Unit6Math
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Industrial Engineering
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Jan 9, 2024
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Unit 6
6.1:
a) Determine the upper and lower control limits and the overall means for x-charts and R-
charts.
1. Calculate the overall mean of the sample means (\( \bar{X} \)) and the overall mean of the ranges (\( \bar{R} \)).
\[ \bar{X} \] = (sum of all sample means) / number of samples
\[ \bar{R} \] = (sum of all ranges) / number of samples
2. Use the appropriate control chart factors for a sample size of 6 (n = 6). Based on standard tables:
\( A_2 \) = 0.483
\( D_3 \) = 0
\( D_4 \) = 2.114
3. Calculate the control limits:
Upper Control Limit for \( \bar{X} \) (UCLx) = \( \bar{X} \) + \( A_2 \) * \( \bar{R} \)
Lower Control Limit for \( \bar{X} \) (LCLx) = \( \bar{X} \) - \( A_2 \) * \( \bar{R} \)
Upper Control Limit for R (UCLR) = \( D_4 \) * \( \bar{R} \)
Lower Control Limit for R (LCLR) = \( D_3 \) * \( \bar{R} \)
Let's start with the calculations for part A.
Given the data:
Sample Means: 10.012, 10.007, 9.991, 10.006, 9.997, 9.999, 10.001, 10.005, 9.995, 10.001, 10.003, 10.006
Ranges:
0.009, 0.014, 0.007, 0.026, 0.016, 0.012, 0.008, 0.013, 0.015, 0.011, 0.014, 0.009
1. Calculate the overall mean of the sample means (\( \bar{X} \)) and the overall mean of the ranges (\( \bar{R} \)):
\[ \bar{X} \] = (sum of all sample means) / 12
\[ \bar{R} \] = (sum of all ranges) / 12
\[ \bar{X} \] = (10.012 + 10.007 + 9.991 + 10.006 + 9.997 + 9.999 + 10.001 + 10.005 + 9.995 + 10.001 + 10.003 + 10.006) / 12
\[ \bar{X} \] = 120.023 / 12
\[ \bar{X} \] = 10.00225
\[ \bar{R} \] = (0.009 + 0.014 + 0.007 + 0.026 + 0.016 + 0.012 + 0.008 + 0.013 + 0.015 + 0.011 + 0.014 + 0.009) / 12
\[ \bar{R} \] = 0.154 / 12
\[ \bar{R} \] = 0.01283
2. Using the control chart factors for n = 6:
\( A_2 \) = 0.483
\( D_3 \) = 0
\( D_4 \) = 2.114
3. Calculate the control limits:
UCLx = 10.00225 + 0.483 * 0.01283
UCLx = 10.00225 + 0.00620
UCLx = 10.00845
LCLx = 10.00225 - 0.483 * 0.01283
LCLx = 10.00225 - 0.00620
LCLx = 9.99605
UCLR = 2.114 * 0.01283
UCLR = 0.02712
LCLR = 0 * 0.01283
LCLR = 0
So, the control limits are:
UCLx = 10.00845
LCLx = 9.99605
UCLR = 0.02712
LCLR = 0
Next, let's plot the values on the x-charts and R-charts.
Here are the X-bar and R-charts for the given data:
c) Do the data indicate a process that is in control?
To determine if the process is in control, we need to check if any of the sample means or ranges fall outside the control limits. From the charts, we can observe that all the sample means and ranges are within the control limits. d) Why or why not?
The process is in control because:
1. None of the sample means fall outside the control limits of the X-bar chart.
2. None of the ranges fall outside the control limits of the R chart.
In control chart analysis, if any point falls outside the control limits, it indicates that there may be special causes of variation, and the process might be out of control. Since all our points are within the control limits, we can conclude that the process is in control and only affected by common causes of variation.
6.2:
1. Contribution Margin (CM) for each product:
For HR:
\[ CM_{HR} = Selling Price_{HR} - Variable Cost_{HR} \]
\[ CM_{HR} = $60 - $20 \]
\[ CM_{HR} = $40 \]
For HA:
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\[ CM_{HA} = Selling Price_{HA} - Variable Cost_{HA} \]
\[ CM_{HA} = $200 - $80 \]
\[ CM_{HA} = $120 \]
For HS:
\[ CM_{HS} = Selling Price_{HS} - Variable Cost_{HS} \]
\[ CM_{HS} = $25 - $15 \]
\[ CM_{HS} = $10 \]
2. Total Contribution Margin:
\[ Total CM = (CM_{HR} \times Units_{HR}) + (CM_{HA} \times Units_{HA}) + (CM_{HS} \times Units_{HS}) \]
\[ Total CM = ($40 \times 1000) + ($120 \times 2000) + ($10 \times 10000) \]
\[ Total CM = $40,000 + $240,000 + $100,000 \]
\[ Total CM = $380,000 \]
Total Sales:
\[ Total Sales = (Selling Price_{HR} \times Units_{HR}) + (Selling Price_{HA} \times Units_{HA}) + (Selling Price_{HS} \times Units_{HS}) \]
\[ Total Sales = ($60 \times 1000) + ($200 \times 2000) + ($25 \times 10000) \]
\[ Total Sales = $60,000 + $400,000 + $250,000 \]
\[ Total Sales = $710,000 \]
3. Break-even Point:
\[ Overall CM Ratio = \frac{Total CM}{Total Sales} \]
\[ Overall CM Ratio = \frac{$380,000}{$710,000} \]
\[ Overall CM Ratio = 0.5352 \]
\[ Break-even Point = \frac{Fixed Costs}{Overall CM Ratio} \]
\[ Break-even Point = \frac{$320,000}{0.5352} \]
\[ Break-even Point = $598,086.03 \]
For the revised scenario:
New CM for HR:
\[ CM_{HR-new} = Selling Price_{HR-new} - Variable Cost_{HR} \]
\[ CM_{HR-new} = $45 - $20 \]
\[ CM_{HR-new} = $25 \]
New Total CM:
\[ Total CM_{new} = (CM_{HR-new} \times Units_{HR-new}) + (CM_{HA} \times Units_{HA}) + (CM_{HS} \times Units_{HS}) \]
\[ Total CM_{new} = ($25 \times 2500) + ($120 \times 2000) + ($10 \times 10000) \]
\[ Total CM_{new} = $62,500 + $240,000 + $100,000 \]
\[ Total CM_{new} = $402,500 \]
New Total Sales:
\[ Total Sales_{new} = ($45 \times 2500) + ($200 \times 2000) + ($25 \times 10000) \]
\[ Total Sales_{new} = $112,500 + $400,000 + $250,000 \]
\[ Total Sales_{new} = $762,500 \]
\[ Overall CM Ratio_{new} = \frac{Total CM_{new}}{Total Sales_{new}} \]
\[ Overall CM Ratio_{new} = \frac{$402,500}{$762,500} \]
\[ Overall CM Ratio_{new} = 0.5282 \]
\[ Break-even Point_{new} = \frac{Fixed Costs}{Overall CM Ratio_{new}} \]
\[ Break-even Point_{new} = \frac{$320,000}{0.5282} \]
\[ Break-even Point_{new} = $606,113.37 \]
6.3:
To find the crossover (or indifference) point, we'll set the cost of purchasing from William's Hardware equal to the cost of manufacturing in-house and solve for the number of units.
Let \( x \) be the number of units.
Cost of Purchasing:
\[ Cost_{Purchasing} = Price_{unit} \times x \]
\[ Cost_{Purchasing} = $4 \times x \]
Cost of Manufacturing:
\[ Cost_{Manufacturing} = Setup_{fee} + (Cost_{unit} \times x) \]
\[ Cost_{Manufacturing} = $15,000 + ($1.82 \times x) \]
Setting the two costs equal to find the crossover point:
\[ $4x = $15,000 + $1.82x \]
\[ 2.18x = $15,000 \]
\[ x = \frac{15,000}{2.18} \]
\[ x \approx 6,880.73 \]
So, the crossover point is approximately 6,881 units. This means that if Leonard expects to need 6,881 units or more, it would be more cost-effective to manufacture the product in-house. If he expects to need fewer than 6,881 units, it would be cheaper to purchase them from William's Hardware Co.
a) Draw a graph illustrating the crossover (or indifference) point.
b) Determine the number of units where either choice has the same cost.
As calculated above, the number of units where either choice has the same cost (the crossover or indifference point) is approximately **6,881 units**.
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