It costs $10 for raw materials and $14 for labor to make one soldier; in carpentry it takes 1 hour, and finishing works 2 hours. Making one train costs $9 for raw materials and $10 for labor; in carpentry it takes 1 hour and finishing works 1 hour.
Activity-based formulation
Profit: maximize (flow out) Activities:. production of a type chair where each is assigned an activity level Chair 1: Production of 1 type 1 chair consumes 1 lb of steel 4 lbs of Wood produces $12 profit. We have already seen the standard form of a maximization problem; this is the same except that we minimize the objective function and switch the signs of the inequalities (this is only done for the sake of convenience when we get to solving the LPs).
Summary and further tricks
If a linear program has an optimal solution, then it also has an optimal solution that is a vertex of the feasible region. Iso value line≡in all points on this line the objective function has the same value.
Fourier-Motzkin Elimination (FME)
To change an inequality into an equation, we add a new non-negative variable called a slack variable.
Simplex method by example
The fit is obtained by expressing the basic variables from the initial set of equations. STOP, current solution is optimal. set non-basic variables to zero. reads the values of the basic variables and the objective function z.
Two phase Simplex method
Special cases
Phase I
This operation does not change the solutions for this system (if we multiply by a non-singular matrix) We can extend it back to the system of linear equations. How do we get an inverse matrix. we perform elementary row operations on the following matrix. multiply a row by a non-zero number. append a row to another row. swap two rows.
Summary
From this we immediately see that the corresponding basic solution whenxN=0 is given by xB=B−1b with the objective function value z=cTBB−1b The non-basic variables in the objective function are. From any dictionary (basis depends on the line) we can always express the objective function under the line z=3x1+2x2=3(.
Changing the right-hand side value
CHANGE OF THE RIGHT SIDE VALUE 37Non-basic variable: If Iisj-the non-basic variable, then only the coefficient of changes inz, it is increased by ∆,.
Detailed example
Adding a variable/activity
Adding a constraint
Adding this to the final dictionary gives a feasible dictionary (since the value 10 in the constraint is nonnegative)→ the optimal solution remains optimal after adding the constraint. Then the original optimal solution becomes infeasible in the modified problem (the constant term works out to -10 instead of 10) and we have to recalculate (or use the simple dual method). Conclusion: as long as we have at least 140 carton units, we do not need to change the production plan.
Modifying the left-hand side of a constraint
For the sake of illustration, we omit the 3rd constraint, and consider the items to be blocks of wood and cans of paint (instead of store hours). He can sell his stock at market prices or buy additional shares at market prices. The maker can produce 1 toy soldier by purchasing 1 block of wood and 2x1 additional cans of paint.
Similarly, there is no profit limit if the cost of making the x2toy trains is lower than the selling price, i.e. Before solving the LP, the producer wants to get a quick rough estimate (upper bound) of the value of the optimal solution. The conclusion is that each production plan will generate no more than $240, which means that the value of any feasible solution (including the optimal one) is no more than $240.
Duality Theorems and Feasibility
General LPs
If x1 is unbounded, we can simply insist that the coefficienty1+2y2 be equal to 3 for the upper bound to work.
Complementary slackness
If Y is a feasible solution in the pair and is a complementary poison, then it is optimal in the pair. If x is a basic feasible primal solution and π are the corresponding shadow prices, then x is optimal if and only if π is a feasible dual solution. Let's remember: for each (basic) Primal solution, we have shadow prices that we can assign to each item (constraint) in such a way that the total value of the items in the shadow prices is exactly the same as the value of the solution.
Shadow prices corresponding to a non-optimal feasible solution of Primal are infeasible in Dual Shadow prices corresponding to an optimal solution of Primal are optimal in Dual. Shadow prices for an unmanageable solution of Primal may or may not be feasible in Dual. A solution of Primal is doubly possible if the corresponding shadow prices are possible in dual.
Upper-Bounded Simplex
Turnxjinto the base,xleave the base. the resulting dictionary is guaranteed to be doubly feasible) 6.
Lower bounds
Dual Simplex with Upper Bounds
Goal Programming
On the other hand, we can allow x5, x6, and x7 to be unbounded, since the respective constraints are soft—we can violate any of them, but incur a penalty. Problem and its solution first described in 1941 by Hitchcock - predates the Simplex Method - discovered independently by Koopmans during World War II used to minimize transit time for cargo ships - guided method from research on linear and non-linear programs in economics. If total demand is greater than total supply for ∆>0, then no feasible solution exists.
Since we are only interested in price differences, we can simply assume that one of the shadow prices is zero. A generalization of the transportation problem where we can send some goods via several intermediate (transshipment) locations. What is the critical path for production ie. the sequence of tasks that delays any one of them delays the entire production.
Minimum Spanning Tree
Maximum Flow problem
MAXIMUM FLOW PROBLEM 71 no route found from Denver to Miami in the remaining network→ no augmenting chain found→ optimal solution. For a subset of verticesA⊆V, the edges that go between the nodes inA and the rest of the graph are called cuts. The ends coming out of the Aare are called the front ends, the ends coming into the Aare are the back ends.
The maximum value of an(s,t)-flow is equal to the minimum capacity of an(s,t)-cut.
Minimum-cost Flow problem
Network Simplex Algorithm
To find shadow prices, we use complementarity – the dual constraints corresponding to fundamental variables must be tight. The solution is thus not optimal, and we can get a better solution by increasing the flow on the edge DH (has positive reduced price). We go around this cycle and increase/decrease the flow on the edges of the cycle by ∆.
That is, if the edge is in the same direction as (i,j), we change the flow at this edge by ∆; if it is in the opposite direction, then we change it with. We then modify the flow at the edges of the loop by adding ∆ to the flow at the edges in the same direction as (i,j) and subtracting ∆ in the rest. That is, the flow at the edge HM increases by ∆, since this edge has the same direction (along the loop) as DH.
Network Simplex Algorithm with capacitites
Complete example
Knowing the shadow price of Slack, we calculate the shadow prices of nodes connected to Slack using basic edges; we use the fact that shadow prices satisfyyi−yj =cijfor each basic edge(i,j); ifj=S, then the value is known and we plug it into getyi. Edge(H,M) has zero flow and positive reduced cost→adding it to base can make the target smaller. Maximum∆ is 0 because the artificial edge(H,S) has no flow on it and therefore the flow on this edge cannot decrease.
The flow at the edge (H,M) has increased to∆(since (H,M) has the same direction as (C,H) in the loop), while at (M,S) it has decreased to 2−∆and on(C ,S)decreases to−∆(both have opposite directions). Note that from now on we use the original edge costs (unlike Phase I where we cost $0 for real edges and $1 for artificial edges). This implies that the largest∆is∆=0, since we cannot increase the flow to (N,M) as it is already saturated.
Summary
This is called a saddle point and the common value of both sides of the equation is called the value of the game. An equilibrium point of the game: choice of strategies for both players so that neither player can improve their outcome by changing their strategy. Please note that this is a zero-sum game as Player 1 pays Player 2 or vice versa (the sum of the players' winnings is zero).
From this we determine that the best mixed strategy for Player 1 is the pointE corresponding to strategy(x1,x2)wherex1=19/24andx2=5/24. So the best strategy for player 2 is to call with a probability of 6623% and fold with a probability of 3313%. This tells us that the optimal solutions for the two programs have the same value by Strong Duality, which defines the value of the game.
Nonconstant-sum Games
Moreover, the optimal strategies for the two players satisfy complementary slackness (verify this for the solutions we found above). In other words, the solutions to the two problems form an equilibrium point (neither game can do better by changing his/her strategy). Important: must make sure that x+andx− are never both basically in the optimum, otherwise this does not work alternatively: 2 constraints f ≤x+ and −f ≤x− (with similar caveats).
The first two constraints ensure that exactly one edge enters and one edge leaves each node.
Cutting Planes
CUTTING PLANES 93 However, let's multiply the first constraint by 3 and add it to the second constraint. We choose the expression forzand and move whole parts to the left, leaving only fractions on the right. We rearrange the expression so that on the right-hand side each variable has a negative fraction coefficient, while the absolute constant is a positive fraction.
Because the coefficients of the variables on the right-hand side are negative, the value of the right-hand side is 2/3 or less (it cannot be more, since the variables are non-negative). Otherwise in the optimal dictionary, select any line with partial constants. a) Rewrite the line by moving whole parts to the left so that. Rewrite by moving integers to the left so that the coefficient of the variables on the right is negative.
Branch and Bound
We can shorten the branching process if we have some heuristic, a way to bind or evaluate the value of the objective function in the following sub-examples. We don't know which of the two situations happens, but if we take the better of the two, we always choose the right one. At the end of the month, the holding charge is $0.50 for each unit on hand.
DYNAMIC PROGRAMMING of the production line enables the production of a maximum of 5 units per month. First, let's consider the number of steps (operations) of each of the methods in rough numbers. For two functions f,g:R →R, we say that f O(g) is of "order g", and we write f =O(g) if there exist constants>0 and N>0, such that.