Airplane Engines

**sadmath** · September 25th 2009, 01:31 PM

"An airplane needs at least half of its engines working to stay airborne. If an engine independently functions with probability p, for what values of p is an airplane with three engines safer than one with five engines?"

Ok, so I'm going to try to work through this. I'm pretty much thinking aloud (atype? is there a word for that?) so please forgive my rambling or tangents.

If we say that an airplane is only as safe as it's likelihood of remaining airborne (which I don't think is an unfair assumption) then I need to find values for p such that the probability for remaining airborne is better with only three engines rather then five engines.

So the chance that the airplane is airborne (for a 3 engine plane) at any point in time is:
Since we have 3 engines we need exactly 1.5 engines running at any point in time. Since this is a discrete problem and there is no concept of a engine half running, we need at least 2 engines running.

So, P(X) = p, where X = 1, 2, 3 (for a 3 engine plane)

So the chances that we're in the air at any given point in time is:
The chances that Engines 1 and 2 are running, the chances that engines 1 and 3 are running and the chances that engines 2 and 3 are running.

P(1) * P(2) + P(1) * P(3) + P(2)* P(3) = p^2 + p^2 + p^2 = 3p^2

I'm not really sure what significance 3p^2 has, but I'm going to try it for the five engine plane now:

So again, P(X) = p, where X = 1,2,3,4,5 (for a 5 engine plane)

We need exactly 2.5 engines running (or in this case 3 engines). Meaning our possibilities are:

P(1) * P(2) * P(3) + P(1) * P(2) * P(4) + P(1) * P(2) * P(5) + P(1) * P(3) * P(4) + P(1) * P(3) * P(5) + P(1) * P(4) * P(5) + P(2) * P(3) * P(4) + P(2) * P(3) * P(5) + P(2) * P(4) * P(5) + P(3) * P(4) * P(5) = 10p^3

So now I guess I need to find values for p where 10p^3 < 3p^2.

10p^3 < 3p^2 = 10p < 3 = p < 3/10.

So any value for p less then 30%, we're safer with 3 engines then with 5 engines? what is the significance of the equations I came up with (10p^3 and 3p^2)?

**Plato** · September 25th 2009, 01:43 PM

As you note, in order for the three engine plane we need at least two engines working.
That probability is $p^3 +3p^2(1-p)$

For the five engine plane we need at least three engines working.
That probability is $p^5 +5p^4(1-p) +10p^3(1-p)^2$

**sadmath** · September 25th 2009, 01:46 PM

Originally Posted by Plato

As you note, in order for the three engine plane we need at least two engines working.
That probability is $p^3 +3p^2(1-p)$

For the five engine plane we need at least three engines working.
That probability is $p^5 +5p^4(1-p) +10p^3(1-p)^2$

I don't understand how you came up with those equations... At what points is my logic flawed? Could you please clarify?

**Plato** · September 25th 2009, 02:03 PM

Originally Posted by sadmath

I don't understand how you came up with those equations... At what points is my logic flawed? Could you please clarify?

From you previous courses, do you recall having studied binomial distributions of independent random variables.

Suppose we have a random variable that acts independently.
Consider tossing a die N times. Each toss has an independent outcome.

If we have independent random variable with probability $p$ then if there are N trials the probability of exactly $K,~1\le K \le N$ successes is $\binom{N}{K}p^K (1-p)^{N-K}$

In N trials the probability of at least $K,~1\le K \le N$ successes is $\sum\limits_{j = K}^N {\binom{N}{j}p^j (1-p)^{N-j}}$

See this web page.

**mr fantastic** · September 25th 2009, 05:31 PM

Of related interest: http://www.mathhelpforum.com/math-he...-question.html

**sadmath** · September 26th 2009, 08:05 AM

Ok so I gave it a shot and this is what I ended up with:

Let's consider the 3 engine plane first. In order for the plane to remain airborne it must have at least 1.5 engines running. Since there is no concept of an engine half-running and 1 engine would be too few, we need 2 engines running at all times. The probability that this plane is in the air at any given point in time is:

The probability that all 3 engines are operational + the probability that 2 of the engines are operational and 1 is nonoperational.

Equivalently:
$p^3 + p^2(1-p)$

Now let's consider the 5 engine plane. In order for this plane to remain airborne it needs 2.5 engines operational. Again since 2 would be too few we need 3 engines running at all times. The probability that this plane is in the air at any given point in time is:

The probability that all 5 engines are operational + the probability that 4 of the engines are operational and 1 is nonoperational + the probability that 3 of the engines are operational and 2 are nonoperational.

Equivalently:
$p^5 + p^4(1-p) + p^3(1-p)^2$

Now let's find values for p where a 3 engine plane is more likely to remain airborne than a 5 engine plane.

$p^3 + p^2(1-p) > p^5 + p^4(1-p) + p^3(1-p)^2$
$p + 1 - p > p^3 + p^2(1-p) + p(1-p)^2$
$1 > p^3 + p^2 - p^3 + p(1 - 2p + p^2)$
$p^3 - p^2 + p < 1$
$p(p^2 - p + 1) < 1$

Now what? I'm not sure how to calculate that inequality.

@Mr Fantastic

I had a look at that post you linked. He calculates the chances that any particular engine will fail. I think I'll try that. Maybe I'll figure out how to do it that way.

**Plato** · September 26th 2009, 08:14 AM

Originally Posted by sadmath

The probability that all 5 engines are operational + the probability that 4 of the engines are operational and 1 is nonoperational + the probability that 3 of the engines are operational and 2 are nonoperational.

Equivalently:
$p^5 + p^4(1-p) + p^3(1-p)^2$

You forgot the binomial coefficients.
${\color{red}\binom{5}{5}}p^5 +{\color{red}\binom{5}{4}} p^4(1-p) + {\color{red}\binom{5}{3}}p^3(1-p)^2$

**sadmath** · September 26th 2009, 08:26 AM

Alright. I think I got it right now, but I still don't know how to proceed. I'm still trying to go about it through the failures instead of the operational cases, but I wouldn't mind knowing how to figure it out this way as well.

Here's what I get now:

$p^3 + 3p^2(1-p) > p^5 + 5p^4(1-p) + 10p^3(1-p)^2$
$p + 3 - 3p > p^3 + 5p^2(1-p) + 10p(1-p)^2$
$3 - 2p > p^3 + 5p^2 - 5p^3 + 10p(1 - 2p + p^2)$
$3 > 12p - 15p^2 + 6p^3$
$p(2p^2 - 5p + 4) < 1$

**sadmath** · September 26th 2009, 08:36 AM

Sorry about double posting but I can't figure it out for the life of me.

In the post Mr. Fantastic references he calculates the rate of failure as:

$1-p$

So the entire (3 engine) airplane should fail with probability:

$(1-p)^2$

But why isn't plato's thing applied? I mean shouldn't the probability that the last engine is still operational be included?

$p(1-p)^2 + (1-p)^3$

Why is it that when considering the failing case of this situation we can ignore the probabilities we had to consider in the operational case?

**mr fantastic** · September 26th 2009, 02:55 PM

Originally Posted by sadmath

Sorry about double posting but I can't figure it out for the life of me.

In the post Mr. Fantastic references he calculates the rate of failure as:

$1-p$

So the entire (3 engine) airplane should fail with probability:

$(1-p)^2$ Mr F says: Let X be the random variable 'number of engines that fail'. For the 3 engine plane, Pr(3 engine plane does not remain airborn) = Pr(X = 2) + Pr(X = 3). You only considered the case X = 3. Do this correctly and you will get an expression that is equivalent to Plato's.

But why isn't plato's thing applied? I mean shouldn't the probability that the last engine is still operational be included?

$p(1-p)^2 + (1-p)^3$

Why is it that when considering the failing case of this situation we can ignore the probabilities we had to consider in the operational case?

The whole problem here is a complete failure to set up the problem in a systematic way.

Step 1: Define the random variable. Let Y be the random variable number of engines that do not fail.

Step 2: Define the distribution that Y follows.

3 engine plane: Y ~ Binomial(n = 3, p = p).

5 engine plane: Y ~ Binomial(n = 5, p = p).

Step 3: Write down probability statements that can be used to solve the question.

5 engine plane: Pr(remains airborn) $= \Pr(Y = 3) + \Pr(Y = 4) + \Pr(Y = 5) = \binom{5}{5}p^5 + \binom{5}{4} p^4(1-p) + \binom{5}{3} p^3(1-p)^2 = f(p)$ .

3 engine plane: Pr(remains airborn) $= \Pr(Y = 2) + \Pr(Y = 3) = ....... = g(p)$ .

Step 4: Solve the problem. Solve the inequality $g(p) > f(p)$ for $p$ .

(In the post I referenced, I defined X as the random variable number of engines that fail. The above structure is applied in exactly the same way.)

**sadmath** · September 26th 2009, 08:49 PM

I feel like a failure since you guys have to hand hold me every step of the way.

Like I stated above this is what I ended up with:

$p^3 + 3p^2(1-p) > p^5 + 5p^4(1-p) + 10p^3(1-p)^2$
$p + 3 - 3p > p^3 + 5p^2(1-p) + 10p(1-p)^2$
$3 - 2p > p^3 + 5p^2 - 5p^3 + 10p(1 - 2p + p^2)$
$3 > 12p - 15p^2 + 6p^3$
$p(2p^2 - 5p + 4) < 1$

How do I simplify $p(2p^2 - 5p + 4) < 1$ or did I mess up somewhere again?

**mr fantastic** · September 27th 2009, 12:43 AM

Originally Posted by sadmath

I feel like a failure since you guys have to hand hold me every step of the way.

Like I stated above this is what I ended up with:

$p^3 + 3p^2(1-p) > p^5 + 5p^4(1-p) + 10p^3(1-p)^2$
$p + 3 - 3p > p^3 + 5p^2(1-p) + 10p(1-p)^2$
$3 - 2p > p^3 + 5p^2 - 5p^3 + 10p(1 - 2p + p^2)$
$3 > 12p - 15p^2 + 6p^3$
$p(2p^2 - 5p + 4) < 1$

How do I simplify $p(2p^2 - 5p + 4) < 1$ or did I mess up somewhere again?

You need to solve $6p^3 - 15p^2 + 12p - 3 < 0$ .

Note that $6p^3 - 15p^2 + 12p - 3 = 3(p - 1)^2(2p - 1)$ .

The factorisation is easily done by noting that p = 1 is a solution to the cubic. Alternatively, see here: Wolfram|Alpha

Therefore the solution to the inequality is ........

**sadmath** · September 27th 2009, 06:13 PM

Ok, I think I finally got it. You guys were an amazing help.

My final solution continuing from the factorization that Mr. Fantastic helped me with:

$6p^3 - 15p^2 + 12p - 3 < 0$
$3(p-1)^2(2p-1) < 0$

Now solve for p

$(p-1)^2 < 0$
$p < 1$

$(2p-1) < 0$
$2p < 1$
$p < {1 \over 2}$

Since ${1 \over 2} < 1$ we can just say that $p < 1$

Right?

Thanks for all your help guys.

**sadmath** · September 27th 2009, 07:07 PM

Sorry again for double posting.

But I think I got it wrong again. if a 3-engine plane is safer then a 5-engine plane whenever p < 1 that means that only when p is 100% are they equivalently safe, and all other values would make the 3 engine plane safer. I tried plugging in values like 0.9999 into the inequality I started with, and it doesn't work.

However putting 0.4999 into the inequality does, so I'm thinking that the p < 1/2 is the right answer but I'm not sure I understand how to get that from the equation.

**mr fantastic** · September 27th 2009, 10:49 PM

Originally Posted by sadmath

Ok, I think I finally got it. You guys were an amazing help.

My final solution continuing from the factorization that Mr. Fantastic helped me with:

$6p^3 - 15p^2 + 12p - 3 < 0$
$3(p-1)^2(2p-1) < 0$

Now solve for p

$(p-1)^2 < 0$
$p < 1$

$(2p-1) < 0$
$2p < 1$
$p < {1 \over 2}$

Since ${1 \over 2} < 1$ we can just say that $p < 1$

Right?

Thanks for all your help guys.

Did you examine the graph in the wolfram link I gave you in my previous post? It should be clear from that graph that $6p^3 - 15p^2 + 12p - 3 < 0$ for $0 < p < \frac{1}{2}$ .

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