Assignment 17.5

A jewel case for a compact disc is made from polycarbonate ($2.20 per lb) by a thermoplastic molding process. Each CD case uses 20 grams of plastic. The parts will be made in a 10-cavity mold that makes 1400 parts per hour at an operating cost of $20 per hour. Manufacturing ovehead is 40%. Since the parts ae sold in large lots, the G&A expenses are a low 15%. Profit is 10%. What is the estimated selling price for each CD case?

Solutions:

jewel case = 20 gram ($ 2.20/lb)
OH = 40%
G&A = 15%
Profit = 10%

1 lb = 454 gram
1 gram = 2.2x10^-3 lb

10(2.2x10^-3) = $ 0.097/unit

$20/1400 = $ 0.014 per unit
Total cost = 0.097 + 0.014 = $ 0.111

Cost/hour = $0.111 x 1400 = $155.40
0.4 x 155.4 = $ 62.16
0.15 x 155.4 = $ 23.31
Total = 155.4 + 62.16 + 23.31 = $ 240.87
Profit = 1.1 x 240.87 = $ 264.96

Estimated selling price/unit = $ 264.96/1400 = $ 0.19/unit

Assignment 17.4: Manufacturing Cost and Selling Price

Assignment 17.4 Manufacturing Cost and Selling Price for a Turbine

A manufacturer of a small hydraulic turbine has the annual cost data given. Calculate the manufacturing cost and the selling price for the turbine.

Raw material and components costs               $2,150,000

Direct labor                                                     $950,000

Direct expenses                                               $60,000

Plant manager and staff                                    $180,000

Utilities for plant                                               $70,000

Taxes and insurance                                         $50,000

Plant and equipment depreciation                    $120,000

Warehouse Expenses                                       $60,000

Office Utilities                                                  $10,000

Engineering expenses (plant)                             $90,000

Engineering staff and salaries (plant)                  $30,000

Administrative staff salaries                              $120,000

Sales staff, salaries and commissions                $100,000

Total Annual Sales                                           60 units

Profit Margin                                                   15%

Variable costs

Raw material and components costs               $2,150,000

Direct labor                                                    $950,000

Direct expenses                                              $60,000

Engineering expenses (plant)                            $90,000

Engineering staff and salaries (plant)                 $30,000

Total Variable Costs                                       $3,280,000

Factory Expenses

Utilities for plant                                             $70,000

Taxes and insurance                                        $50,000

Plant and equipment depreciation                    $120,000

Warehouse Expenses                                      $60,000

Total Factory Expenses                                  $300,000

General and Administrative Expenses (G & A)

Plant manager and staff                                  $180,000

Office Utilities                                                $10,000

Administrative staff salaries                             $120,000

Total G & A                                                   $310,000

Manufacturing Cost = Variable costs + Factory Expenses + General and Administrative Expenses

                                  = $3,280,000 + $300,000 + $310,000

                                  = $3,890,000

Sales, staff, salaries and commissions      $100,000

Total Cost = Manufacturing costs + Sales, staff, salaries and commissions

                  = $3,990,000

Profit Margin = Profit / Sales = 15%

Profit (P) = Selling Price (S) – Total Cost (C_T)

S = P + C_T

S = 0.15S + C_T

S – 0.15S = C_T

0.85S = C_T

S = C_T / 0.85

S = $3,990,000 / 0.85

S = $4,694,117.65

Total Annual Sales = 60 units

Selling Price for One Turbine = S / 60

                                                 = $4,694,117.65 / 60

Selling Price                           = $78,235.29 per unit         

Assignment 17.1

Solution:

Adjustment for the cost of the different size or capacity of the cyclone dust collector need to be purchased can not be calculated directly. Note that while the new units has 10 times the capacity as the unit purchased in 1985 will cost 10X more, because of economy of scale. However, the purchase cost will have increased because of inflation in the 27 years since it was purchased in 1985. Assuming cost inflation or 5 % per year, the original cost of $35,000 is now equivalent to

new cost for the 100 ft^3/min = [$35,000 x 5%] + $35,000 = $36,750

if 100 ft^3/min = X

1000 ft^3/min = 10X

Thus, the estimate of the cost to purchase 1000 ft^3/min in 2012 = 10X = $367,500

Assignment 11.3: Materials Selection

Which material property would be selected as guide in material selection if the chief performance characteristics of the component was:

PERFORMANCE CHARACTERISTICS	MATERIAL PROPERTY
Strength in Bending	Tensile Strength
Resistance to Twisting	Hardness
The Ability of a Sheet Material to be Stretched into a Complex Curvature	Ductility
Ability to Resist Fracture from Cracks at Low Temperatures	Malleability
Ability to Resist Shattering if Dropped On the Floor	Toughness
Ability to Resist Alternating Cycles of Rapid Heating and Cooling	Thermal Shock Resistance

Assignment 9.9: Power Point Presentation for the first design review [PROJECT]

Assignment Failure Mode and Effects Analysis (FMEA)

Failure Mode and Effects Analysis (FMEA)

                Failure mode and effects analysis (FMEA) is a team – based methodology for identifying potential problems with the new or existing designs. It is the first step of a system reliability study. FMEA is the core task in reliability engineering, safety engineering, and quality engineering. FMEA involves reviewing as many components, assemblies, and subsystem as possible to identify failure modes, and their causes and effects. In order to determine the components of the process that are most in need of change, FMEA includes the following steps:

1.      Steps in the process.

2.      Failure modes (What could go wrong?)

3.      Failure causes (Why would the failure happen?)

4.      Failure effects (What would be the consequences of each failure?)

            Three factors that are considered in developing a FMEA:

1.      The severity of the failure.

2.      The probability of occurrence of failure.

3.      The likelihood of detecting the failure in either design or manufacturing, before the product is used by the customer.

            Different types of FMEA analysis:

1.      Functional,

2.      Design, and

3.      Process FMEA.

            Major benefits derived from a properly implemented FMEA are as follows:

1.      It provides documentation in selecting design with high probability of successful operation and safety.

2.      A uniform documentation method of assessing potential failure mechanism, failure modes and their impact on system operation, resulting in a list of failure modes ranked according to seriousness of their system impact and likelihood of occurrence.

3.      Early identification of single failure points (SFPS) and system interface problems which may be critical to mission success and/or safety.

4.      An effective method for evaluating the effect of proposed changes t the design and/or operational procedures on mission success and safety.

5.      A basis for in – flight troubleshooting procedures and for locating performance monitoring and fault – detection devices.

6.      Criteria for early planning of tests.

Example of FMEA analysis for car tire:

Keywords:

SEV – severity

OCC – occurrence

DET – Detection

RPN – Risk Priority Number

Function or Process Step	Failure Type	Potential Impact	SEV	Potential Causes	OCC	Detection Mode	DET	RPN
Briefly outline function, step or item being analyzed	Describe what has gone wrong	What is the impact on the key output variables or internal requirements?	How severe is the effect to the customer?	What causes the key input to go wrong?	How frequently is this likely to occur?	What are the existing controls that either prevent the failure from occurring or detect it should it occur?	How easy is it to detect?	Risk priority number
Tire function: support weight of car, traction, comfort	Flat tire	Stops car journey, driver and passengers stranded	10	Puncture	2	Tire checks before journey. While driving, steering pulls to one side, excess noise	3	60

Recommended Actions	Responsibility	Target Date	Action Taken	SEV	OCC	DET	RPN
What are the actions for reducing the occurrence of the cause or improving the detection?	Who is responsible for the recommended action?	What is the target date for the recommended action?	What were the actions implemented? Now recalculate the RPN to see if the action has reduced the risk.
Carry spare tire and appropriate tools to change tire	Car owner	From immediate effect	Spare tire and appropriate tools permanently carried in trunk	4	2	3	24

Criteria for FMEA Analysis

            FMEA is analyzed based on three criteria:

1.      Severity effect on customer.

2.      Occurrence of failure.

3.      Easy to detect.

            It is then ranked from 1 (low) to 10 (high) for each criterion.

Table 1: Severity, Occurrence and Detection Ratings

	Description	Low Number	High Number
Severity	Severity ranking encompasses what is important to the industry, company or customers (e.g., safety standards, environment, legal, production continuity, scrap, loss of business, damaged reputation)	Low impact	High impact
Occurrence	Rank the probability of a failure occuring during the expected lifetime of the product or service	Not likely to occur	Inevitable
Detection	Rank the probability of the problem being detected and acted upon before it has happened	Very likely to be detected	Not likely to be detected

Formula for RPN is:

RPN = severity x occurrence x detection

 

 

 

 

Setting Priorities

 

            After failure has been identified, the FMEA list of failure is adjusted in descending RPN order. This can ensure that the corrective action can be focused.

 

Making Corrective Action

            Once the priorities have been agreed, it’s time to generate appropriate corrective actions for reducing the occurrence of the failure modes. When the corrective actions have been done, it is better to rescore and reassess the severity, probability of occurrence and likelihood of detection for the top failure modes. This enables to determine the effectiveness of the corrective action taken.

Assignment 8.3: New fingernail clipper

Normal or conventional fingernail clipper is consisting with the supporting shaft, upper and lower steel plate, and pressurizing bar. However, for some modification the design had been improvised for more aesthetic value. Casing and magnifying glass had been the additional parts into the new design of fingernail clipper. The casing is purposely to protect the nail cutting edge from tearing of thorn the clothes especially when keep it into the bag. The magnifying glass is to help for elderly especially to cut their nail and toe nail which is a bit far.