,4,7,1,5)
> n = length(f.obj)
> f.con <- rbind(c(3,5,11,2,7), diag(n), -diag(n))
> f.dir <- rep('<=',2 * n+1 )
> f.rhs <- c(10,rep(1,n),rep(0,n))
> optimum <- lp(direction = "max", objective.in = f.obj, const.mat = f.con, const.dir = f.dir, const.rhs = f.rhs)
> optimum
Success: the objective function is 7.571429
> optimum$solution
[1] 0.0000000 1.0000000 0.0000000 0.0000000 0.7142857
answered 2 days ago
Siong Thye Goh
93.1k1462114
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let's write down our simplex table, remembering our upper bounds for each variables.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -2 & -4 & color{blue}{-7} & -1 & -5 & 0 & 1 & 0 & - & - \ hline
s & 3 & 5 & 11 & 2 & 7 & 1 & 0 & 10 & frac{10}{11} & 1 \ hline
end{array}
At the beginning, we are at $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$. The entering variable would be $x_3$. since the minimum ratio is less than the upper bound, we will do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & color{blue}{-frac9{11}} & 0 & frac3{11} & -frac6{11} & frac7{11} & 1 & frac{70}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{10}{11} & 2 & 1 \ hline
end{array}
Upon doing the simplex update, we reach $(0,0,frac{10}{11},0,0)$. Our next entering variable is $x_2$. The difference from the previous step this time is the ratio is more than the upper bound. We will increment the $x_2$ by the upperbound amount, which is $1$. We will also update the RHS and mark $x_2$ as $bar{x}_2$ to indicate that it is now non-basic.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & -frac9{11} & 0 & frac3{11} & color{blue}{-frac6{11}} & frac7{11} & 1 & frac{79}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{5}{11} & frac57 & 1 \ hline
end{array}
Now, we arrive at $(0,1,frac5{11},0,0)$. Our next entering variable is $x_5$, since the ratio is less than the upper bound, we do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & frac1{7} & -frac3{7} & frac67 & frac3{7} & 0 & frac5{7} & 1 & frac{53}{7} & - & - \ hline
x_5 & frac3{7} & frac5{7} & frac{11}7 & frac2{7} & 1 & frac1{7} & 0 & frac{5}{7} & - & - \ hline
end{array}
Now, we arrive at $(0,1,0,0,frac57)$. Upon checking the reduced cost, we can see that we are optimal with objective value $frac{53}{7}$.
R code to check the final solution:
> library(lpSolve)
> f.obj <- c(2,4,7,1,5)
> n = length(f.obj)
> f.con <- rbind(c(3,5,11,2,7), diag(n), -diag(n))
> f.dir <- rep('<=',2 * n+1 )
> f.rhs <- c(10,rep(1,n),rep(0,n))
> optimum <- lp(direction = "max", objective.in = f.obj, const.mat = f.con, const.dir = f.dir, const.rhs = f.rhs)
> optimum
Success: the objective function is 7.571429
> optimum$solution
[1] 0.0000000 1.0000000 0.0000000 0.0000000 0.7142857
answered 2 days ago
Siong Thye Goh
93.1k1462114
add a comment |
up vote
3
down vote
Let's write down our simplex table, remembering our upper bounds for each variables.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -2 & -4 & color{blue}{-7} & -1 & -5 & 0 & 1 & 0 & - & - \ hline
s & 3 & 5 & 11 & 2 & 7 & 1 & 0 & 10 & frac{10}{11} & 1 \ hline
end{array}
At the beginning, we are at $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$. The entering variable would be $x_3$. since the minimum ratio is less than the upper bound, we will do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & color{blue}{-frac9{11}} & 0 & frac3{11} & -frac6{11} & frac7{11} & 1 & frac{70}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{10}{11} & 2 & 1 \ hline
end{array}
Upon doing the simplex update, we reach $(0,0,frac{10}{11},0,0)$. Our next entering variable is $x_2$. The difference from the previous step this time is the ratio is more than the upper bound. We will increment the $x_2$ by the upperbound amount, which is $1$. We will also update the RHS and mark $x_2$ as $bar{x}_2$ to indicate that it is now non-basic.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & -frac9{11} & 0 & frac3{11} & color{blue}{-frac6{11}} & frac7{11} & 1 & frac{79}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{5}{11} & frac57 & 1 \ hline
end{array}
Now, we arrive at $(0,1,frac5{11},0,0)$. Our next entering variable is $x_5$, since the ratio is less than the upper bound, we do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & frac1{7} & -frac3{7} & frac67 & frac3{7} & 0 & frac5{7} & 1 & frac{53}{7} & - & - \ hline
x_5 & frac3{7} & frac5{7} & frac{11}7 & frac2{7} & 1 & frac1{7} & 0 & frac{5}{7} & - & - \ hline
end{array}
Now, we arrive at $(0,1,0,0,frac57)$. Upon checking the reduced cost, we can see that we are optimal with objective value $frac{53}{7}$.
R code to check the final solution:
> library(lpSolve)
> f.obj <- c(2,4,7,1,5)
> n = length(f.obj)
> f.con <- rbind(c(3,5,11,2,7), diag(n), -diag(n))
> f.dir <- rep('<=',2 * n+1 )
> f.rhs <- c(10,rep(1,n),rep(0,n))
> optimum <- lp(direction = "max", objective.in = f.obj, const.mat = f.con, const.dir = f.dir, const.rhs = f.rhs)
> optimum
Success: the objective function is 7.571429
> optimum$solution
[1] 0.0000000 1.0000000 0.0000000 0.0000000 0.7142857
answered 2 days ago
Siong Thye Goh
93.1k1462114
add a comment |
up vote
3
down vote
up vote
3
down vote
Let's write down our simplex table, remembering our upper bounds for each variables.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -2 & -4 & color{blue}{-7} & -1 & -5 & 0 & 1 & 0 & - & - \ hline
s & 3 & 5 & 11 & 2 & 7 & 1 & 0 & 10 & frac{10}{11} & 1 \ hline
end{array}
At the beginning, we are at $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$. The entering variable would be $x_3$. since the minimum ratio is less than the upper bound, we will do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & color{blue}{-frac9{11}} & 0 & frac3{11} & -frac6{11} & frac7{11} & 1 & frac{70}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{10}{11} & 2 & 1 \ hline
end{array}
Upon doing the simplex update, we reach $(0,0,frac{10}{11},0,0)$. Our next entering variable is $x_2$. The difference from the previous step this time is the ratio is more than the upper bound. We will increment the $x_2$ by the upperbound amount, which is $1$. We will also update the RHS and mark $x_2$ as $bar{x}_2$ to indicate that it is now non-basic.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & -frac9{11} & 0 & frac3{11} & color{blue}{-frac6{11}} & frac7{11} & 1 & frac{79}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{5}{11} & frac57 & 1 \ hline
end{array}
Now, we arrive at $(0,1,frac5{11},0,0)$. Our next entering variable is $x_5$, since the ratio is less than the upper bound, we do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & frac1{7} & -frac3{7} & frac67 & frac3{7} & 0 & frac5{7} & 1 & frac{53}{7} & - & - \ hline
x_5 & frac3{7} & frac5{7} & frac{11}7 & frac2{7} & 1 & frac1{7} & 0 & frac{5}{7} & - & - \ hline
end{array}
Now, we arrive at $(0,1,0,0,frac57)$. Upon checking the reduced cost, we can see that we are optimal with objective value $frac{53}{7}$.
R code to check the final solution:
> library(lpSolve)
> f.obj <- c(2,4,7,1,5)
> n = length(f.obj)
> f.con <- rbind(c(3,5,11,2,7), diag(n), -diag(n))
> f.dir <- rep('<=',2 * n+1 )
> f.rhs <- c(10,rep(1,n),rep(0,n))
> optimum <- lp(direction = "max", objective.in = f.obj, const.mat = f.con, const.dir = f.dir, const.rhs = f.rhs)
> optimum
Success: the objective function is 7.571429
> optimum$solution
[1] 0.0000000 1.0000000 0.0000000 0.0000000 0.7142857
answered 2 days ago
Siong Thye Goh
93.1k1462114
Let's write down our simplex table, remembering our upper bounds for each variables.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -2 & -4 & color{blue}{-7} & -1 & -5 & 0 & 1 & 0 & - & - \ hline
s & 3 & 5 & 11 & 2 & 7 & 1 & 0 & 10 & frac{10}{11} & 1 \ hline
end{array}
At the beginning, we are at $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$. The entering variable would be $x_3$. since the minimum ratio is less than the upper bound, we will do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & x_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & color{blue}{-frac9{11}} & 0 & frac3{11} & -frac6{11} & frac7{11} & 1 & frac{70}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{10}{11} & 2 & 1 \ hline
end{array}
Upon doing the simplex update, we reach $(0,0,frac{10}{11},0,0)$. Our next entering variable is $x_2$. The difference from the previous step this time is the ratio is more than the upper bound. We will increment the $x_2$ by the upperbound amount, which is $1$. We will also update the RHS and mark $x_2$ as $bar{x}_2$ to indicate that it is now non-basic.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & -frac1{11} & -frac9{11} & 0 & frac3{11} & color{blue}{-frac6{11}} & frac7{11} & 1 & frac{79}{11} & - & - \ hline
x_3 & frac3{11} & frac5{11} & 1 & frac2{11} & frac7{11} & frac1{11} & 0 & frac{5}{11} & frac57 & 1 \ hline
end{array}
Now, we arrive at $(0,1,frac5{11},0,0)$. Our next entering variable is $x_5$, since the ratio is less than the upper bound, we do a regular simplex pivot updating.
begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}hline
& x_1 & bar{x}_2 & x_3 & x_4 & x_5 & s & -z & RHS & text{ratio} &UB \ hline
-z & frac1{7} & -frac3{7} & frac67 & frac3{7} & 0 & frac5{7} & 1 & frac{53}{7} & - & - \ hline
x_5 & frac3{7} & frac5{7} & frac{11}7 & frac2{7} & 1 & frac1{7} & 0 & frac{5}{7} & - & - \ hline
end{array}
Now, we arrive at $(0,1,0,0,frac57)$. Upon checking the reduced cost, we can see that we are optimal with objective value $frac{53}{7}$.
R code to check the final solution:
> library(lpSolve)
> f.obj <- c(2,4,7,1,5)
> n = length(f.obj)
> f.con <- rbind(c(3,5,11,2,7), diag(n), -diag(n))
> f.dir <- rep('<=',2 * n+1 )
> f.rhs <- c(10,rep(1,n),rep(0,n))
> optimum <- lp(direction = "max", objective.in = f.obj, const.mat = f.con, const.dir = f.dir, const.rhs = f.rhs)
> optimum
Success: the objective function is 7.571429
> optimum$solution
[1] 0.0000000 1.0000000 0.0000000 0.0000000 0.7142857
answered 2 days ago
Siong Thye Goh
93.1k1462114
edited 2 days ago
answered 2 days ago
Siong Thye Goh
93.1k1462114
answered 2 days ago
Siong Thye Goh
93.1k1462114
answered 2 days ago
Siong Thye Goh
93.1k1462114
93.1k1462114
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Here is a algorithm for Bounded Variables with examples.
– callculus
Nov 16 at 12:24
What more do you expect from an answer than the algorithm callculus linked to?
– LinAlg
Nov 18 at 13:57
LinAlg The example in the link is different than my problem. I dont see a way to apply what is on this link to this actual problem.
– Jimmy Sabater
2 days ago
Could you add to your question where you get stuck? The examples on pages 5-8 seem to cover your needs.
– LinAlg
2 days ago