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kotykmax [81]
3 years ago
7

Problem 3.10 One/half million parts of a certain type are to be manufactured annually on dedicated production machines that run

16 hours per day, five days per week, 50 weeks per year. (a) The cycle time to produce one part is 1.8 min. If availability and utilization are each 100%, and setup time is neglected, how many of the machines will be required to satisfy demand? (b) Solve part (a) except that availability = 95%.
Engineering
2 answers:
goldenfox [79]3 years ago
8 0

Answer:

  • 4 machines
  • 4 machines

Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

  .5 million/(.1333... million/machine) = 3.75 machines

The demand can be satisfied by 4 machines.

__

(b) If availability is 95% then 3.75/0.95 ≈ 3.95 machines will be required.

The demand can be satisfied by 4 machines.

sesenic [268]3 years ago
7 0

Answer:

  • 4 machines
  • 4 machines

A.

B

➡ are both in 4 machines used in

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Answer:

The heat from the sun melted it

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In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh
astraxan [27]

( 12 17 18 19 25 )

<u>Explanation:</u>

<u>First Pass:</u>

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

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( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

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3 years ago
Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele
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3 years ago
2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
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Answer:

coupling is in tension

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= 32 ( 25 / 35 )^2

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