If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt. Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement.
We trust you find the Supplement a useful teaching tool. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. When C changes from 1 to 0; A'C also goes from 1 to 0. Gate delays are assumed to be 10ns in the timing diagram below.
Avoid 1-hazard by adding product term bc. Use NAND gates as asked in the question. Or, one can design a product of sums POS circuit with no hazards. A properly designed 2-level POS circuit has no 1-hazards. Data is clocked into D flip-flop at the rising edge of the clock. X can change as early as 2 ns after the clock edge. However, because the rising clock edge of flip-flop 2 is delayed 5 ns from the rising edge of flip-flop 1, then X can change -3 ns after and 29 ns before the rising clock edge of flip-flop 2.
In other words, X cannot change between 29 and 3 ns before the rising clock edge of flip-flop 2. However, because the rising clock edge of flip-flop 2 is advanced 5 ns from the rising edge of flip-flop 1, then X can change 7 ns after and 19 ns before the rising clock edge of flip-flop 2.
Therefore, the minimum clock period should be 41 ns. Critical path: A-D-F-H b. No, it is a critical activity d. Yes, 2 weeks e. A-D-E-H c. Activity F: 4 weeks d. Project Completion Time 49 weeks. A-C-E-H-I b. Project completion 24 weeks. The park can open within the 6 months 26 weeks after the project is started. Critical activities: B-D-F Expected project completion time: 9. Variance of projection completion time: 0.
Activity Expected Time Variance A 4. Activity Expected Time Variance A 5 0. Activity Expected Time Variance A 8 1. It is very unlikely that the project can be completed in 10 months. But it is also unlikely the project will take longer than 13 months. Therefore, Davison Construction might prefer to tell the owner that the best estimate is a completion time of one year. Activity Expected Time Variance A 2 0. Critical path is the longest path and generally will have the lowest probability of being completed by the desired time.
The noncritical paths should have a higher probability of being completed on time. It may be desirable to consider the probability calculation for a noncritical path if the path activities have little slack, if the path completion time is almost equal to the critical path completion time, or if the path activity times have relatively high variances.
When all of these situations occur, the noncritical path may have a probability of completion on time that is less than the critical path. Activity Expected Time Variance A 1.
Compute the probability of project completion in 13 weeks or less. Activity Expected Time Variance A 4 0. This problem involves the formulation of a linear programming model that will determine the length of the critical path in the network.
Since xI, the completion time of activity I, is the project completion time, the objective function is: Min xI Constraints are needed for the completion times for all activities in the project. The optimal solution will determine xI which is the length of the critical path. Activities A, B, C, and F are critical. The expected project completion time is 31 weeks.
See solution to problem Information needed: 1. Maximum crash time for each activity Mi 2. Crashing cost for each activity Ki 3. Learn where inventory costs occur and why it is important for managers to make good inventory policy decisions. Learn the economic order quantity EOQ model. Know how to develop total cost models for specific inventory systems. Be able to use the total cost model to make how-much-to-order and when-to-order decisions.
Extend the basic approach of the EOQ model to inventory systems involving production lot size, planned shortages, and quantity discounts. Be able to make inventory decisions for single-period inventory models. Know how to make order quantity and reorder point decisions when demand must be described by a probability distribution. Learn about lead time demand distributions and how they can be used to meet acceptable service levels.
Be able to develop order quantity decisions for periodic review inventory systems. Thus the order quantity of is acceptable. This is a production lot size model. However, the operation is only six months rather than a full year. The basis for analysis may be for periods of one month, 6 months, or a full year. The inventory policy will be the same.
In the following analysis we use a monthly basis. Allowing backorders reduces the total cost. Backorder case since the maximum wait is only 5. From the graph we can see that all TC values of category 1 exceed the minimum cost solution of category 2. However, this added cost may be worth the advantage of coordinating orders for multi products.
Go with the periodic review system. For 2. Be able to identify where waiting line problems occur and realize why it is important to study these problems. Know the difference between single-channel and multiple-channel waiting lines. Understand how the Poisson distribution is used to describe arrivals and how the exponential distribution is used to describe services times.
Learn how to use formulas to identify operating characteristics of the following waiting line models: a. Single-channel model with Poisson arrivals and exponential service times b. Multiple-channel model with Poisson arrivals and exponential service times c. Single-channel model with Poisson arrivals and arbitrary service times d. Multiple-channel model with Poisson arrivals, arbitrary service times, and no waiting e.
Single-channel model with Poisson arrivals, exponential service times, and a finite calling population 5. Know how to incorporate economic considerations to arrive at decisions concerning the operation of a waiting line.
Average one customer in line with a 50 second average wait appears reasonable. The average waiting times are identical, but there is a higher probability of waiting and the number waiting increases with the new system.
Average waiting time goal: 5 minutes or less. The two counter system has better service, but has the added cost of installing a new counter. Ocala needs to add more consultants to meet its service guidelines. This service is probably much better than necessary with average waiting time only 24 seconds.
The average number of passengers in the waiting line is 7. The average time for a passenger to move through security screening is 1. To see if 3 screening stations are adequate, we must compute Lq. The average time required for a passenger to pass through security screening is Lq 7. Thus, one postal clerk cannot handle the arrival rate. Average time in system of 7. Post office expansion to allow at least four postal clerks should be considered. Service Cost per Channel System A: 6.
Design B is slightly better due to the lower variability of service times. The company should consider a second channel or other ways of improving the emergency repair service. Four lines will be necessary. The probability of denied access is 0. Five administrative assistants x Understand what simulation is and how it aids in the analysis of a problem. Learn why simulation is a significant problem-solving tool. Understand the difference between static and dynamic simulation.
Identify the important role probability distributions, random numbers, and the computer play in implementing simulation models. Realize the relative advantages and disadvantages of simulation models. Simulation will provide probability information about the various profit levels possible.
What if scenarios show possible profit outcomes but do not provide probability information. Risk analysis would be helpful in evaluating the probability of a loss.
Random Number Direct Labor Cost 0. Sales Interval 0. Stock Price Change Probability Interval For the 10 trials Gustin opened 37 new accounts. Because the seminars are a very profitable way of generating new business, Gustin should continue running the seminars.
Using the random numbers in column 6 beginning with 0. Thus, Atlanta wins the 6-game World Series 4 games to 2 games. Repeat the simulation many times. In each case, record who wins the series and the number of games played, 4, 5, 6 or 7. Count the number of times Atlanta wins. Divide this number by the total number of simulation runs to estimate the probability that Atlanta will win the World Series. Count the number of times the series ends in 4 games and divide this number by the total number of simulation runs to estimate the probability of the World Series ending in 4 games.
This can be repeated for 5-game, 6-game and 7-game series. Base case using most likely completion times. Simulation will provide a distribution of project completion time values. Calculating the percentage of simulation trials with completion times of 35 weeks or less can be used to estimate the probability of meeting the completion time target of 35 weeks.
Hand Value Probability Interval On the basis of these results, we would not recommend the player take a hit on 16 when the dealer is showing a 6. Conducting a risk analysis would require multiple simulations of the eight-quarter, two-year period.
For each simulation, the price per share at the end of two years would be recorded. The distribution of the ending price per share values would provide an indication of the maximum possible gain, the maximum possible loss and other possibilities in between.
Simulation will provide a distribution of the profit per unit values. Use the PortaCom spreadsheet. Block rows 21 to 18 2. On the Insert menu, clickRows 19 3. Copy row 14 Trial 2 to fill rows 15 to Thus, the probability of a loss should be between 0. This project appears too risky. More precise estimates of the variable cost per unit and the demand could help determine a more precise profit estimate.
Random No. Block rows 24 to 21 2. On the Insert menu, click Rows 22 3. Copy row 17 Trial 3 to fill rows 18 to The probability of a 7 should be approximately 0. Target Answers: a.
Simulation runs will vary. Again, simulation results will vary. Copy row 14 Tire 6 to fill rows 15 to Most simulations will provide between and tires exceeding 40, miles. Of mileages considered, 30, miles should come closest to meeting the tire guarantee mileage guideline. On the Insert menu, click Rows 19 3. Copy row 14 Trial 4 to fill rows 15 to 20 Trial will appear in row of the spreadsheet.
The probability of winning the bid should be between 0. Butler Inventory simulation spreadsheet. Trial replenishment levels of , , and can be entered in cell C7. Since the shortage cost has been eliminated, Butler can be expected to reduce the replenishment level.
This will allow more shortages. However, since the cost of a stockout is only the lost profit and not the lost profit plus a goodwill shortage cost, Butler can permit more shortages and still show an improvement in profit. The replenishment levels of , and all provide near optimal results. Block rows 22 to 19 2. On the Insert menu, click Rows 20 3. Copy row 15 Trial 2 to fill rows 16 to The probability of a stockout is about 0.
When a 50, unit production quantity is used, the probability of a stockout should be approximately 0. This is a relative high probability indicating that Mandrell has a good chance of being able to sell all the dolls it produces for the holiday season.
As a result, a shortage of dolls is likely to occur. However, this production strategy will enable the company to avoid the high cost associated with selling excess dolls for a loss after the first of the year. On the Insert menu, clickRows 22 3. Copy row 17 Trial 2 to fill rows 18 to The overbooking strategy appears worthwhile. The simulation spreadsheet indicates a service level of approximately This indicates that only 0.
The overbooking strategy up to a total of 32 reservations is recommended. The same spreadsheet design can be used to simulate other overbooking strategies including accepting 31, 33 and 34 passenger reservations. In each case, South Central would need to obtain data on the passenger demand probabilities.
Changing the passenger demand table and rerunning the simulation model would enable South Central to evaluate the other overbooking alternatives and arrive at the most beneficial overbooking policy. Use the Hammondsport Savings Bank spreadsheet.
Changing the interarrival times to a uniform distribution between 0 and 4 is the only change needed for each spreadsheet. The mean time between arrivals is 2 minutes and the mean service time is 2 minutes. On the surface it appears that there is an even balance between the arrivals and the services. However, since both arrivals and services have variability, simulated system performance with 1 ATM will probably be surprisingly poor.
Simulation results can be expected to show some waiting times of 30 minutes or more near the end of the simulation period. One ATM is clearly not acceptable. The interarrival times and service times section of the spreadsheet will need to be modified. Assume that the mean interarrival time of 0. The following cell formulas would be required. The service time mean and standard deviation would be entered in cells B8 and B9 as in the original Hammondsport 1 ATM spreadsheet.
Again simulation results will vary. The lower variability of the normal probability distribution should improve the performance of the waiting line by reducing the average waiting time. An average waiting time in the range 1. The interarrival times section of the spreadsheet will need to be modified.
Assume that the mean interarrival time of 4 is placed in cell B4. Both the mean interarrival time and the mean service time should be approximately 4 minutes. Simulation results should provide a mean waiting time of approximately. Simulation results should predict approximately to customers had to wait.
Learn how to describe a problem situation in terms of decisions to be made, chance events and consequences. Be able to analyze a simple decision analysis problem from both a payoff table and decision tree point of view. Be able to develop a risk profile and interpret its meaning.
Be able to use sensitivity analysis to study how changes in problem inputs affect or alter the recommended decision. Be able to determine the potential value of additional information. Learn how new information and revised probability values can be used in the decision analysis approach to problem solving. Understand what a decision strategy is. Learn how to evaluate the contribution and efficiency of additional decision making information.
Be able to use a Bayesian approach to computing revised probabilities. Be able to use TreePlan software for decision analysis problems. Decision Maximum Profit Minimum Profit d1 25 d2 75 Optimistic approach: select d1 Conservative approach: select d2 Regret or opportunity loss table: s1 s2 s3 d1 0 0 50 d2 0 0 Maximum Regret: 50 for d1 and for d2; select d1 2. The choice of which approach to use is up to the decision maker.
Since different approaches can result in different recommendations, the most appropriate approach should be selected before analyzing the problem. Decision Minimum Cost Maximum Cost d1 5 14 d2 7 11 d3 9 11 d4 8 13 Optimistic approach: select d1 Conservative approach: select d2 or d3 Regret or Opportunity Loss Table s1 s2 s3 s4 Maximum Regret d1 6 0 2 0 6 d2 3 1 0 2 3 d3 1 1 2 6 6 d4 0 1 3 8 8 Minimax regret approach: select d2 3.
The decision to be made is to choose the best plant size. There are 2 alternatives to choose from: a small plant or a large plant. The chance event is the market demand for the new product line. It is viewed as having 3 possible outcomes states of nature : low, medium and high. To use the conservative approach, compute the minimum annual return for each mutual fund. With this fund, the investor is guaranteed an annual return of at least 6.
The minimum annual return is 6. Using this recommendation, the minimum annual return is The expected annual return for the Small-Cap Stock mutual fund is The Technology Sector mutual fund recommended in part b has a larger expected annual return. The difference is The annual return for the Technology Sector mutual fund ranges from The annual return for the Technology Sector mutual fund shows the greater variation in annual return. It is considered the investment with the more risk.
It does have a higher expected annual return, but only by 1. This is a judgment recommendation and opinions may vary. But the investor is described as being conservative. The higher risk Technology Sector only has a 1. We believe the lower risk, Small-Cap Stock mutual fund is the preferred recommendation for this investor. The decision is to choose the best lease option; there are three alternatives.
The chance event is the number of miles Amy will drive per year. There are three possible outcomes. The payoff table for Amy's problem is shown below. To illustrate how the payoffs were computed, we show how to compute the total cost of the Forno Saab lease assuming Amy drives 15, miles per year.
There is a probability of 0. However, if the probability of driving 18, miles per year goes up any further, the Hopkins Automotive lease will be the best. The risk profile in tabular form is shown. Cost Probability 0. The decision to be made is to choose the type of service to provide. The chance event is the level of demand for the Myrtle Air service.
The consequence is the amount of quarterly profit. There are two decision alternatives full price and discount service. There are two outcomes for the chance event strong demand and weak demand.
For values of p below 0. For values of p greater than 0. High 0. Risk Profile for Space Pirates Outcome: 0. In order for d3 to remain optimal, the expected value of d2 must be less than or equal to There is only one decision to be made: whether or not to lengthen the runway.
There are only two decision alternatives. The payoffs and probabilities for the chance event depend on the decision alternative chosen. The probabilities and payoffs corresponding to these outcomes are given in the tables of the problem statement.
The approximate probabilities and payoffs for this case are given in the last paragraph of the problem statements. The consequence is the estimated annual revenue. The town should not lengthen the runway. The decision is to choose what type of grapes to plant, the chance event is demand for the wine and the consequence is the expected annual profit contribution.
There are three decision alternatives Chardonnay, Riesling and both. For instance, W,S denotes the outcomes corresponding to weak demand for Chardonnay and strong demand for Riesling. In constructing a decision tree, it is only necessary to show two branches when only a single grape is planted.
But, the branch probabilities in these cases are the sum of two probabilities. This changes the expected value in the case where both grapes are planted and when Riesling only is planted. The optimal decision is sensitive to this change in probabilities. Only the expected value for node 2 in the decision tree needs to be recomputed. It is now best to plant both types of grapes. The optimal decision is sensitive to a change in the payoff of this magnitude. If s1 then d1 ; if s2 then d1 or d2; if s3 then d2 b.
Note that using the expected value approach, the Town Council would be indifferent between building a medium-size community center and a large-size center. Risk profile for medium-size community center: 0.
The large-size center has a probability of 0. The Town's optimal decision strategy based on perfect information is as follows: If the worst-case scenario, build a small-size center If the base-case scenario, build a medium-size center If the best-case scenario, build a large-size center Using the consultant's original probability assessments for each scenario, 0.
If the promotional campaign is conducted, the probabilities will change to 0. That is, it eliminates the risk of a loss, which appears to be a significant factor in the mayor's decision-making process. Compare Expected Values at nodes 4 and 7. Shown below is the reduced decision tree showing only the sequence of decisions and chance events for Dante's optimal decision strategy.
If Dante follows this strategy, only 3 outcomes are possible with payoffs of , , and The probabilities for these payoffs are found by multiplying the probabilities on the branches leading to the payoffs. No agency; sell the pilot. Success 0. The company should accept the manuscript.
The manuscript review cannot alter the decision to accept the manuscript. Do not do the manuscript review. Perfect Information. A better procedure for assessing the market potential for the textbook may be worthwhile. The decision tree is as shown in the answer to problem 16a. The calculations using the decision tree in problem 16a with the probabilities and payoffs here are as follows: a,b.
The following decision tree applies. The low efficiency of 3. The ability of the consultant to forecast market conditions should be considered. The revised probabilities are shown on the branches of the decision tree. If rain, take Queen City Avenue. Expected time: Optimal decision strategy with perfect information: If s1 then d2 If s2 then d2 If s3 then d1 Expected value of this strategy is 0.
Assuming the test market study is used, a portion of the decision tree is shown below. Understand the concept of multicriteria decision making and how it differs from situations and procedures involving a single criterion. Be able to develop a goal programming model of a multiple criteria problem. Know how to use the goal programming graphical solution procedure to solve goal programming problems involving two decision variables.
Understand how the relative importance of the goals can be reflected by altering the weights or coefficients for the decision variables in the objective function. Know how to develop a solution to a goal programming model by solving a sequence of linear programming models using a general purpose linear programming package.
Know what a scoring model is and how to use it to solve a multicriteria decision problem. Understand how a scoring model uses weights to identify the relative importance of each criterion. Know how to apply the analytic hierarchy process AHP to solve a problem involving multiple criteria. Understand how AHP utilizes pairwise comparisons to establish priority measures for both the criteria and decision alternatives. In the graphical solution, point A minimizes the sum of the deviations from the goals and thus provides the optimal product mix.
In the graphical solution, point A provides the optimal solution. Refer to the graphical solution in part b. Although this solution achieves the original department B labor goal and the profit goal, this solution uses 1 For instance, there must be at least pumps in inventory at the end of May to meet the June requirement of shipping pumps.
The inventory variables are constrained to be nonnegative so we only need to be concerned with positive deviations. Production capacity constraints are needed for each month. Note in part c that the inventory deviation variables are equal to the ending inventory variables. In this case the inventory variables themselves represent the deviations from the goal of zero. Note, however, that rounding down results in not achieving goals 1 and 2.
The overall scores for the accountant position in Denver and the auditor position in Houston are the same. There is no clear second choice between the two positions. However, Georgetown, Kentucky is a close second choice Kenyon Management may want to review the relative advantages and disadvantages of these two locations one more time before making a final decision.
However Tecumseh State is the second choice and is less expensive than Handover. If cost becomes a constraint, Tecumseh State may be the most viable alternative. The individual's judgements could not be inconsistent since there are only two programs being compared. Criteria: Yield and Risk Step 1: Column totals are 1. Overall Priorities: CCC 0. The same calculations for the leadership, personal and administrative pairwise comparison matrices provide the following.
Overall Priorities: Jacobs 0. Overall Priorities: System A 0. Understand that the long-run success of an organization is often closely related to how well management is able to predict future aspects of the operation.
Know the various components of a time series. Be able to use smoothing techniques such as moving averages and exponential smoothing. Be able to use the least squares method to identify the trend component of a time series. Understand how the classical time series model can be used to explain the pattern or behavior of the data in a time series and to develop a forecast for the time series.
Be able to determine and use seasonal indexes for a time series. Know how regression models can be used in forecasting. For the limited data provided, the 5-week moving average provides the smallest MSE. You could always find a weighted moving average at least as good as the unweighted one. Actually the unweighted moving average is a special case of the weighted ones where the weights are equal.
The more recent data receives the greater weight or importance in determining the forecast. The moving averages method weights the last n data values equally in determining the forecast. However, exponential smoothing was penalized by including month 2 which was difficult for any method to forecast.
The exponential smoothing forecast is preferred because it has a smaller MSE. A linear trend model is not appropriate. A nonlinear model would provide a better approximation. A linear trend appears to be reasonable. Given the uncertainty in market conditions this far out, making a prediction for July using only time is not recommended. Capital expenditures are increasing by 1. A graph of these data shows a linear trend. Yes, a linear trend appears to exist. A linear trend appears to be appropriate.
Trend projection provides much better forecasts because it has the smallest MSE. The reason MSE is smaller for trend projection is that sales are increasing over time; as a result, exponential smoothing continuously underestimates the value of sales. If you look at the forecast errors for exponential smoothing you will see that the forecast errors are positive for periods 2 through Note: Results were obtained using the forecasting module of The Management Scientist.
Forecast for July is Exponential smoothing provides the same forecast for every period in the future. Solutions manual to accompany organic chemistry - The solutions manual to accompany Organic Chemistry provides fully explained solutions to all the problems that are featured in the second edition of Organic [PDF] Great Expectations.
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