Problem 9.1:
Examine the detail drawing for a product designed by your team. Be sure you identify the actual shape, dimensions, and tolerances. What other information contained in the drawing?
Tuesday, 23 December 2014
Tuesday, 16 December 2014
problem 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
problem 17.2:
Many consumers item today are designed in the United State and manufactured over-seas where labor cost is much lower. A middle range athletic shoe from a name brand manufacturer sells for $70 in the U.S. The shoe company buys the shoe from an off-shore supplier for $20 and sells to the retailer for $36. The profit margin for each unit in the chain is: supplier-9%, shoe company-17%, retailer-13%. Estimate major categories of cost breakdown for each unit of chain. Do this as a team problem and compare the results for the entire class.
SOLUTION:
Estimation of the major categories
of cost breakdown for each unit in the chain :
Shoe Company
Production
labor
$2.75
Materials
$9.00
Rent,
equipment
$3.00
Supplier’s operating
profit
$1.75
Duties
$3.00
Shipping
$0.50
Total
cost
$20.00
Retailer
Research and
development
$20.50
Promotion and
advertising
$4.00
Sales, distribution,
admin
$5.00
Company’s operating profit
$6.50
Total
cost
$36.00
Consumer
Retailer’s
rent $20.00
Personnel
$15.00
Others
$15.00
Retailer’s operating
profit
$20.00
Total
cost
$70.00
problem 17.4:
SOLUTION:
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 (CT)
S = P + CT
S = 0.15S + CT
S – 0.15S = CT
0.85S = CT
S = CT / 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
problem 17.10:
A company has received an order for
four sophisticated space widgets. The buyer will take delivery of one unit at
the end of the first year and one unit at the end of each of the succeeding
three years. He will pay for a unit immediately upon receipt and not before.
However, the manufacturer can make the units ahead of time and store them at
negligible cost for future delivery.
The chief component of cost of the space widget is labor at $25 per h. All units made in the same year can take advantage of an 80% learning curve. The first unit requires 100,000 h of labor. Learning occurs only in one year and is not carried over from year to year. If money is worth 16% after a 52% tax rate, decide whether it would be more economical to build four units the first year and store them, or build one unit in each of the four years.
The chief component of cost of the space widget is labor at $25 per h. All units made in the same year can take advantage of an 80% learning curve. The first unit requires 100,000 h of labor. Learning occurs only in one year and is not carried over from year to year. If money is worth 16% after a 52% tax rate, decide whether it would be more economical to build four units the first year and store them, or build one unit in each of the four years.
SOLUTION:
For an 80% learning curve, n = -0.322. The first unit requires y1 = 100,000 h to build.
The cost of making four units in one year is:
(100,000 + 80,000 +70,175 +63,979) x $25/h = $7,853,850
No Advantage from Learning Curve: If we do not take advantage of the learning curve we would make one space widget each year for a total cost over four years of
(100,000) x $25 x 4 = $10,000,000. This neglects the cost of living escalation that would occur each year, but it will be assumed that this would equal the cost of maintaining widgets in storage for up to four years.
In terms of engineering economy, the problem becomes which of the two alternatives results in the lowest present value. The present value (P) after taxes, assuming that labor costs are paid at the end of the year (an unrealistic but conventional assumption in engineering economy), for the use of the learning curve situation is given as:
If the four units are made in four successive years,
= 0.48(2,155,172 + 1,857,907 + 1,601,644 + 1,380,728) =0.48(6,995,451)
= $3,357,816
Taking advantage of the learning curve saves over $100,000 in cost compared with making one widget each year and deferring the costs into the future. Making the widgets in a batch also frees up the work force to take on other profitable work
Sunday, 14 December 2014
ASSIGNMENT 8
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
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
Formula for RPN is:
RPN = severity x occurrence x detection
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