CH2 HW

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Mercer University *

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560

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Industrial Engineering

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Apr 3, 2024

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docx

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11

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ISE 460/560 Chapter 2 Homework: all six problems I1 =Demand of Component/((1−d1) ∗ (1−d2) ∗ (1−d3)) =1,000(1−3%)×(1−5%)×(1−7%) =1,167 units I2 =Demand of Component/((1−d2) ∗ (1−d3)) =1,000(1−5%)×(1−7%) =1,132 units I3 =Demand of Component/((1−d3)) =1,0001−7% =1,075 units The scrap values and scrap cost at each step: Process 3 =I3−Demand of component =1,075−1,000 =75 units Scrap cost =Scrap cost at Process 3 ∗ Process 3 scrap value =$15×75 =$1,125
Process 2 =I2−I3 =1,132−1,075 =57 Scrap cost =Scrap cost at Process 2 ∗ Process 2 scrap value =$10×57 =$570 Process 1 =I1−I2 =1,167−1,132 =35 Scrap cost =Scrap cost at Process 1 ∗ Process 1 scrap value =$5×35 =$175 Total scrap cost =Scrasp cost for process 1+ Scrap Cost for process 2+Scrap cost for process 3 =$175+$570+$1,125 =$1,870 Reversed scrap rates system: I1 =1167 units I2 =Demand for component/((1−d1) ∗ (1−d2)) =1,000(1−3%)×(1−5%) =1,085 I3 =Demand for component/(1−d1) =1,0001−3% =1,031 The scrap values and scrap cost at each step: Process 3 =I3−Demand for component =1,031−1,000 =31 units
Scrap cost for process 3 =Scrap value of process 3 ∗ scrap cost at process 3 =$15×31 =$465 Process 2 =I2−I3 =1,085−1,031 =54 Scrap cost for process 2 =Scrap value of process 2 ∗ scrap cost at process 2 =$10×54 =$540 Process 1 =I1−I2 =1,167−1,085 =82 Scrap cost for process 1 =Scrap value of process 1 ∗ scrap cost at process 1 =$5×82 =$410 Total scrap cost =Scrasp cost for process 1+ Scrap cost for process 2+Scrap cost for process 3=$465+ $540+$410=$1,415 Conclusion: I would prefer the second system where the scrap rates are reversed as we can see that the scrap cost in this system is less than the scrap cost in first system
Given: d1=5% d2=5% d3=10% d4=2% First, convert failure rate to reliability. Reliability of Component 1 is: R1 =1−d1R1 =1−5%R1 =1−5/100R1 =0.95 Reliability of Component 2 is: R2 =1−d2R2 =1−5%R2 =1−5/100R2 =0.95
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