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In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is?

Using the standard equation f(x) = a * b^x, I figured that when a = 1000000 and b = 8912509381337455, the answer would be 46.02.

Is this correct a fine approach? However it involves using calculator.

Further, this was a multiple choice question in some exam, and the options given were:

a) 40.00 b) 46.02 c) 60.01 d) 92.02.

I saw an explanation using the process of elimination as follows:

(down) load cycles for failure (increase) exponentially

Load = x
Cycles for failure = y
x/2 = y^2, x/3 = y^3, x/4 = y^4

Hence,

Option a) 80/2 = 100 ^ 2 = 10,000 cycles
Option d) Load is >80 hence it is not possible
Option c)
Load = 60.01 = 3/4 (80)
From the given relation,
3/4 (80) = 10 ^ 4/3 = 464 cycles which is not right

So the answer is option b)

Is this right?

I mean is the relation x/2 = y^2, x/3 = y^3, x/4 = y^4 correct?

I would have accepted if it is x/2 = y^1/2, X/3 = y^1/2, x/4 = y^1/4.

Nishant
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  • So basically the derivation is not correct right? It uses that assumption and also 464 cycles is not right for a load of 60.02. Btw you also mean x/8 = y^8, x/16 = y^16 etc. Sorry still learning to use math overflow. – Nishant Aug 14 '16 at 11:05
  • From the wording of the sentences, it may be wrong, but I realized that the first statement is more in general characterizing the exponential decrease and I would like to take it back, sorry!! – Satish Ramanathan Aug 14 '16 at 11:14
  • Hence the process of elimination sounds correct – Satish Ramanathan Aug 14 '16 at 11:15
  • I still think the real relation should be x/2 = y^1/2, X/3 = y^1/2, x/4 = y^1/4. Also I think the value 464 cycles is not correct for 60 units. I mean I need to try if it works for the value mentioned. Notice that regardless of decrease or increase it depends on x as is and not the inverse. – Nishant Aug 14 '16 at 11:17

2 Answers2

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The derivation is right $\frac{3x}{4} = \frac{\frac{3x}{2}}{2} = y^{\frac{4}{3}}$. Hence it is valid.

The general relationship is as below:

x = number of loads

y = cycles of failure

The realtionship is

$\frac{x}{b} = y^b$

according to the first and third statement. The first statement is the opposite way of expressing the explanation given in the third statement if you notice. That is why you are confused.

Edit:

The correct way:

$100 = N_0 e^{-80\lambda}\tag{1}$

$10000= N_0 e^{-40\lambda}\tag{2}$

From the above we get, $\lambda = \frac{ln(100)}{40}=0.11515$ $

$$N_0 = 100*e^{2ln(100)} = 1000000$$

$$N(t) = 1000000*e^{\left(-46.02*0.11515\right)} = 5000$$

Thus this is the right model for this

The earlier one is done to skin the cat without working out the details.

Satish Ramanathan
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  • And by that you mean x/2 = y^1/2, X/3 = y^1/2, x/4 = y^1/4 is a wrong relation right? – Nishant Aug 14 '16 at 11:25
  • Yes it is wrong, See further explanation – Satish Ramanathan Aug 14 '16 at 11:34
  • But if you take 46.02 case, it is 46.02/80=0.57525. Take the inverse ~1.74 and 100^1.74 != 5000. Am I doing it right? Somehow I am not thinking that relaitionship is correct. – Nishant Aug 14 '16 at 14:05
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    See the updated model – Satish Ramanathan Aug 14 '16 at 15:45
  • This is 100% right, agree with this. It follows the exponential decay function as well. Don't you think the earlier one is actually incorrect, not just even close to right? I proved that by finding the equivalent factor when you have 46.02 and raise the value of y against that. It won't come close to 5000! That should be a good proof right? Btw, I got that from some coaching institute trying to solve this problem which actually appeared in a GATE Exam so I could just sent this above proof right? This is a question that needs to be sold by intuition so the method of elimination is right. – Nishant Aug 14 '16 at 15:57
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What you did is good. Set up f(x) = a * b^x.

If no calculator is available then one can exclude a) c) and d) using simple estimations.

But you are right. This is crazy: x/2 = y^2, x/3 = y^3, x/4 = y^4.

Full calculation for reference.

h(80) = 100
h(40) = 10000

h = 10^(A*x+B)

2 = 80A + B
4 = 40A + B

A = -2/40 = -1/20
B = 6

h = 10^(6-x/20)

h = 5000

lg(5000) = 6-x/20
x = 20*(6-lg(5000)) = 20*(3-lg(5)) = 46.020599913279625
Joel Sjögren
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