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Estimate the bandwidth (bits / s) of a 747 crossing the Atlantic filled with CDs.

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Estimate the bandwidth (bits / s) of a 747 crossing the Atlantic filled with CDs.

 

Clarifying questions?

1.       I think this is a multi part estimation question that is trying to determine if a plane were completely filled with CD’s

Estimation 1 the number of CDs that fit in the plane

Estimation 2 the amount of data contained on those CDs

Estimation 3 the time of travel of the plane

Estimation 4 the bandwidth of the plane data moved over duration of the flight

I have a few other questions, but I want to make sure I have the framework, right?  Let us assume that is the framework.

2.       Across the Atlantic

a.       LA to London, NYC to London or other destination?  Which direction - I believe going is faster.  Lets assume NYC to London

3.       747 should I assume that the plane is totally empty cargo plane (no seats) and that we have access to the entire storage area?  Yes

4.       Can I Assume that the CDs are all in Jewel Cases? Sure

Size of a 747 is really big – I want to compare it to something that I know like a school bus in length I think it is about 4-5 school buses long.  A school bus is 35-40 feet. 

Length 140-175 160-200  lets go with an average I rounded to 170 feet *12 = 2040 inches

Width I believe the traditional config of a 747 in economy is 3 asile 5 asile 3 the seat that I’m sitting in I estimate to be two feet wide and it’s comfortable – so an economy seat is less than that I’m going to say 20 inches each and the aisle is 3 feet

60+36+100+36+60 = 300 inches

300 inches

Lets assume that a plane is a cylinder so volume  is pie*(r)2 *length

3.14*(150)2 = 3.14*22500*2040 = 144,126,000in3 (just so everyone knows we are already pretty far off the actual cargo capacity of a 747 set up for Cargo is 42,270,336 in3 but this is an estimation question so keep moving forward.)

 

A jewel case is 4.25 inches by 4.25 inches by .25 inches.  So the volume occupied by a jewel case is 4.5in3  (thin jewel case is actually 5.59 in × 4.92 in × 0.20 in or 5.5in3)

 

Now I am sure that there is an optimal configuration of jewel cases to maximize fit but we are deep into an estimation question that still has multiple sections to go so lets just do a straight divide at this point.

144,126,000in3 /4.5in3  = 32,028,000 lets just call that 32 million cd’s at this point

Each CD holds 800mb of data (its been a long time but that is about right)

32,028,000 *800 =  25,622,400,000.00

*1000 to get to KB

*8 to get to bits = 205 trillion

 

Time of the flight NYC to London 7 hours

420 min

25200 seconds

205,000,000,000,000/25200 = 8.1 billion bits per second

 

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Clarifying Questions 

Are we talking about CDs, DVDs or Blu-ray Discs?  CD-ROMs

Are we going to remove all seats and everything from the plane to hold CDs?  Yes

Calculations

1 CD can hold up to 700 MB of data

 

Let’s convert CDs into some unit which can be measurable later compared to a 747.

Let’s assume one human body can hold upto 500 CDs

One row of 747 can hold upto 9 people. There are approximate 75 rows in a plane of that size. Total = 675 people.

Considering the height of the plane, we can hold upto 4 times that. 

If we remove cargo also, we can double that. So 675*4*2 people can be fit in that space

Total number of people = 675*8 = 5400 people can fit in empty 747 without seats, cargo and all.

 

Assume number of hours to cross the Atlantic = 8 hours 

 

Now let’s look at what is bandwidth? 

Bandwidth = memory capacity of CDROM in 747 / number of hours to cross the Atlantic 

  = 5400 * 700*500 MB / 8 hours

  = 5400 * 700 * 500 / 8*1000*60*60 GB/s

  = 65.625 GB/s

 

 

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The estimate of 500 cds per person seems random.  It would have been nice if you provided some idea about how you got to that estimate.  A cd jewel case is 5.59 in × 4.92 in × 0.20.  I am six feet tall.  So a stack 6 feet tall would be 360 cd's.  Lets just round it to 5x5x.2.  even if we laid down four in a square pattern it would be 10x10x.2.  I think most people are more than 10 inches square.  But assuming I fit into a 10inch box then I represent closer to 360 *4 cd’s =1440.  So, your per person estimate is low and was made without any suggestion of how you got there.

Then there is the cargo space estimation of 5400 people.  At 500cds per person = 2,700,000 cd’s, each with a volume of 5*5*.2= 5 cubic inches your total predicted space is 5400 people * 500 cds * 5 cubic inches = 13,500,000 cubic inches or 7,812 cubic feet.  Which is the equivalent of 100 feet in length, 10 feet wide and 7.8 feet high.  Here again a quick gut check would have shown you that you’re that the cargo space of a 747 is bigger than that.    

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