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Downloads and Support for POM-QM for Windows.Download POM-QM for Windows POM for Windows QM for Windows – File Type Advisor

When a graph is opened it will be displayed covering the entire area below the extra data. The Save option will save the graph as a BMP file. The Copy option can be used to copy the graph in order to paste it into another Windows document. The Print image will send the graph to your printer and print on one page. The Windows Snipping Tool can be opened with the last option. The input required for each module, the options available for modeling and solving, and the different output screens and reports that can be seen and printed are explained.

For all modules that are in included in the POM menu, we display the POM for Windows icon and for all modules that are in the QM menu we display the QM for Windows icon: For example, in the first module, aggregate planning, which begins on the next page, the POM icon is displayed because aggregate planning is typically a topic in Operations Management courses but not in Management Science courses and thus appears in the POM-only menu. In addition, examples from your textbook have been included in a folder with the name of the first author of the textbook.

For these modules, the names used by KRM follow the names used by the software. Aggregate planning refers to the fact that the production planning is usually carried out across product lines. The terms aggregate planning and production planning are used interchangeably.

The main planning difficulty is that demands vary from month to month. Production should remain as stable as possible, yet it should maintain minimum inventory and experience minimum shortages. The costs of production, overtime, subcontracting, inventory, shortages, and changes in production levels must be balanced. In some cases, aggregate planning problems might require the use of the transportation or linear programming modules.

The second submodel in the aggregate planning module creates and solves a transportation model of aggregate planning for cases where all of the costs are identical. The transportation model is also available as one of the methods for the first submodel. The Aggregate Planning Model Production planning problems are characterized by a demand schedule, a set of capacities, various costs, and a method for handling shortages.

Consider the following example. Example 1: Smooth Production Consider a situation where demands in the next four periods are for , , , and units. Current inventory is 0 units.

Suppose that regular time capacity is units per month and that overtime and subcontracting are not considerations. The screen for this example follows. In addition to the data, there are two considerations — shortage handling and the method to use for performing the planning. These appear in the area above the data. Shortage handling. In production planning there are two models for handling shortages. In one model, shortages are backordered.

In another model, the shortages become lost sales. That is, if you cannot satisfy the demand in the period in which it is requested, the demand disappears. This option is above the data table. Six methods are available, which will be demonstrated. Please note that smooth production accounts for two methods. Smooth production will have equal production in every period. This yields two methods because the production can be set according to the gross demand or the net demand gross demand minus initial inventory.

Produce to demand will create a production schedule that is identical to the demand schedule. Constant regular time production, followed by overtime and subcontracting if necessary. The lesser cost method will be selected first. Any production schedule is available in which case the user must enter the amounts to be produced in each period. The transportation model. Quantities Demand. The demands are the driving force of aggregate planning and these are given in the second column.

Capacities — regular time, overtime, and subcontracting. The program allows for three types of production — regular time, overtime, and subcontracting, — and capacities for these are given in the next three columns. When deciding whether to use overtime or subcontracting, the program will always first select the one that is less expensive.

Costs The costs for the problem are all placed in the far right column of the data screen. Production costs — regular time, overtime, and subcontracting. These are the per- unit production costs depending on when and how the unit is made. Inventory holding cost. This is the amount charged for holding 1 unit for 1 period. The total holding cost is charged against the ending inventory. Be careful; although most textbooks charge against the ending inventory, some textbooks charge against average inventory during the period.

Shortage cost. This is the amount charged for each unit that is short in a given period. Whether it is assumed that the shortages are backlogged and satisfied as soon as stock becomes available in a future period or are lost sales is indicated by the option box above the data table. Shortage costs are charged against end-of- month levels. Cost to increase production. This is the cost that results from having changes in the production schedule. It is given on a per-unit basis.

The cost for increasing production entails hiring costs. It is charged against the changes in the amount of regular time production but not charged against any overtime or subcontracting production volume changes. If the units produced last period see other considerations below is zero, there will be no charge for increasing production in the first period.

Cost to decrease production. This is similar to the cost of increasing production and is also given on a per-unit basis. However, this is the cost for reducing production. It is charged only against regular time production volume changes.

Other Considerations Initial inventory. Oftentimes we have a starting inventory from the end of the previous period. The starting inventory is placed in the far right column towards the bottom. Units last period. These units appear in the far right column at the bottom. The Solution In the first example, shown in the following screen, the smooth production method and backorders have been chosen.

The demands are , , , and , and the regular time capacity of exceeds this demand. There is no initial inventory. The numbers represent the production quantities. The costs can be seen toward the bottom of the columns. The screen contains information on both a period-by-period basis and on a summary basis.

Notice the color coding of the data black , intermediate computations magenta and results blue. Regular time production. This amount is determined by the program for all options except User Defined. In this example, because the gross or net demand is , there are units produced in regular time in each of the 4 periods.

If the total demand is not an even multiple of the number of periods, extra units will be produced in as many periods as necessary in order to meet the demand. For example, had the total demand been , the production schedule would have been in the first and second periods and in the other two periods. The accumulated inventory appears in this column if it is positive. If there is a shortage, the amount of the shortage appears in this column.

In the example, the in the shortage column for Period 3 means that units of demand have not been met. Because the backlog option has been chosen, the demands are met as soon as possible, which is in the last period. In this example, no increase or decrease from month to month occurs, so these columns do not appear in this display.

The total numbers of units demanded, produced, in inventory, short, or in increased and decreased production are computed. In the example, units were demanded and units were produced, and there were a total of unit- months of inventory, unit-months of shortage, and 0 increased or decreased production unit-months.

The totals of the columns are multiplied by the appropriate costs, yielding the total cost for each of the cost components. Total cost.

The overall total cost is computed and displayed. Graph Two graphs are available in this module. It is possible to display a bar graph of production in each period not shown , and it is also possible to display a graph of the cumulative production versus the cumulative demand shown.

These modifications can be seen in the following screen. In addition, the method has been changed to use the net demand. Thus only units per month need to be produced. This can be seen as follows. Because there is not enough regular time capacity, the program looks to overtime and subcontracting. It first chooses the one that is less expensive.

Example 4: When subcontracting is less expensive than overtime The following screen shows a case where subcontracting is less expensive than overtime. This time, the program first chooses subcontracting and, because there is sufficient capacity, overtime is not used at all.

The output shows a shortage of units at the end of Period 3. In the next period, we produce units even though we need only units. These extra units are not used to satisfy the Period 3 shortage, because these have become lost sales. The units go into inventory, as can be seen from the inventory column in Period 4. It does not make sense to use the smooth production model and have lost sales. In the end, the total demand is not actually , because of the sales were lost.

Example 6: The produce to demand no inventory strategy From the first example the method has been toggled to produce to demand or chase strategy. The inventory is not displayed because it is always 0 under this option. With production equal to demand and no starting inventory, there will be neither changes in inventory nor shortages. In this example, production in Period 1 was and production in Period 2 was Therefore, the increase column has a in it for Period 2.

The program will not list any increase in Period 1 if no initial production is given. The total increases have been ; decreases The change in production from the previous period to this period occurs in this column if the change represents an increase.

Notice that the program assumes that no change takes place in the first period in this example because the initial data not displayed indicated that 0 units were produced last month. In this example, there is no change in other periods because production is constant under the smooth production option. If production decreases, the decrease appears in this column.

Example 7: Increase and decrease charging The previous example had increases and decreases in production. These increases and decreases are accounted for by regular time production. In the following screen, the regular time capacity is reduced in order to force production through regular time and overtime. Notice that the increase column only has a value in it in the second period when regular time production went from to units.

The regular time production remains at ; even though overtime increases, this does not show up in the increase columns. There are no charges against overtime or subcontracting increases. The only difference is that the transportation model does not consider changes in production levels, so there is no data entry allowed for increase and decrease costs or for units last period. The creation screen will ask for the number of periods and whether shortages are allowed. The similarity to the previous input screens can be seen as follows.

Notice that there is only one entry for each of the costs. Thus, this model can not be used for situations where the costs change from period to period. You must formulate these problems yourself using the transportation model from the Module menu rather than this transportation submodel of aggregate planning. Note: The transportation model that is the second submodel in the New menu can also be accessed as the last method in the first submodel, The solution screen is displayed next.

The window on the right summarizes the production quantities, unit-months of holding and shortage if applicable , and the costs. It is even more obvious that this is a transportation problem if the second window of output which is the transportation model itself is examined. The large numbers 9, have been entered in order to preclude the program from backordering. If you like, this table could be copied; you could then open the Transportation model, create a new empty table that is 13 by 4 and paste this data in to that table.

Five heuristic rules can be used for performing the balance. The cycle time can be given explicitly or the production rate can be given and the program will compute the cycle time. This model will not split tasks. Task splitting is discussed in more detail in a later section. The Model The general framework for assembly-line balancing is dictated by the number of tasks that are to be balanced.

These tasks are partially ordered, as shown, for example in the precedence diagram that follows. The five heuristic rules that can be chosen are as follows: 1. Longest operation time 2. Most following tasks 3. Ranked positional weight 4. Shortest operation time 5. Note that tie breaking can affect the final results. The remaining parameters are as follows: Cycle time computation.

The cycle time can be given in one of two ways. One way involves giving the cycle time directly as shown in the preceding screen.

Although this is the easiest method, it is more common to determine the cycle time from the demand rate. The cycle time is converted into the same units as the times for the tasks. See Example 2. Task time unit. The time unit for the tasks is given by this drop-down box. You must choose seconds, hours, or minutes. Notice that the column heading for the task times will change as you select different time units.

Task names. The task names are essential for assembly-line balancing because they determine the precedences. Case does not matter. Task times. The task times are given. Enter the precedences, one per cell. If there are two precedences they must be entered in two cells. In fact, a comma will not be accepted.

Notice that in the precedence list in the previous screen both a and A have been typed. As mentioned previously, the case of the letters is irrelevant. Example 1 In this example there are six tasks, a through f.

The precedence diagram for this problem appears previously. The time to perform each task is above the task. Also, note that the tasks that are ready at the beginning of the balance are tasks a and b.

Finally, in this first example, we use a cycle time of Solution The following screen contains the solution to the first example. The solution screen consists of two windows as shown in the following screen. The window on the left gives the complete results for the method chosen whereas the window on the right gives the number of stations required not the theoretical number when using each balancing rule.

The solution screen will always have the same appearance and contain the same information regardless of the rule that is chosen for the balance. This is not always the case as is demonstrated later in this section.

Station numbers. The station numbers appear in the far left column. They are displayed only for the first task that is loaded into each station. In this example, three stations are required. The tasks that are loaded into the station are listed in the second column. In this example, Tasks b, e, and a are in Station 1; Tasks d and c are in Station 2; and Task f is in station 3.

The length of time for each task appears in the third column. Time left. The length of time that remains at the station is listed in the fourth column. The last number at each station is, of course, the idle time at that station. The idle times are colored in red.

For example, there is 1 second of idle time at Station 1, 1 second of idle time at Station 2, and 2 seconds of idle time at Station 3, for a total of 4 seconds of idle time per cycle. Ready tasks. The tasks that are ready appear here. A ready task is any task that has had its precedences met. This is emphasized because some books do not list a task as ready if its time exceeds the time remaining at the station.

Also, if the number of characters in the ready task list is very long, you might want to widen that column. The cycle time that was used appears below the balance. This cycle time was either given directly or computed. In this example, the cycle time was given directly as 10 seconds. Time allocated. The total time allocated for making each unit is displayed. This time is the product of the number of stations and the cycle time at each station.

In this example there are three stations, each with a cycle time of 10 seconds, for a total work time of 30 station-seconds. The time needed to make one unit. This is simply the sum of the task times.

Idle time. This is the time needed subtracted from the time allocated. Efficiency is defined as the time needed divided by the time allocated. Balance delay.

The balance delay is the percentage of wasted time or percent minus the efficiency. Minimum theoretical number of stations. This is the total time to make 1 unit divided by the cycle time and rounded up to the nearest integer.

In this example, 26 seconds are required to make 1 unit divided by a second cycle time for an answer of 2. In addition, a second window opens that displays the number of stations required using each of the different balancing rules.

In this particular case, each rule led to the same number of stations, 3. This is not always the case as shown in Example 4. The precedence graph can be displayed see the end of this section , as well as a bar graph indicating how much time was used at each station. These are shown at the end of this section. In addition, if there is idle time at every station, a note will appear at the top indicating that the balance can be improved by reducing the cycle time.

For example, because there are idle times of 1, 1, and 2 seconds at the three stations, we could reduce the cycle time by 1 second. Example 2: Computing the cycle time Suppose that for the same data a production rate of units in 7. Other Rules Other rules that may be used are mentioned although the results are not displayed.

Please note that this is one of the modules where if you change the method using the drop-down box from the solution screen, the problem will immediately be resolved. That is, you do not need to use the EDIT button and return to the data. Most Following Tasks A common way to choose tasks is by using the task with the most following tasks.

Notice from the diagram at the beginning of the section that a has three tasks following it, and b also has two tasks following it. Therefore, there is a tie for the first task. If Task a is chosen then the next task chosen will be Task b because Task b has 3 following tasks whereas Task c has only one.

The task with the largest weight is scheduled first if it will fit in the remaining time. Notice that e has a higher ranked positional weight than c. Least Number of Followers The last rule that is available is the least number of followers. Example 3: What to do if longest operation time will not fit Some books and some software do not apply the longest operation time rule properly. If the task with the longest time will not fit into the station, the task with the second longest time should be placed in the station if it will fit.

In the following screen data is presented for eight tasks. Notice that Tasks b, c, e, and f immediately follow Task a. The balance appears in the following screen for a cycle time of 5 seconds.

After Task a is completed, tasks b, c, d, and e are ready. Task b is longest but will not fit in the 4 seconds that remain at Station 1. Therefore, Task c is inserted into the balance. Example 4: Splitting tasks If the cycle time is less than the amount of time to perform a specific task, there is a problem. We perform what is termed task splitting but which in reality is actually duplication. For example, suppose that the cycle time is 2 minutes and some task takes 5 minutes.

The task is performed 3 times by three people at three machines independent of one another. The effect is that 3 units will be done every 5 minutes, which is equivalent to 1 unit every 1. Now, the actual way that the three people work may vary.

Although other programs will split tasks, the assumptions vary from program to program. Rather than making assumptions, you should split the tasks by dividing the task time appropriately.

Suppose that in Example 1 a cycle time of 5 seconds was used. Then it is necessary to replicate both Tasks d and f because they will not fit in the cycle time. The approach to use is to solve the problem by dividing the task times by 2, because this replication is needed. The results are presented in the following screen. Notice that different rules lead to different minimum numbers of stations! The first is a precedence graph, as shown in the following figure.

Please note that there may be several different ways to draw a precedence graph. The second graph not displayed here is of time used at each station. In a perfect world these would all be the same a perfect balance. The model is a special case of the transportation method. In order to generate an assignment problem, it is necessary to provide the number of jobs and machines and to indicate whether the problem is a minimization or maximization problem.

The number of jobs and number of machines do not have to be equal but usually they are. Objective function. The objective can be to minimize or to maximize.

This is set at the creation screen but can be changed in the data screen. Example 1 The following table shows data for a 7-by-7 assignment problem. The goal is to assign each salesperson to a territory at minimum total cost.

There must be exactly one salesperson per territory and exactly one territory per salesperson. The data structure is nearly identical to the structure for the transportation model. The basic difference is that the assignment model does not display supplies and demands because they are all equal to one.

Note: To try to preclude an assignment from being made, such as Bruce to Pennsylvania in this example, enter a very large cost. The assignments can also be given in list form, as shown in the following screen. The marginal costs can be displayed also. Cost-volume analysis is used to find the point of indifference between two options based on fixed and variable costs. A breakeven point is computed in terms of units or dollars. Breakeven analysis is simply a special case of cost-volume analysis where there is one fixed cost, one variable cost, and revenue- per-unit.

Cost-Volume Analysis In cost-volume analysis, two or more options are compared to determine what option is least costly at any volume. The costs consist of two types – fixed costs and variable costs, but there may be several individual costs that comprise the fixed costs or the variable costs. In the example that follows, there are five different individual costs and two options. Data Cost type. Each type of cost must be identified as either a fixed cost or a variable cost.

The default is that the first cost in the list is fixed and that all other costs are variable. These values can be changed by using the drop-down box in that cell.

The specific costs for each option are listed in the two right columns in the table. If a volume analysis is desired, enter the volume at which this analysis should be performed. The volume analysis will compute the total cost revenue at the chosen volume.

If the volume is 0, no volume analysis will be performed other than for the breakeven point. Volume analysis is at units. In the preceding screen there are five costs with some fixed and some variable. The program displays the following results: Total fixed costs. For each of the two options, the program takes the fixed costs, sums them, and lists them in the table. Total variable costs. The program identifies the variable costs, sums them, and lists them.

Breakeven point in units. The breakeven point is the difference between the fixed costs divided by the difference between the variable costs, and this is displayed in units. In the example, it is units.

Breakeven point in dollars. The breakeven point can also be expressed in dollars. A volume analysis has been performed for a volume of units. The total fixed costs and total variable costs have been computed for each option and these have been summed to yield the total cost for each option. A graph is available, as follows. Data entry for this option is slightly different in that the program creates a column for costs and a column for revenues. The fixed and variable costs get entered in the cost column and the revenue per unit is placed in the revenue column.

This model requires exactly three inputs. This example could also have been solved using the cost-volume submodel.

Select two options and let one be the costs and one be the revenues. Place the fixed costs and variable costs in their obvious cells; use no fixed cost for the revenue and use the revenue per unit as a variable cost, displayed as follows.

The following screen demonstrates the output for a three-option breakeven. The screen indicates that there are three breakeven points as it makes comparisons for Computer 1 versus Computer 2, Computer 1 versus Computer 3, and Computer 2 versus Computer 3. Of course, even though there are three breakeven points, only two of them are relevant.

This is seen a little more easily by looking at the following breakeven graph. The breakeven point at 40, units does not matter because at 40, units the two computers that break even have higher costs than the Computer 2 option. The data for this example consist of a stream of inflows and a stream of outflows. In addition, for finding the net present value an interest rate must be given. Net Present Value Consider the following example.

The company would like to know the net present value using an interest rate of 10 percent. The data screen follows. The screen has two columns for data. One column is labeled Inflow and the other column is labeled Outflow.

At the time of problem creation a six-period problem was created and the data table includes the six periods plus the current period 0. The six savings in the second column are inflows, and they are placed in the inflow column for Periods 1 through 6. The salvage value could be handled two ways, and we have chosen the way that we think gives a better display.

Instead, it is represented as a negative outflow. This keeps the meaning of the numbers clearer. The last item to be entered is the interest rate in the text box above the data. To the right of this, the inflows and outflows are multiplied by these present value factors, and the far right column contains the present values for the net inflow inflow minus outflow on a period-by-period basis.

Internal Rate of Return The computation of the internal rate of return is very simple. The data is set up the same way but the method box is changed from net present value to internal rate of return.

The results appear as follows. You can see that the internal rate of return for the same data is The Decision Table Model The decision table can be used to find the expected value, the maximin minimax , or the maximax minimin when several decision options are available and there are several scenarios that might occur. Also, the expected value under certainty, the expected value of perfect information, and the regret opportunity cost can be computed.

The general framework for decision tables is given by the number of options or alternatives that are available to the decision maker and the number of scenarios or states of nature that might occur. In addition, the objective can be set to either maximize profits or to minimize costs. Scenario probabilities. For each scenario it is possible but not required to enter a probability.

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Explanation of graphs in POM-QM for Windows. Oct 26, · POM is a small cross-platform tool that is used to merge several PDF files into one file. It is written in Javascript. Selection: POM has an integrated browser that allows the user to easily find PDF files and display them on the screen. The 1/5(1). POM for Windows 3 adalah software yan digunakan untuk memcahkan masalah produksi dan manajemen produksi dan manajemen yang bersifat kuantitatif dan juga membantu dalam praktikum SIM (Sintem Informasi Managemen) Software ini menyediakan 20 Menu/Modul yang berbeda penggunaannya yaitu: Silakan Bagi yang berminat download silakan download.

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Be careful; although most textbooks charge against the ending inventory, some textbooks charge against average inventory during the period.

Shortage cost. This is the amount charged for each unit that is short in a given period. Whether it is assumed that the shortages are backlogged and satisfied as soon as stock becomes available in a future period or are lost sales is indicated by the option box above the data table.

Shortage costs are charged against end-of- month levels. Cost to increase production. This is the cost that results from having changes in the production schedule.

It is given on a per-unit basis. The cost for increasing production entails hiring costs. It is charged against the changes in the amount of regular time production but not charged against any overtime or subcontracting production volume changes.

If the units produced last period see other considerations below is zero, there will be no charge for increasing production in the first period. Cost to decrease production. This is similar to the cost of increasing production and is also given on a per-unit basis. However, this is the cost for reducing production. It is charged only against regular time production volume changes. Other Considerations Initial inventory.

Oftentimes we have a starting inventory from the end of the previous period. The starting inventory is placed in the far right column towards the bottom. Units last period. These units appear in the far right column at the bottom. The Solution In the first example, shown in the following screen, the smooth production method and backorders have been chosen.

The demands are , , , and , and the regular time capacity of exceeds this demand. There is no initial inventory. The numbers represent the production quantities. The costs can be seen toward the bottom of the columns.

The screen contains information on both a period-by-period basis and on a summary basis. Notice the color coding of the data black , intermediate computations magenta and results blue. Regular time production. This amount is determined by the program for all options except User Defined.

In this example, because the gross or net demand is , there are units produced in regular time in each of the 4 periods. If the total demand is not an even multiple of the number of periods, extra units will be produced in as many periods as necessary in order to meet the demand.

For example, had the total demand been , the production schedule would have been in the first and second periods and in the other two periods. The accumulated inventory appears in this column if it is positive. If there is a shortage, the amount of the shortage appears in this column.

In the example, the in the shortage column for Period 3 means that units of demand have not been met. Because the backlog option has been chosen, the demands are met as soon as possible, which is in the last period. In this example, no increase or decrease from month to month occurs, so these columns do not appear in this display. The total numbers of units demanded, produced, in inventory, short, or in increased and decreased production are computed.

In the example, units were demanded and units were produced, and there were a total of unit- months of inventory, unit-months of shortage, and 0 increased or decreased production unit-months. The totals of the columns are multiplied by the appropriate costs, yielding the total cost for each of the cost components.

Total cost. The overall total cost is computed and displayed. Graph Two graphs are available in this module. It is possible to display a bar graph of production in each period not shown , and it is also possible to display a graph of the cumulative production versus the cumulative demand shown.

These modifications can be seen in the following screen. In addition, the method has been changed to use the net demand. Thus only units per month need to be produced. This can be seen as follows. Because there is not enough regular time capacity, the program looks to overtime and subcontracting.

It first chooses the one that is less expensive. Example 4: When subcontracting is less expensive than overtime The following screen shows a case where subcontracting is less expensive than overtime. This time, the program first chooses subcontracting and, because there is sufficient capacity, overtime is not used at all.

The output shows a shortage of units at the end of Period 3. In the next period, we produce units even though we need only units. These extra units are not used to satisfy the Period 3 shortage, because these have become lost sales. The units go into inventory, as can be seen from the inventory column in Period 4. It does not make sense to use the smooth production model and have lost sales. In the end, the total demand is not actually , because of the sales were lost.

Example 6: The produce to demand no inventory strategy From the first example the method has been toggled to produce to demand or chase strategy. The inventory is not displayed because it is always 0 under this option. With production equal to demand and no starting inventory, there will be neither changes in inventory nor shortages. In this example, production in Period 1 was and production in Period 2 was Therefore, the increase column has a in it for Period 2.

The program will not list any increase in Period 1 if no initial production is given. The total increases have been ; decreases The change in production from the previous period to this period occurs in this column if the change represents an increase. Notice that the program assumes that no change takes place in the first period in this example because the initial data not displayed indicated that 0 units were produced last month. In this example, there is no change in other periods because production is constant under the smooth production option.

If production decreases, the decrease appears in this column. Example 7: Increase and decrease charging The previous example had increases and decreases in production. These increases and decreases are accounted for by regular time production.

In the following screen, the regular time capacity is reduced in order to force production through regular time and overtime. Notice that the increase column only has a value in it in the second period when regular time production went from to units. The regular time production remains at ; even though overtime increases, this does not show up in the increase columns. There are no charges against overtime or subcontracting increases. The only difference is that the transportation model does not consider changes in production levels, so there is no data entry allowed for increase and decrease costs or for units last period.

The creation screen will ask for the number of periods and whether shortages are allowed. The similarity to the previous input screens can be seen as follows. Notice that there is only one entry for each of the costs. Thus, this model can not be used for situations where the costs change from period to period.

You must formulate these problems yourself using the transportation model from the Module menu rather than this transportation submodel of aggregate planning. Note: The transportation model that is the second submodel in the New menu can also be accessed as the last method in the first submodel, The solution screen is displayed next. The window on the right summarizes the production quantities, unit-months of holding and shortage if applicable , and the costs.

It is even more obvious that this is a transportation problem if the second window of output which is the transportation model itself is examined. The large numbers 9, have been entered in order to preclude the program from backordering. If you like, this table could be copied; you could then open the Transportation model, create a new empty table that is 13 by 4 and paste this data in to that table.

Five heuristic rules can be used for performing the balance. The cycle time can be given explicitly or the production rate can be given and the program will compute the cycle time. This model will not split tasks. Task splitting is discussed in more detail in a later section. The Model The general framework for assembly-line balancing is dictated by the number of tasks that are to be balanced. These tasks are partially ordered, as shown, for example in the precedence diagram that follows.

The five heuristic rules that can be chosen are as follows: 1. Longest operation time 2. Most following tasks 3. Ranked positional weight 4. Shortest operation time 5. Note that tie breaking can affect the final results. The remaining parameters are as follows: Cycle time computation. The cycle time can be given in one of two ways.

One way involves giving the cycle time directly as shown in the preceding screen. Although this is the easiest method, it is more common to determine the cycle time from the demand rate. The cycle time is converted into the same units as the times for the tasks. See Example 2. Task time unit. The time unit for the tasks is given by this drop-down box.

You must choose seconds, hours, or minutes. Notice that the column heading for the task times will change as you select different time units. Task names. The task names are essential for assembly-line balancing because they determine the precedences. Case does not matter. Task times. The task times are given. Enter the precedences, one per cell.

If there are two precedences they must be entered in two cells. In fact, a comma will not be accepted. Notice that in the precedence list in the previous screen both a and A have been typed.

As mentioned previously, the case of the letters is irrelevant. Example 1 In this example there are six tasks, a through f. The precedence diagram for this problem appears previously. The time to perform each task is above the task.

Also, note that the tasks that are ready at the beginning of the balance are tasks a and b. Finally, in this first example, we use a cycle time of Solution The following screen contains the solution to the first example. The solution screen consists of two windows as shown in the following screen.

The window on the left gives the complete results for the method chosen whereas the window on the right gives the number of stations required not the theoretical number when using each balancing rule.

The solution screen will always have the same appearance and contain the same information regardless of the rule that is chosen for the balance. This is not always the case as is demonstrated later in this section. Station numbers. The station numbers appear in the far left column. They are displayed only for the first task that is loaded into each station. In this example, three stations are required. The tasks that are loaded into the station are listed in the second column.

In this example, Tasks b, e, and a are in Station 1; Tasks d and c are in Station 2; and Task f is in station 3. The length of time for each task appears in the third column.

Time left. The length of time that remains at the station is listed in the fourth column. The last number at each station is, of course, the idle time at that station. The idle times are colored in red. For example, there is 1 second of idle time at Station 1, 1 second of idle time at Station 2, and 2 seconds of idle time at Station 3, for a total of 4 seconds of idle time per cycle. Ready tasks. The tasks that are ready appear here. A ready task is any task that has had its precedences met.

This is emphasized because some books do not list a task as ready if its time exceeds the time remaining at the station. Also, if the number of characters in the ready task list is very long, you might want to widen that column.

The cycle time that was used appears below the balance. This cycle time was either given directly or computed. In this example, the cycle time was given directly as 10 seconds. Time allocated. The total time allocated for making each unit is displayed. This time is the product of the number of stations and the cycle time at each station.

In this example there are three stations, each with a cycle time of 10 seconds, for a total work time of 30 station-seconds. The time needed to make one unit. This is simply the sum of the task times. Idle time. This is the time needed subtracted from the time allocated. Efficiency is defined as the time needed divided by the time allocated. Balance delay.

The balance delay is the percentage of wasted time or percent minus the efficiency. Minimum theoretical number of stations. This is the total time to make 1 unit divided by the cycle time and rounded up to the nearest integer. In this example, 26 seconds are required to make 1 unit divided by a second cycle time for an answer of 2. In addition, a second window opens that displays the number of stations required using each of the different balancing rules.

In this particular case, each rule led to the same number of stations, 3. This is not always the case as shown in Example 4. The precedence graph can be displayed see the end of this section , as well as a bar graph indicating how much time was used at each station. These are shown at the end of this section.

In addition, if there is idle time at every station, a note will appear at the top indicating that the balance can be improved by reducing the cycle time. For example, because there are idle times of 1, 1, and 2 seconds at the three stations, we could reduce the cycle time by 1 second.

Example 2: Computing the cycle time Suppose that for the same data a production rate of units in 7. Other Rules Other rules that may be used are mentioned although the results are not displayed. Please note that this is one of the modules where if you change the method using the drop-down box from the solution screen, the problem will immediately be resolved. That is, you do not need to use the EDIT button and return to the data. Most Following Tasks A common way to choose tasks is by using the task with the most following tasks.

Notice from the diagram at the beginning of the section that a has three tasks following it, and b also has two tasks following it. Therefore, there is a tie for the first task. If Task a is chosen then the next task chosen will be Task b because Task b has 3 following tasks whereas Task c has only one. The task with the largest weight is scheduled first if it will fit in the remaining time.

Notice that e has a higher ranked positional weight than c. Least Number of Followers The last rule that is available is the least number of followers. Example 3: What to do if longest operation time will not fit Some books and some software do not apply the longest operation time rule properly. If the task with the longest time will not fit into the station, the task with the second longest time should be placed in the station if it will fit.

In the following screen data is presented for eight tasks. Notice that Tasks b, c, e, and f immediately follow Task a. The balance appears in the following screen for a cycle time of 5 seconds. After Task a is completed, tasks b, c, d, and e are ready. Task b is longest but will not fit in the 4 seconds that remain at Station 1. Therefore, Task c is inserted into the balance. Example 4: Splitting tasks If the cycle time is less than the amount of time to perform a specific task, there is a problem.

We perform what is termed task splitting but which in reality is actually duplication. For example, suppose that the cycle time is 2 minutes and some task takes 5 minutes. The task is performed 3 times by three people at three machines independent of one another. The effect is that 3 units will be done every 5 minutes, which is equivalent to 1 unit every 1. Now, the actual way that the three people work may vary. Although other programs will split tasks, the assumptions vary from program to program.

Rather than making assumptions, you should split the tasks by dividing the task time appropriately. Suppose that in Example 1 a cycle time of 5 seconds was used. Then it is necessary to replicate both Tasks d and f because they will not fit in the cycle time.

The approach to use is to solve the problem by dividing the task times by 2, because this replication is needed. The results are presented in the following screen. Notice that different rules lead to different minimum numbers of stations!

The first is a precedence graph, as shown in the following figure. Please note that there may be several different ways to draw a precedence graph.

The second graph not displayed here is of time used at each station. In a perfect world these would all be the same a perfect balance. The model is a special case of the transportation method.

In order to generate an assignment problem, it is necessary to provide the number of jobs and machines and to indicate whether the problem is a minimization or maximization problem. The number of jobs and number of machines do not have to be equal but usually they are.

Objective function. The objective can be to minimize or to maximize. This is set at the creation screen but can be changed in the data screen. Example 1 The following table shows data for a 7-by-7 assignment problem. The goal is to assign each salesperson to a territory at minimum total cost. There must be exactly one salesperson per territory and exactly one territory per salesperson. The data structure is nearly identical to the structure for the transportation model.

The basic difference is that the assignment model does not display supplies and demands because they are all equal to one. Note: To try to preclude an assignment from being made, such as Bruce to Pennsylvania in this example, enter a very large cost. The assignments can also be given in list form, as shown in the following screen. The marginal costs can be displayed also. Cost-volume analysis is used to find the point of indifference between two options based on fixed and variable costs.

A breakeven point is computed in terms of units or dollars. Breakeven analysis is simply a special case of cost-volume analysis where there is one fixed cost, one variable cost, and revenue- per-unit. Cost-Volume Analysis In cost-volume analysis, two or more options are compared to determine what option is least costly at any volume.

The costs consist of two types – fixed costs and variable costs, but there may be several individual costs that comprise the fixed costs or the variable costs.

In the example that follows, there are five different individual costs and two options. Data Cost type. Each type of cost must be identified as either a fixed cost or a variable cost. The default is that the first cost in the list is fixed and that all other costs are variable. These values can be changed by using the drop-down box in that cell. The specific costs for each option are listed in the two right columns in the table. If a volume analysis is desired, enter the volume at which this analysis should be performed.

The volume analysis will compute the total cost revenue at the chosen volume. If the volume is 0, no volume analysis will be performed other than for the breakeven point. Volume analysis is at units. In the preceding screen there are five costs with some fixed and some variable. The program displays the following results: Total fixed costs.

For each of the two options, the program takes the fixed costs, sums them, and lists them in the table. Total variable costs.

The program identifies the variable costs, sums them, and lists them. Breakeven point in units. The breakeven point is the difference between the fixed costs divided by the difference between the variable costs, and this is displayed in units.

In the example, it is units. Breakeven point in dollars. The breakeven point can also be expressed in dollars. A volume analysis has been performed for a volume of units. The total fixed costs and total variable costs have been computed for each option and these have been summed to yield the total cost for each option. A graph is available, as follows. Data entry for this option is slightly different in that the program creates a column for costs and a column for revenues.

The fixed and variable costs get entered in the cost column and the revenue per unit is placed in the revenue column. This model requires exactly three inputs. This example could also have been solved using the cost-volume submodel.

Select two options and let one be the costs and one be the revenues. Place the fixed costs and variable costs in their obvious cells; use no fixed cost for the revenue and use the revenue per unit as a variable cost, displayed as follows. The following screen demonstrates the output for a three-option breakeven. The screen indicates that there are three breakeven points as it makes comparisons for Computer 1 versus Computer 2, Computer 1 versus Computer 3, and Computer 2 versus Computer 3. Of course, even though there are three breakeven points, only two of them are relevant.

This is seen a little more easily by looking at the following breakeven graph. The breakeven point at 40, units does not matter because at 40, units the two computers that break even have higher costs than the Computer 2 option.

The data for this example consist of a stream of inflows and a stream of outflows. In addition, for finding the net present value an interest rate must be given.

Net Present Value Consider the following example. The company would like to know the net present value using an interest rate of 10 percent. The data screen follows. The screen has two columns for data. One column is labeled Inflow and the other column is labeled Outflow. At the time of problem creation a six-period problem was created and the data table includes the six periods plus the current period 0. The six savings in the second column are inflows, and they are placed in the inflow column for Periods 1 through 6.

The salvage value could be handled two ways, and we have chosen the way that we think gives a better display. Instead, it is represented as a negative outflow. This keeps the meaning of the numbers clearer.

The last item to be entered is the interest rate in the text box above the data. To the right of this, the inflows and outflows are multiplied by these present value factors, and the far right column contains the present values for the net inflow inflow minus outflow on a period-by-period basis. Internal Rate of Return The computation of the internal rate of return is very simple. The data is set up the same way but the method box is changed from net present value to internal rate of return.

The results appear as follows. You can see that the internal rate of return for the same data is The Decision Table Model The decision table can be used to find the expected value, the maximin minimax , or the maximax minimin when several decision options are available and there are several scenarios that might occur.

Also, the expected value under certainty, the expected value of perfect information, and the regret opportunity cost can be computed. The general framework for decision tables is given by the number of options or alternatives that are available to the decision maker and the number of scenarios or states of nature that might occur.

In addition, the objective can be set to either maximize profits or to minimize costs. Scenario probabilities. For each scenario it is possible but not required to enter a probability. The expected value measures expected monetary value, expected value under certainty, and expected value of perfect information require probabilities, whereas the maximin minimax and maximax minimin do not.

Profits or costs. The profit cost for each combination of options and scenarios is to be given. Hurwicz alpha. The Hurwicz value is used to give a weighted average of the best and worst outcomes for each strategy row.

Please note that the Hurwicz value is not in every textbook. The possible scenarios states of nature are that demand will be low, normal, or high; or that there will be a strike or a work slowdown.

The table contains profits as indicated. The first row in the table represents the probability that each of these states will occur. The remaining three rows represent the profit that we accrue if we make that decision and the state of nature occurs. For example, if we select to use overtime and there is high demand, the profit will be Solution The results screen that follows contains both the data and the results for this example. Expected values. Row minimum.

For each row, the minimum element has been found and listed. This element is used to find the maximin or minimin. For each row, the maximum element in the row has been found and listed. This number is used for determining the maximax or minimax. These represent 40 percent multiplied by the best outcome plus 60 percent multiplied by the worst outcome for each row.

For example, for subcontracting the Hurwicz is. Maximum expected value. Because this is a profit problem finding the maximum values is of importance.

The maximum expected value is the largest number in the expected value column, which in this example is In this example, the maximin is The maximax is the largest value in the table or the largest value in the maximum column. In this example, it is Perfect Information A second screen of results presents the computations for the expected value of perfect information as follows. Perfect information. In this row, the best outcome for each column is listed. For example, for the low demand scenario the best outcome is the given by using overtime.

The expected value under certainty is computed as the sum of the products of the probabilities multiplied by the best outcomes. Expected value of perfect information. The expected value of perfect information EVPI is the difference between the best expected value Table values. The values in the table are computed for each column as the cell value subtracted from the best value in the column in the data. For example, under low demand the best outcome is The two columns on the right yield two sets of results.

There also is a window not displayed in this manual that yields Hurwicz values for alpha ranging from 0 to 1 by. Decision Trees Decision trees are used when sequences of decisions are to be made.

The trees consist of branches that connect either decision points, points representing chance, or final outcomes. All decision tables can be put in the form of a decision tree. The converse is not true. Note: Version 4 of the software includes two different input styles for decision trees. The first model has tabular data entry whereas the second model is easier to use because it has graphical data entry.

The first model has been maintained in the software for consistency with previous versions. Example 2: A decision tree — Graphical user interface One of the models allows for decision trees to be entered graphically rather than in the table as given previously.

This model can be used to examine the same example just completed. After selecting the model, the interface will be displayed as follows. This is the only model in the software that has an input interface that is not the usual data table interface.

The graph is displayed in the large area on the left and created using the tools on the right. In the beginning, there is only one node. 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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