Minimize subject to the constraints and :

Minimize subject to the constraints and :

Minimize subject to and bounds :

Use VectorLessEqual to express several LessEqual inequality constraints at once:

Use v<= to enter the vector inequality in a compact form:

An equivalent form using scalar inequalities:

Use VectorGreaterEqual to express several GreaterEqual inequality constraints at once:

Use v>= to enter the vector inequality in a compact form:

An equivalent form using scalar inequalities:

Use Equal to express several equality constraints at once:

An equivalent form using several scalar equalities:

Specify constraints using a combination of scalar and vector inequalities:

Minimize subject to . Use vector variable :

An equivalent form using scalar variables:

Use vector variables and vector inequalities to specify the problem:

Use Indexed to access components of a vector variable, e.g. :

Use constant parameter equations to specify the coefficients of the objective and constraints:

Use constant parameter equations to specify coefficients of several constraints:

Minimize subject to :

Use Vectors[n] to specify the dimension of vector variables when needed:

Specify non-negative constraints using NonNegativeReals ():

An equivalent form using vector inequalities:

Specify non-positive constraints using NonPositiveReals ():

An equivalent form using vector inequalities:

Minimize subject to by specifying the scalars, vectors and matrices:

Solve using the matrix-vector form:

The equivalent form using variables and inequalities:

Try to find a vector that minimizes the function subject to the constraints :

The minimum value is unbounded because there is a singularity across the feasible region:

Specify integer domain constraints using Integers:

Specify integer domain constraints on vector variables using Vectors[n,Integers]:

Specify non-negative integer domain constraints using NonNegativeIntegers ():

An equivalent form using vector inequalities:

Specify non-positive integer domain constraints using NonPositiveIntegers ():

An equivalent form using vector inequalities:

Specify complex variables using Complexes:

Specify the complex domain on vector variables using Vectors[2,Complexes]:

Minimize subject to :

Get the primal minimizer as a vector:

Get the minimal value:

Solve an optimization problem using symbolic inputs:

Extract the matrix-vector inputs of the optimization problem:

Recover the symbolic input result using the matrix-vector inputs:

Find the slack associated with a minimization problem:

Get the primal minimizer and the constraint matrices:

The slack for inequality constraints is a vector such that :

The slack for the equality constraints is a vector such that :

The equality slack is typically a zero vector:

Minimize subject to and :

The dual problem is to maximize subject to :

The primal minimum value and the dual maximum value coincide because of strong duality:

The duality gap of zero. In general, at optimal points:

Construct the dual problem using constraint matrices extracted from the primal problem:

Extract the objective vectors and constraint matrices:

The dual problem is to maximize subject to . Use Inactive[Times] to avoid threading in equality constraint:

Get the dual maximum value directly using solution properties:

Get the dual maximizer directly using solution properties:

Use "ConstraintSensitivity" to find the change in optimal value due to constraint relaxations:

The first vector is inequality sensitivity and the second is equality sensitivity:

Consider new constraints where is the relaxation. The approximate new optimal value is:

Compare directly by solving the relaxed problem:

Each sensitivity is associated with inequality or equality constraints:

Extract the constraints:

The inequality constraints and their associated sensitivity:

The equality constraints and their associated sensitivity:

The change in optimal value due to constraint relaxation is proportional to the sensitivity value:

Compute the minimal value and constraint sensitivity:

A zero sensitivity will not change the optimal value if the constraint is relaxed:

A negative sensitivity will decrease the optimal value:

A positive sensitivity will increase the optimal value:

The "ConstraintSensitivity" is related to the dual maximizer of the problem:

The inequality sensitivity satisfies , where is the dual inequality maximizer:

The equality sensitivity satisfies , where is the dual equality maximizer:

The default method for MachinePrecision is "CLP":

The default method for arbitrary or infinite WorkingPrecision is "Simplex":

All methods work for MachinePrecision input:

"Simplex" and "RevisedSimplex" can be used for arbitrary- and infinite-precision inputs:

Different methods have different strengths, which is typically problem and implementation dependent:

"Simplex" and "RevisedSimplex" are good for small dense problems:

"InteriorPoint" and "CLP" are good for large sparse problems:

Different methods may give different results for problems where the optimal solution set is not unique:

"InteriorPoint" may return a solution in the middle of the optimal solution set:

"Simplex" always returns a solution at a corner of the optimal solution set:

A smaller Tolerance setting gives a more precise result:

Compute minimum value with very small tolerance to serve as reference:

Compute the solution using different Tolerance settings:

Visualize the change in error with respect to tolerance:

A smaller Tolerance setting gives a more precise answer but typically takes longer to compute:

A smaller tolerance takes longer:

The tighter tolerance gives a more precise answer:

Exact input gives exact output:

Use MachinePrecision to get an inexact result:

LinearFractionalOptimization can compute results using a higher working precision:

If the precision is less than the precision of the input arguments, a message is issued:

Maximize subject to . Solve a maximization problem by negating the objective function:

Negate the primal minimal value to get the corresponding maximum value:

Minimize subject to . Let , and convert into a linear function using :

The problem can also be converted using :

Minimize subject to . Since , letting , the objective is transformed to :

Minimize subject to . Since , using , , the objective is transformed to :

Minimize subject to , where is a nondecreasing function, by instead minimizing . The primal minimizer will remain the same for both problems. Consider minimizing subject to :

The true minimum value can be obtained by applying the function to the primal minimum value:

Minimize , subject to :

Since , solve the problem with and :

The optimal solution is the transformed minimum of the two solutions:

Choose a number of units to transport for five goods by minimizing cost and maximizing profit. The cost and profit of transporting one unit of each good are:

The weight of each unit of goods and the maximum units available for transport are:

The minimum operating cost of running the truck is $2000 for the trip. The total transportation cost is:

The truck makes a minimum profit of $1000 if it rides empty. The total profit is:

The truck can carry a maximum of 8000 lbs. The constraints on the truck are:

The optimal mix of goods to transport if found by minimizing the ratio of cost to profit:

Find the mix of four products to manufacture that minimize cost and maximize profit. The cost and profit of manufacturing one unit of each product are:

To manufacture each product, the company needs a combination of five resources, given by:

The maximum resources available are:

The constraints on the resources and products are:

The company has a minimum operating cost of $100. The total manufacturing cost is:

The total profit made by selling the products is:

To find out the maximum units to manufacture, minimize the ratio of cost to profit:

Find the amount of electricity a company must send from its three power plants to four cities so as to maximize profit and minimize cost while meeting the cities' peak demands. The cost of transporting the 1 million kWh of electricity from each plant to the various cities is:

The profit that each power plant generates by selling 1 million kWh to each city is:

Let represent the amount of electricity sent by plant to city . The total cost of transporting electricity is and constructed using Inactive[Times]:

The total profit made by the power company is and constructed using Inactive[Times]:

The cities have a peak demand of 45, 20, 30, 30 million kWh, respectively:

The power plants can supply a minimum of 35, 50 and 40 million kWh of electricity, respectively:

The plants can only give and not receive power from the cities:

The optimal amount of electricity to send each city for each plant can be found by minimizing the ratio of cost to profit:

The breakdown of electricity supplied is:

Find the mix of old alloys needed to make a new alloy containing at most 60% lead and at most 35% tin while minimizing cost and maximizing profit. The foundry has four existing old alloys containing tin and lead:

Let be the weight of the old alloy . The new alloy is produced by a combination of the old alloys:

The cost to acquire one pound of each of the four alloys is:

The cost to produce the new alloy is:

The foundry sells the new alloy for $200. The total profit that the foundry makes is:

The foundry must produce at least 15 lbs of the new alloy containing at most 60% lead and 35% tin:

The foundry has access to only 12, 15, 16 and 10 lbs of the four old alloys respectively:

The optimal mix of the old alloys to use can be found by minimizing the ratio of cost to profit:

Find the allocation of a maximum of $250,000 of capital to purchase two stocks and a bond such that the return on investment is maximized while reducing the cost of purchasing the stocks/bond. Let be the amount to invest in the two stocks and let be the amount to invest in the bond:

The amount invested in the utilities stock cannot be more than $40,000 and pay a 9% dividend:

The amount invested in the bond must be at least $70,000 and pay 5% interest:

The total amount invested in the two stocks must be at least half the total amount invested:

The electronics stock pays a 4% dividend annually. The total return on investment is:

The cost associated with the investment is:

The optimal amount to be invested can be found by minimizing the ratio of cost to profit:

The total amount invested and the annual dividends received from the investments are:

Find the optimal combination of investments that yields maximum profit while minimizing cost. The net present value and the costs associated with each investment are:

Let be a decision variable such that if investment is selected. The objective is to minimize costs while maximizing profits:

There is a maximum $14,000 available for investing:

Solve the maximization problem to get the optimal combination of investments:

Find the optimal combination of doctors a hospital ER must keep on call so that the ER can perform a list of procedures. Each doctor can only perform a certain number of procedures:

The cost (per hour) for keeping each doctor on call is:

Let be a decision vector such that if doctor is on call. The objective is to minimize cost while maximizing the total doctors present in the ER:

At least one doctor must perform procedure :

Find the combination of doctors to keep on call:

Find the optimal number of fire stations that a city containing six districts must build such that there is at least one fire station within 15 minutes of each district. The time required to drive between districts is:

The cost (in millions of dollars) to build a fire station in each city is:

Let be a decision vector such that if a fire station is built in district . The objective is to minimize the cost for the largest number of fire stations:

At least one fire station must be within 15 minutes of each district:

Find the districts in which a fire station will be built:

The total cost is:

Find the optimal number of storage depots a company needs to build to distribute to six of its retail stores. The company has selected five possible sites. The cost to supply each store from each depot is:

The construction cost (in millions) for each depot is:

Let be a decision variable such that if depot is built. Let represent the fraction of goods that depot supplies to store . The total cost is:

The company expects to make a profit of $1 million from each depot by renting out unused space:

Each store must receive all the goods:

Only the depots that are built can supply the goods to the store:

Find which of the five depots should be constructed:

Find the depots that will be built:

The depots that are supplying the retail stores are:

Find the path that a salesman should take through cities such that each city is only visited once and that minimizes the distance and maximizes travel cost savings. Generate the locations:

Let be the distance between city and city . Let be a decision variable such that if , the path goes from city to city :

The travel budget to go from one city to another is $15. If two cities are are within 50 miles, the salesman pays $5, else pays $10 flat rate. The total savings are:

The objective is to minimize distance while maximizing the savings:

The salesman can arrive from exactly one city and can depart to exactly one other city:

The salesman cannot arrive to one city and depart to the same city:

The salesman must travel to all the locations in a single tour:

The decision variable is a binary variable and the dummy variable is :

Find the path that minimizes the distance:

Extract the path:

The distance traveled is:

The total savings is: