Professor of Anthropology Bruce Hardy makes headlines for his study on the cognitive abilities of Neanderthals. A senior English major talks with her advisor about his philosophy on creative writing and teaching. Kenyon alumni working in food and drink industries are sharing their go-to happy hour pairings. For instance, a 3-week schedule for the first of these divisions is shown below: Week 6 R1 vs.
W1 R2 vs. W2 R3 vs. A similar 3-week schedule can be easily set up for the other two new divisions. This will provide us with a schedule for the first 8 weeks of the season. For the final 9 weeks, we continue to create new divisions by pairing 3 teams from the original Red, White and Blue divisions with 3 teams from the other divisions that they have not yet been paired with.
Then a 3week schedule is developed as above. Shown below is a set of divisions for the next 9 weeks. This Red, White and Blue scheduling procedure provides a schedule with every couple playing every other couple over the week season.
If one of the couples should cancel, the schedule can be modified easily. Designate the couple that cancels, say R4, as the Bye couple. Then whichever couple is scheduled to play couple R4 will receive a Bye in that week. With only 17 couples a Bye must be scheduled for one team each week. This same scheduling procedure can obviously be used for scheduling sports teams and or any other kinds of matches involving 17 or 18 teams. Modifications of the Red, White and Blue algorithm can be employed for 15 or 16 team leagues and other numbers of teams.
Management would not implement this solution because no units of the DI would be produced. Time spent on Line 1: 3 This might be a concern to management if no other work assignments were available for the employees on Line 2. Line 1 is scheduled for minutes and Line 2 for minutes.
The total profit contribution is 42 But, this solution results in perfect workload balancing because the total time spent on each line is minutes. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as The Management Scientist.
In the short run the requirement reduces profits. Since the optimal solution uses all of the available machining and welding time, management should try to obtain additional hours of this resource. Security Systems Market Analysis Total.
The total cost is the sum of the purchase cost and the transportation cost. We show the calculation for Division 1 - Supplier 1 and present the result for the other Division-Supplier combinations. The linear programming formulation and solution as printed by The Management Scientist is shown. Three arcs must be added to the network model in problem 23a. The new network is shown. There is now excess capacity of units at plant 1. The negative numbers by nodes indicate the amount of demand at the node.
Origin — Node 1 Transshipment Nodes 2 to 5 Destination — Node 7 The linear program will have 14 variables for the arcs and 7 constraints for the nodes. The linear program has 13 variables for the arcs and 6 constraints for the nodes.
Use same six constraints for the Gorman shortest route problem as shown in the text. The objective function changes to travel time as follows. Origin — Node 1 Transshipment Nodes 2 to 5 Destination — Node 6 The linear program will have 13 variables for the arcs and 6 constraints for the nodes. Allowing 8 minutes to get to node 1 and 69 minutes to go from node 1 to node 6, we expect to make the delivery in 77 minutes.
It is p. Guarantee delivery by p. Origin — Node 1 Transshipment Nodes 2 to 5 and node 7 Destination — Node 6 The linear program will have 18 variables for the arcs and 7 constraints for the nodes. Origin — Node 1 Transshipment Nodes 2 to 9 Destination — Node 10 Identified by the subscript 0 The linear program will have 29 variables for the arcs and 10 constraints for the nodes.
Origin — Node 0 Transshipment Nodes 1 to 3 Destination — Node 4 The linear program will have 10 variables for the arcs and 5 constraints for the nodes. The capacitated transshipment problem to solve is given: Max x61 s. The maximum number of messages that may be sent is 10, Flow reduced to 9, gallons per hour; Five gallons will flow from node 3 to node 5. Modify the problem by adding two nodes and two arcs. Let node 0 be a beginning inventory node with a supply of 50 and an arc connecting it to node 5 period 1 demand.
Let node 9 be an ending inventory node with a demand of and an arc connecting node 8 period 4 demand to it. Let R1, R2, R3 represent regular time production in months 1, 2, 3 O1, O2, O3 represent overtime production in months 1, 2, 3 D1, D2, D3 represent demand in months 1, 2, 3 Using these 9 nodes, a network model is shown. Use the following notation to define the variables: first two letters designates the "from node" and the second two letters designates the "to node" of the arc.
For instance, R1D1 is amount of regular time production available to satisfy demand in month 1, O1D1 is amount of overtime production in month 1 available to satisfy demand in month 1, D1D2 is the amount of inventory carried over from month 1 to month 2, and so on.
The values of the slack variables for constraints 1 through 6 represent unused capacity. The only nonzero slack variable is for constraint 2; its value is Thus, there are 75 units of unused overtime capacity in month 1. Be able to recognize the types of situations where integer linear programming problem formulations are desirable. Know the difference between all-integer and mixed integer linear programming problems. Be able to solve small integer linear programs with a graphical solution procedure.
Be able to formulate and solve fixed charge, capital budgeting, distribution system, and product design problems as integer linear programs. See how zero-one integer linear variables can be used to handle special situations such as multiple choice, k out of n alternatives, and conditional constraints. Be familiar with the computer solution of MILPs. This is a mixed integer linear program.
This is an all-integer linear program. Its LP Relaxation just requires dropping the words "and integer" from the last line. Its value is This is not the same solution as that found by rounding down. It provides a 3 unit increase in the value of the objective function.
Its value is 5. The optimal integer solution is the same as the optimal solution to the LP Relaxation. This is always the case whenever all the variables take on integer values in the optimal solution to the LP Relaxation. Since we have all less-than-or-equal-to constraints with positive coefficients, the solution obtained by "rounding down" the values of the variables in the optimal solution to the LP Relaxation is feasible.
Thus a lower bound on the value of the optimal solution is given by this feasible integer solution with value An upper bound is given by the value of the LP Relaxation, Actually an upper bound of 36 could be established since no integer solution could have a value between 36 and The solution found by "rounding down" the solution to the LP relaxation had a value of Optimal solution to 6 LP relaxation 0,5.
These two values provide an upper bound of In this case rounding the LP solution down does provide the optimal integer solution. This solution is clearly not optimal. Its value is 7. Thus an upper bound on the value of the optimal is given by 7. Thus a lower bound on the value of the optimal solution is given by 6. The following mutually exclusive constraint must be added to the model.
The following co-requisite constraint must be added to the model in b. No change in optimal solution. Choose locations B and E. We introduce a variable yi that is one if any quantity of product i is produced and zero otherwise. We must add a constraint requiring 60 tons to be shipped and an objective function. One just needs to add the following multiple choice constraint to the problem. Since one plant is already located in St.
Let 1 denote the Michigan plant 2 denote the first New York plant 3 denote the second New York plant 4 denote the Ohio plant 5 denote the California plant It is not possible to meet needs by modernizing only one plant.
The following table shows the options which involve modernizing two plants. Modernize plants 1 and 3 or plants 4 and 5. A population of , cannot be served by this solution.
Counties and 10 will not be served. The only change necessary in the integer programming model for part a is that the right-hand side of the last constraint is increased from 1 to 2. The optimal solution has principal places of business in counties 3 and 11 with an optimal value of 76, Only County 10 with a population of 76, is not served. It is not the best location if only one principal place of business can be established; 1,, customers in the region cannot be served. However, , can be served and if there is no opportunity to obtain a principal place of business in County 11, this may be a good start.
Perhaps later there will be an opportunity in County The solution to the LP Relaxation is integral therefore it is the optimal solution to the integer program.
A difficulty with this solution is that only part-time employees are used; this may cause problems with supervision, etc. The large surpluses from 5, 4 employees , and 9 employees indicate times when the tellers are not needed for customer service and may be reassigned to other tasks.
Add the following constraints to the formulation in part a. The new solution uses 5 full-time employees and 12 part-time employees; the previous solution used no full-time employees and 21 part-time employees. Locating a principal place of business in Ashland county will permit Ohio Trust to do business in all 7 counties.
Add the part-worths for Antonio's Pizza for each consumer in the Salem Foods' consumer panel. This calls for a pizza with a thick crust, a cheese blend, a chunky sauce, and medium sausage. The coefficients for the yi variable must be changed to in constraints and to in constraints Four children will prefer this design: 1, 2, 4, and 5.
Thus, cameras should be located at 4 openings: 1, 5, 8, and Thus, cameras should be located at openings 3, 6, 9, 11, and A mixed integer linear program can be set up to solve this problem. Binary variables are used to indicate whether or not we setup to produce the subassemblies. There are 11 constraints.
Constraints 1 to 5 are to satisfy demand. Constraint 6 reflects the limitation on manufacturing time. Finally, constraints 7 - 11 are constraints not allowing production unless the setup variable equals 1. This part can be solved by changing appropriate coefficients in the formulation for part a. The new optimal solution is shown below. Variable for movie 1: x, x, x b.
Only 2-screens are available at the theater. Note: alternate optimal solution is New York, Richmond and Tampa. Note: alternate optimal solution: Boston, Philadelphia, Florence and Tampa. Learn about applications of nonlinear programming that have been encountered in practice. Develop an appreciation for the diversity of problems that can be modeled as nonlinear programs.
Obtain practice and experience in formulating realistic nonlinear programming models. Understand nonlinear programming applications such as: Allocating a media budget Markowitz portfolio selection Minimizing value at risk The Black-Scholes option pricing model Selecting the smoothing parameter used in the exponential smoothing model Selecting the parameters in the Bass forecasting model The quadratic assignment problem The Cobb-Douglass production function Production scheduling between multiple plants or facilities Note to Instructor The application problems of Chapter 8 have been designed to give the student an understanding and appreciation of the broad range of problems that can be modeled as a nonlinear program.
Each problem will give the student an opportunity to practice formulating a nonlinear programming model. The Management Scientist software cannot solve nonlinear programming problems. If we resolve the problem with a new right-hand-side of 9 the new optimal objective function value is 4.
This formulation involves 6 variables and 4 constraints. With the previous formulation the solution only provided the values of the prices. The other information was computed from that. This also shows that there can be more than one correct formulation of a problem. The formulation in part c is an unconstrained problem. The formulation in part d is a constrained problem. We simply add a budget constraint to the sales function that is to be maximized.
This is a local minimum. The optimization model is Max 5 L. Let OT be the number of overtime hours scheduled. Define the variables to be the dollars invested the in mutual fund. Objective value: Below is a screen capture of an Excel Spreadsheet Solver model for this problem. In Cell E16 is the sum of squared errors. This is what we minimize by using solver. In the solver parameters dialog box, we use cell E16 as the target cell and cell B1 as the changing cell.
Here are the returns calculated from the Yahoo stock data. Objective value: 0. In order to measure the semi-variance it is necessary to measure only the downside deviation below the mean. Do this by introducing two new variables for each scenario. For example, for scenario 1 define D1P as the deviation of return 1 above the mean, and D1N as the deviation of return 1 below the mean.
Then in the objective function, we minimize the average of the negative deviations squared, i. Objective value: 7. Excel Solver may have trouble with this problem, depending upon the starting solution that is used. A starting solution of each fund at. The selection of stocks that go into the portfolio determine the mean return of the portfolio and the standard deviation of portfolio returns. The number 2. Although people commonly talk about minimizing the Value at Risk of a portfolio, this terminology is misleading.
Objective value: 2. No, minimizing risk is not the same thing as minimizing VaR. This problem may also be solved with Excel Solver.
Black Scholes Model! BID - 1. ASK - 1. First consider the constraints. Every tanker must be assigned to a loading dock. These constraints are as follows. DOCK 3; The constraints that require each tanker to be assigned a loading dock, and each loading dock assigned a tanker form the constraint set for the assignment problem. The assignment problem was introduced in Chapter 6. However, unlike the assignment problem the objective function in this problem is nonlinear.
Consider, for example, the result of assigning tanker 1 to dock 2 and tanker 3 to dock 1. The distance between loading docks 1 and 2 is meters. Also, tanker 1 must transfer 80 tons of goods to tanker 3. This means that 80 tons must be moved meters. Depending on the starting point, Excel Solver will likely get stuck at a local optimum and not the find the optimal solution that LINGO finds. The objective is to minimize total production cost. To minimize total product cost minimize the production cost at Aynor plus the production cost at Spartanburg.
Minimize the production cost at the two plants subject to the constraint that total production of kitchen chairs is equal to Duplicate the objective function and constraints given in Section 8. We recommend a start value for P of at least. Learn how to define a project in terms of activities such that a network can be used to describe the project. Know how to compute the critical path and the project completion time. Know how to convert optimistic, most probable, and pessimistic time estimates into expected activity time estimates.
With uncertain activity times, be able to compute the probability of the project being completed by a specific time. Understand the concept and need for crashing.
As a student, completing homework assignments can be challenging. Sometimes you forget the material that you previously learned in class. Other times, the subject matter is very complex and leaves you feeling confused.
On the other hand, maybe you have a very busy schedule and frequently miss the deadline to hand in your homework. B A0 , , 7. There are four extreme points: 0,0 , 4,0 , 3,1,5 , and 0,3. The extreme points are 5, 1 and 2, 4. Material 2: 4 tons are used, 1 ton is unused.
No redundant constraints. Similar to part a : the same feasible region with a different objective function. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem. Department Hours Used Max. Optimal solution occurs at the intersection of constraints 1 and 2. Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints.
Constraint 3 is the nonbinding constraint. Because exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints. The binding constraints are the manufacturing time and the assembly and testing time.
The graphical solution is shown below. There are four extreme points: , ; , ; , ; , d. There are four extreme points: 15,10 ; Extreme Point Cost 15, 15 7. However, the value of the optimal solution is reduced to 7. The value of the optimal solution is reduced to 7.
Invest everything in the stock fund. Feasible Region Constraint Value of Slack Variable Interpretation 1 0 All available grade A crude oil is used 2 0 Total production capacity is used 3 10, Premium gasoline production is 10, gallons less than the maximum demand d. Download instructor resources.