363561 business - decision analysis
The network diagram above provides a blueprint of how tasks are executed and the time taken to complete the tasks. For instance, task A takes 10 minutes while B and c take 9 minutes to complete. The diagram also shows which stations performing the tasks.
The sum of the time needed for each task before the tasks were combined at workstations 2 and 4. By combining tasks so that (a) the product does not need to pass from workstation to workstation Ten Minutes have been saved . Therefore 36 minutes is equal to the new cycle time. 46 minutes originally minus 10 minutes combing tasks = total cycle time of 36 minutes
Workstations 2 and 4 have multiple production processes and hence the production time will be reduced. This arrangement provides the most efficient operation plan leading cuts in production line costs.
Slide 10 and Speaker notes
In order to optimize costs (efficiency) and meet the demand efficiently POM fro Windows was used. The linear programming and cost optimization tools in POM were used. Resource allocation is aided by the linear programming tool.
The constraints: The company needs to meet a production goal of at least 25 batches of loafers with tassels and 10 batches of penny loafers as a starting point. (See column RHS) Loafers with tassels are more expensive to produce at $2000 and the penny loafers are produced at a cost of $1500 per batch.
The results are that 25 batches of both types of shoes should be produced.
The cost would be $87,500.
The tasselled loafers are assembled (in the Assembly department) for 2 hours and then move to the finishing department for 8 hours to be finished. The total profit = $800
The penny loafers need 6 hours in the assembly department and 4 hours in the finishing department. The total profit $1,200
Minimum hours of capacity for tasselled loafers = 1,200 hours & minimum production level for tasselled loafers = 50 tasselled loafers
Minimum hours of capacity for penny loafers = 1,600 hours & minimum production levle for penny loafers = 100 pairs penny loafers
The POM for Windows was used. The result was that if 120 batches of tasselled loafers and 160 batches of penny loafers are produced, profits will be maximized to $288,000. The company does not need to run a consumer survey in order to understand their demands exactly because they are a well-established name in the market, and all they require to do is to open new stores at locations where they have a legitimate customer base. The social media advertisement will give the management a crude idea about the demand in the local market. The company will be suggested to open the store in the location that has the highest demand of its products. Additionally, the organization is recommended not to waste resources on running a survey because if the response is low then, the relevant expenditure will represent sunk cost.
Slide 11 and Speaker Notes
A decision tree can be used to make money or to save money. $20,000 was invested in a survey to determine the expected monetary value EMV of the following scenarios dependent on favorable or unfavorable marketing conditions.
Trade-offs need to be made but which ones are the best to choose? The market research team that provided the survey recommended that the best scenario was to build a stand-alone building to house a new store. The highest profits are predicted by the results of the survey to be made using the stand alone scenario. The predicted profits of carrying out the plan are $210,000.
Excel and POM for Windows were used to input the data and better understand a favorable probability of success. The use of the two tools was helpful because the decision is a difficult one to make. The Stand-alone store has a possibility of adding profits of $700,000 if the market is favorable. With unfavorable markets no profits were predicted, but instead a loss of $400,000. With the renting of a commercial space (in a mall) the profits are expected to be 43 percent less than with a stand alone store, at the same time, during unfavorable markets the losses would also be less; approximately 14 percent less losses. [($50,000/$400,000) * 100]
The $20,000 for the market research is an investment because the uncertainty in which scenario to chose is minimized. The market research tells me that the best scenario is to build a stand alone store. The odds are that a 70% increase in profits is possible. The favorable outcome will depend on other factors that were considered in the decision tree, but the uncertainties of proceeding with the project are less.
Slide 12 and Speaker notes
- Preparing a Gnatt chart (red) at the beginning stages of the project, that is during the planning stages, is time-consuming. But the pay-off comes during the project when time is being saved and stress is relieved because a basis for the long-term planning has been laid. Gnatt allows enough flexibility to make necessary changes in the chart as the situation on the ground changes. The Critical Path (blue) is automatically generated and aids in the carrying out the project to its end. On the one hand the Gnatt chart aids in deciding how soon the project can be started, whereas the Critical Path predicts the longest amount of time it will take to finish the project. By using the Gantt chart and Critical Path simultaneously the project manager can determine which components along the critical path can be ‘crashed’ or not used because they are unnecessary. In that way, the project can be shortened in the most realistic manner.
Factor Rating Method of Choosing a Location
The visit to Shanghai remained a success after all, and the demand of our products is substantial over there as well. The following table compares two cities where we can open our new stores.
The company must select Beijing over Delhi because it is leading over its competitor in key areas of the business such as productivity and attitude. The Chinese will ensure that they provide their partners with high quality products and supplies in order to preserve FDI.
CPM adds some flexibility in choosing components in the timeline to crash, thereby shortening the project duration. But to make CPM more flexible, using CPM with PERT is a better choice. Project Evaluation and Review Technique (PERT). The reason is because with PERT a project director has the option of evaluating three different project end times. Plus, in order to help the project manager judge uncertainty, PERT calculates standard value and standard deviation.
Slide 13 and Speaker notes
Critical path sets one length of duration of the project giving one end time but the option to shorten the project using critical path is the opportunity to crash portions of the project that have been deemed unnecessary. The slide depicts the critical path form making up-grades at the Bellevue store. The length of duration is 108 2 days (about 3.5 days). By looking at the chart only, without even knowing what D component is about, we know that D does not have an effect on the length of the project because it runs in parallel to the critical path, not on the critical path.
The project manager must determine the 95 percent probability of completing the plan on time. The equation needed is Due Date = Expected Completion Time + (appropriate area under the normal curve * the project standard deviation) Z-values can be looked up on Appendix 1. Z is equal to the area under the normal curve. In this case the 95 percent level correlates with 1.64 (without going above the 95 percent level of probability) . 108.2 + (1.64 * 8.46) = 122.07 days, therefore 123 days are needed to finish the project. (95% C.I is given)
The project manager also must determine what component of the project can be crashed if unavoidable delays occur. In this case if 5 days are needed, then Activity one is the best choice for meeting the 5 days time table and will only cost of $500 per day (Total $2500). The CPM needs to be integrated with the PERT because the former one will help the management in driving down the time of the project whereas; the latter one can handle the complexity of the endeavour as well.
The Schuzworld Project Expansion Option needs to meet target dates.
A simple calculation can be used to find a target day and the equation includes the standard deviation (Where Z = 95% and
σp = standard deviation for the project)
Z = (Target Time – Expected Time) \ σp, so rearranging the equation gives
Expected Time = Project Completion Time + (Z * σp)
Target Time = (Z * σp) + Expected Time
= (1.645 * 8.46) + 108.20
= 122.12 days
Crashing by 5 days
The lowest crash cost (run by Gnatt programming and Excel) ranges from $2400.50 to $2500 and it is generated by crushing the activity I because it represents the lowest cost and highest benefit possible in terms of shortening the project.
Crashing by 5 days
Kidshuz shoes and sneakers:Production Mix
Linear programming was used to Linear Programming
Optimizing a mixed production line for Kidshuz shoes and sneakers
Was calculated on the Excel according to the linear programming
25 batches of Kidshuz shoes and 25 batches of Kidshuz sneakers
For the minimum cost achievable $87,500
Linear Programming
When X1 = # Kiltie Tassel Loafers
X2 = # Classic Penny Loafers
Optimize 800X1 + 1200X2
Constraints are
2X1+ 6X2 ≤ 1200
8X1 + 4X2 ≤ 1600
X1 ≥ 50, and
X2 ≥ 100
120 batches of Kiltie Tassel Loafers and 160 batches of Classic Penny Loafers would optimize the production of loafers and maximize profits.
The maximum profit possible is $288,000.
Slide 21
Whole page is on the slide like this:
Close up of number used for calculations is on the next page
The simulation inputs are
order quantity = 30 and reorder = 12
The simulation results (outputs) are demand = 9.85 units every 20 days
ending inventory = 10.25 every 20 days
lost sales = 28 every 20 days
The first Monte Carlo simulation called for an average daily ending inventory of 10.25 units/day
The simulation started with an order quantity of 30 and a reorder quantity of 12. The first Monte Carlo simulation calculated that when the demand is 9.85 units per day, then the ending inventory needs to be 10.25 units/day. An average lost sales of 1.4 units/day (28 lost sales/20 days = 1.4)
An average number of orders placed equal to one order approximately every 2.0 days (20 days / 9.85 orders = 2.03)
SLIDE 22
The next Monte Carlo simulation assumed an order quantity = to 35 units and reorder = 15 units
The results (outputs) are demand = 9.85 units/20 days
Ending inventory = 14.2 units/20 days
And lost sales = 1 lost sale/ days
The results show that the demand is approximately 9.85 units every 20 days, so 9.85 units/20 days = 0.4925 units/day or ~ 0.5 units per day. Average number of orders placed is 9.85 units every 20 days or 0.479 which can be rounded up to about 1 sale every 2 days
For reorders based on the outputs
9.58 units = (35 units) / (x days)
x days = 35 units / 9.85 units
x = 3.6 days,
when x equals approximately 3.6; it would be good to send 35 units every 3.6 days to maintain the end-of-day inventory of 14.2 units each day (predicted from the simulation)
The simulation predicts 1 lost sale every 20 days
References
Cardy, R.L., Miller, J.S., Ellis, A.D. (2007) “Employee equity: Toward a person-based approach to HRM, Human Resource Management Review, Volume 17(2): 140-151) http://dx.doi.org/10.1016/j.hrmr.2007.03.006
De, V.B. (2006) Decision trees for business intelligence and data mining. UAS Enterprise Miner. North Carolina: SAS Institute. http://support.sas.com/publishing/pubcat/chaps/57587.pdf
Excel Introduction to Monte Carlo Simulation, Microsoft http://office.microsoft.com/en-us/excel-help/introduction-to-monte-carlo-simulation-HA001111893.aspx
Ikeda, Y., Kubo, O., Kobayashi, Y. 2004) “Forecast of business performance using an agent-based model and its application to a decision tree Monte Carlo business valuation,” Physica A: Statistical Mechanics and its Applications, 344(1–2): 87-94, http://dx.doi.org/10.1016/j.physa.2004.06.093
Kendra, K. A. (2003). An organization design approach to project management. (Order No. 3092851, Benedictine University). ProQuest Dissertations and Theses, 180-180
