Introduction to management science solution manual

Machine Hours Schedule: Machine 1 The objective coefficient range for H is No Lower Limit to 0. Since 0. The dual price for constraint 1 is 0.

The right-hand-side range for constraint 1 is 0 to No Upper Limit. The second constraint now becomes RoundTree should accept SuperSaver reservations, 60 Deluxe reservations and 50 Business reservations. Yes, the effect of a person upgrading is an increase in demand for Deluxe accommodations from 60 to Convert to a Type I room.

We would need the forecast of demand for each rental class on the next night. Using the demand forecasts, we would modify the right-hand sides of the first three constraints and resolve. The dual price for constraint 5 is 0. Therefore, additional hours for Lisa will not change the solution. The dual price for constraint 3 is 0. Because there is No Lower Limit on the right-hand-side range, the optimal solution will not change.

Resolving the problem without this constraint will also show that the solution obtained in b does not change. Constraint 3, therefore, is really a redundant constraint.

Nothing, since there are minutes of slack time on the grinder at the optimal solution. No, since at that price it would not be profitable to produce any of component 3. To get data on a per share basis multiply price by rate of return or risk measure value. Obtaining a reasonable return with a lower risk is a preferred strategy in many financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk.

For example, for O1 is the product of 50 and 36, for O2 is the product of 42 and 42 and so on. No, since cabinetmaker 1 has a slack of Alternatively, since the dual price for constraint 1 is 0, increasing the right hand side of constraint 1 will not change the value of the optimal solution. The dual price for constraint 2 is 1. The upper limit on the right-hand-side range is The solution to Hartmann's portfolio mix problem is given.

These are given by the objective coefficient ranges. The portfolio above will be optimal as long as the yields remain in the following intervals: Growth stock 0. The dual price for the first constraint provides this information. A change in the risk index from 0. This is within the right- hand-side range, so the dual price of 1. The value of the optimal solution would increase by 1. Hartmann's yield with a risk index of 0. This change is outside the objective coefficient range so we must re-solve the problem.

The solution is shown below. The client's risk index and the amount of funds available. With the new yield estimates, Pfeiffer would solve a new linear program to find the optimal portfolio mix for each client. Then by summing across all 50 clients he would determine the total amount that should be placed in a growth fund, an income fund, and a money market fund. Pfeiffer then would make the necessary switches to have the correct total amount in each account.

There would be no actual switching of funds for individual clients. Relevant cost since LaJolla Beverage Products can purchase wine and fruit juice on an as - needed basis. No; only gallons of the rose are currently being used. Note to instructor: Although this explanation is technically correct, it does not provide an explanation that is especially useful in the context of the problem.

Note that in this case, the first constraint would change to 0. This shows the student that the coefficients on the left-hand side are changing; note that this is beyond the scope of sensitivity analysis discussed in this chapter. These interpretations are valid for increase up to 10 minutes and decreases up to 2 minutes from the current level of 20 minutes. Increasing the sports by one minute will have no effect for this constraint since the dual price is 0.

The new optimal solution would increase the copies assigned to Benson printing to 30, In this case, the additional copies assigned to Benson Printing would reduce on a one-for-one basis the number assigned to Johnson Printing. Learn about applications of linear programming that have been encountered in practice. Develop an appreciation for the diversity of problems that can be modeled as linear programs. Obtain practice and experience in formulating realistic linear programming models.

Understand linear programming applications such as: media selection production scheduling portfolio selection work force assignments blending problems Note to Instructor The application problems of Chapter 4 have been designed to give the student an understanding and appreciation of the broad range of problems that can be approached by linear programming.

While the problems are indicative of the many linear programming applications, they have been kept relatively small in order to ease the student's formulation and solution effort. Each problem will give the student an opportunity to practice formulating a linear programming model. This information can be obtained from The Management Scientist as follows. The dual price for the budget constraint is The right-hand-side range for the budget constraint will show this interpretation is correct.

The dual price for Dept. Therefore we would attempt to schedule overtime in Departments A and B. A 10 hrs. No funds are placed in savings for years 3, 4 and 6. To reformulate this problem, one additional variable needs to be added, the right-hand sides for the original constraints need to be shifted ahead by one, and the right-hand side of the first constraint needs to be set equal to zero.

The constraints for the six four-hour periods are as follows: Time of Day a. Let each decision variable, A, P, M, H and G, represent the fraction or proportion of the total investment placed in each investment alternative.

Marginal rate of return is. Changing the right-hand-side value for constraint 2 to 0. Note that a maximum risk of 0. Thus, this investor is willing to take more risk than the solution shown above provides.

For the data given here, the investor should ask the investment advisor to relax either or both of these constraints. Defining the decision variables as proportions means the investment advisor can use the linear programming model for any investor, regardless of the amount of the investment. All the investor advisor needs to do is to establish the maximum total risk for the investor and resolve the problem using the new value for maximum total risk. Max 6.

During the 4-week period, ASC has taken profits to reinvest and build inventory by pounds in anticipation of future higher prices. The dual prices tell us how much it would cost if demand were to increase by one additional unit. The dual price of 0 for Quarter 4 tells us we have excess capacity in Quarter 4. The positive dual prices in Quarters tell us how much increasing the production capacity will improve the objective function.

Only the objective function needs to be changed. An objective function minimizing waste production and the new optimal solution are given.

Perhaps these can be inventoried for future use. Minimizing waste may cause you to over-produce. Alternative b might be preferred on the basis that the surplus could be held in inventory for later demand. However, in some trim problems, excess production cannot be used and must be scrapped. Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used.

Nothing, there are already more hours available than are being used. The last two constraints in the formulation above must be deleted and the problem resolved. Actually this is an integer programming problem, since partial tankers can't be purchased. We were fortunate in part a that the optimal solution turned out integer.

To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to , , 50, and 50 respectively. The revised optimal solution is given. This seems to be a small expense to have less drastic labor force fluctuations. The new labor levels are , , and You may want to experiment yourself to see what happens. Note that constraints guarantee that next week's schedule will be met and constraints enforce machine capacities. Machine 3 has minutes of idle capacity.

Each yi appears in three constraints because each refers to a three hour shift. The optimal solution is shown below. Still have 20 part-time shifts, but 14 are 3-hour shifts. The surplus has been reduced by a total of 14 hours. Learn about applications of linear programming that are solved in practice. Understand linear programming applications such as: data envelopment analysis revenue management portfolio selection game theory 5.

Know what is meant by a two-person, zero-sum game. Be able to identify a pure strategy for a two-person, zero-sum game. Be able to use linear programming to identify a mixed strategy and compute optimal probabilities for the mixed strategy games. Understand the following terms: game theory two-person, zero-sum game saddle point pure strategy mixed strategy Note to Instructor The application problems of Chapter 5 are designed to give the student an understanding and appreciation of the broad range of problems that can be approached by linear programming.

Min E s. The composite hospital is General Hospital. For any hospital that is not relatively inefficient, the composite hospital will be that hospital because the model is unable to find a weighted average of the other hospitals that is better.

D is relatively inefficient Composite requires Hospitals A, C, and E. Make the following changes to the model in problem Hospital E is the only hospital in the composite. If a hospital is not relatively inefficient, the hospital will make up the composite hospital with weight equal to 1. Consider the Bardstown, Jeffersonville, and New Albany restaurants. Thus, it would be better to locate the larger plane in Newark. The bid prices for each ODIF are provided by the deal prices in the following output.

The formulation and output obtained using The Management Scientist is shown below. Max The solution obtained using The Management Scientist is shown. The expected return for this portfolio is One constraint must be added to the model in part a. The expected return for this portfolio is 1. Max 9. The expected portfolio return is The portfolio model is modified by changing the right-hand side of the first 5 constraints from 2 to 0.

This is an increase of. Most investors would conclude that the small increase in the portfolio return is not enough to justify the increased risk. Player B b1 b2 b3 Minimum Player A a1 8 5 7 5 a2 2 4 10 2 Maximum 8 5 7 The maximum of the row minimums is 5 and the minimum of the column maximums is 5.

The game has a pure strategy. Player A should take strategy a1 and Player B should take strategy b2. The value of the game is 5. By definition, a pure-strategy solution means each player selects a single strategy with probability 1. Thus, if a linear programming solution has a probability decision variable equal to 1, the game has a pure-strategy solution.

The row minimums are -5, 6, and 4. Station A prefers the maximin strategy a2 to obtain a gain of at least 6. The column maximums are 10, 8 and 7. Station B prefers the minimax strategy b3 to limit its maximum loss to no more than 7.

However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed-strategy solution with a value of the game between 6 and 7 exists. The linear programming formulation and solution for Station A follows. Using the absolute value of the dual prices, we see that it is optimal for Station B to implement strategy b1 with probability 0.

The expected value of the game is 6. This is an expected increase of viewers for Station A. The row minimums are , , , and The Republican candidate prefers the maximin strategy a4 to obtain a gain of at least The column maximums are 30, 20, 10, and Station B prefers the minimax strategy b3 to limit its maximum loss to no more than The maximum of the row minimums is equal to the minimum of the row maximums.

The value of the game shows a 10, voter increase for the Republican candidate. The row minimums are -1, -3, and Player A prefers the maximin strategy Red to obtain a payoff of at least The column maximums are 5, 4 and 2. Player B prefers the minimax strategy Blue to limit its maximum loss to no more than 2. However, because the maximum of the row minimums is not equal to the minimum of the column maximums, the game does not have a pure strategy. A mixed-strategy solution with a value of the game between -1 and 2 exists.

The linear programming formulation and solution for Player A is as follows. Using the absolute value of the dual prices, the optimal mixed strategy for Player B is to select a white chip with a 0. The value of the game is 0.

This is an expected gain of 50 cents for Player A. Player A is the preferred player. Over the long run, Player A average winning 50 cents per game. To make the value of the game 0 and thus, fair for both players, Player A should pay Player B to play the game. The game would be considered fair if Player A pays Player B 50 cents per game. The row minimums are 0, -2, 2 and Company A prefers the maximin strategy a3 to obtain a payoff of at least 2. The column maximums are 4, 6, 5 and 6.

Player B prefers the minimax strategy b1 to limit its maximum loss to no more than 4. A mixed-strategy solution with a value of the game between 2 and 4 exists.

Using the dual prices, the optimal mixed strategy for Company A is to select strategy a3 with a probability 0. The expected gain for Company A is 2. The Packers prefer the minimax strategy pass defense to limit its maximum loss to no more than 6 yards.

A mixed-strategy solution with a value of the game between 2 and 6 exists. The linear programming formulation and solution for the Bears is as follows. Using the absolute value of the dual prices, the optimal mixed strategy for the Packers is to use a run defense with a 0. The expected value of the game shows that with the mixed-strategy solution, the Bears average 4. Be able to identify the special features of the transportation problem.

Become familiar with the types of problems that can be solved by applying a transportation model. Be able to develop network and linear programming models of the transportation problem. Know how to handle the cases of 1 unequal supply and demand, 2 unacceptable routes, and 3 maximization objective for a transportation problem. Be able to identify the special features of the assignment problem. Become familiar with the types of problems that can be solved by applying an assignment model.

Be able to develop network and linear programming models of the assignment problem. Be familiar with the special features of the transshipment problem. Become familiar with the types of problems that can be solved by applying a transshipment model. Know the basic characteristics of the shortest route problem. Be able to develop a linear programming model and solve the shortest route problem. Know the basic characteristics of the maximal flow problem.

Be able to develop a linear programming model and solve the maximal flow problem. Know how to structure and solve a production and inventory problem as a transshipment problem. The network model is shown. Atlanta 2 Phila. Hamilton 1 10 Southern 20 1 15 Butler 12 2 15 Northwest 2 18 Clermont 3 b. To answer this question the simplest approach is to increase the Butler County demand to and to increase the supply by at both Southern Gas and Northwest Gas.

The linear programming formulation and optimal solution as printed by The Management Scientist are shown. It is summarized. A 1 Avery 1 B 75 2 Baker 2 C 3 Campbell 3 D 85 4 b.

The network model, the linear programming formulation, and the optimal solution are shown. Note that the third constraint corresponds to the dummy origin.

The variables x31, x32, x33, and x34 are the amounts shipped out of the dummy origin; they do not appear in the objective function since they are given a coefficient of zero. Paul 80 b. There are alternative optimal solutions. Solution 1 Solution 2 Denver to St. Paul: 10 Denver to St. Paul: 70 Chicago to St.

There will be idle capacity for 90 motors at Denver. If solution 2 is used, Forbelt should adopt the same production schedule but a modified shipping schedule. The linear programming formulation and optimal solution are shown. This can be formulated as a linear program with a maximization objective function.

There are 10 constraints, 6 for the potential programs and 4 for the time slots. Hollywood Briefings — p.

World News — p. This is the variation of the assignment problem in which multiple assignments are possible. Each distribution center may be assigned up to 3 customer zones. The linear programming model of this problem has 40 variables one for each combination of distribution center and customer zone. It has 13 constraints. The problem can also be solved using the Transportation module of The Management Scientist. The optimal solution is given below. The Nashville distribution center is not used.

All the distribution centers are used. Columbus is switched from Springfield to Nashville. A linear programming formulation and the optimal solution are given.

For the decision variables, xij, we let the first subscript correspond to the supplier and the second subscript correspond to the distribution hub. A linear programming formulation of this problem can be developed as follows. Let the first letter of each variable name represent the professor and the second two the course.

Note that a DPH variable is not created because the assignment is unacceptable. Max 2. The solution will not change, but the total hours required will increase by 5. This is the extra time required for statistician 4 to complete the job for client A.

The solution will not change, but the total time required will decrease by 20 hours. The solution will not change; statistician 3 will not be assigned. Note that this occurs because increasing the time for statistician 3 makes statistician 3 an even less attractive candidate for assignment.

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. The work is protected by local and international copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.

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Download Resources. Previous editions. Pharmacists Technicians. We encourage our students to proceed in a systematic step-by-step fashion when formulating these types of constraints.

If we think of this situation as an on-going continuous production process, the fractional values simply represent partially completed products. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible it had an optimal solution. Every solution that was feasible is still feasible when we change the constraint to lessthan-or-equal-to, since the new constraint is satisfied at equality as well as inequality.

In summary, we have relaxed the constraint so that the previous solutions are feasible and possibly more satisfying the constraint as strict inequality. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-thanor-equal to constraint as shown below. Constraints 2,3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible. It makes no sense to add this constraint.

Then the new constraint makes the problem infeasible. Millions discover their favorite reads on issuu every month. Give your content the digital home it deserves.