Chapter 4.S, Problem 12P

An electronic chess game has a useful life that is exponential with a mean of 30 month. Determine each of the following:

a. The probability that any given unit will operate for at least (1) 39 months. (2) 48 men the (3) 60 months.

b. The probability that any given unit will fail sooner than (1) 33 months. (2) 15 months. (3) 6 months.

c. The length of service tune after which the percentage of tilled units will approximately equal (1) 50 percent, (2) 85 percent, (3) 95 percent, (4) 99 percent.

Summary Introduction

To determine: The probability the unit will operate at least for the following times.

Introduction:

Mean time between failures (MTBF):

The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.

Answer to Problem 12P

The probability the unit will operate at least for the following times are:

1) 39 months = 0.2725

2) 48 months = 0.2019

3) 60 months = 0.1353

Explanation of Solution

Given information:

MTBF = 30 months

Formula to calculate the probability of no failure before a time period:

$\text{Probability}=e^{\frac{-T}{MTBF}}$

Calculation of probability:

1) 39 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{39}{30}\\ =1.3\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ =e^{-1.3}\\ =0.2725\end{array}$

Hence, the probability the unit will operate at least for the 39 months is 0.2725.

2) 48 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{48}{30}\\ =1.6\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ =e^{-1.6}\\ =0.2019\end{array}$

Hence, the probability the unit will operate at least for the 48 months is 0.2019.

3) 60 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{60}{30}\\ =2\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ =e^{-2}\\ =0.1353\end{array}$

Hence, the probability the unit will operate at least for the 60 months is 0.1353.

Summary Introduction

To determine: The probability the unit will fail before the following times.

Introduction:

Mean time between failures (MTBF):

The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.

Answer to Problem 12P

The probability the unit will fail before the following times are:

1) 33 months = 0.6671

2) 15 months = 0.3935

3) 6 months = 0.1813

Explanation of Solution

Given information:

MTBF = 30 months

Formula to calculate the probability of failure before a time period:

$\text{Probability}=1-e^{\frac{-T}{MTBF}}$

Calculation of probability:

1) 33 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{33}{30}\\ =1.1\end{array}$

$\begin{array}{c}\text{Probability}=1-e^{\frac{-T}{MTBF}}\\ =1-e^{-1.1}\\ =1-0.3329\\ =0.6671\end{array}$

Hence, the probability the unit will fail before 33 months is 0.6671.

2) 15 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{15}{30}\\ =0.5\end{array}$

$\begin{array}{c}\text{Probability}=1-e^{\frac{-T}{MTBF}}\\ =1-e^{-0.5}\\ =1-0.6065\\ =0.3935\end{array}$

Hence, the probability the unit will fail before 15 months is 0.3935.

3) 6 months:

$\begin{array}{c}\frac{\text{T}}{\text{MTBF}}=\frac{6}{30}\\ =0.2\end{array}$

$\begin{array}{c}\text{Probability}=1-e^{\frac{-T}{MTBF}}\\ =1-e^{-0.2}\\ =1-0.8187\\ =0.1813\end{array}$

Hence, the probability the unit will fail before 6 months is 0.1813.

Summary Introduction

To determine: The length of service time for the percentage of failed units.

Introduction:

Mean time between failures (MTBF):

The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.

Answer to Problem 12P

The length of service time for the percentage of failed units is:

1) 50 percent = 21 months.

2) 85 percent = 57 months.

3) 95 percent = 90 months.

4) 99 percent = 138 months.

Explanation of Solution

Given information:

MTBF = 30 months

Formula to calculate the probability of no failure before a time period:

$\text{Probability}=e^{\frac{-T}{MTBF}}$

Calculation of probability:

The different probabilities are obtained from the above table.

1) 50 percent:

$\begin{array}{c}\text{Probability of no failure}=1-0.50\\ =0.50\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ 0.50=e^{\frac{-T}{MTBF}}\end{array}$

From the table for the value of 0.50 is equivalent to:

$\frac{T}{MTBF}=0.70$

Therefore,

$\begin{array}{c}\frac{T}{30}=0.70\\ T=0.70\times 0.30\\ =21\text{ months}\end{array}$

Hence, the length of service time for the percentage of failed units is 21 months.

2) 85 percent:

$\begin{array}{c}\text{Probability of no failure}=1-0.85\\ =0.15\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ 0.15=e^{\frac{-T}{MTBF}}\end{array}$

From the table for the value of 0.15 is equivalent to:

$\frac{T}{MTBF}=1.90$

Therefore,

$\begin{array}{c}\frac{T}{30}=1.90\\ T=1.90\times 0.30\\ =57\text{ months}\end{array}$

Hence, the length of service time for the percentage of failed units is 57 months.

3) 95 percent:

$\begin{array}{c}\text{Probability of no failure}=1-0.95\\ =0.05\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ 0.05=e^{\frac{-T}{MTBF}}\end{array}$

From the table for the value of 0.05 is equivalent to:

$\frac{T}{MTBF}=3.00$

Therefore,

$\begin{array}{c}\frac{T}{30}=3.00\\ T=3.00\times 0.30\\ =90\text{ months}\end{array}$

Hence, the length of service time for the percentage of failed units is 90 months.

4) 99 percent:

$\begin{array}{c}\text{Probability of no failure}=1-0.99\\ =0.01\end{array}$

$\begin{array}{c}\text{Probability}=e^{\frac{-T}{MTBF}}\\ 0.01=e^{\frac{-T}{MTBF}}\end{array}$

From the table for the value of 0.01 is equivalent to:

$\frac{T}{MTBF}=4.60$

Therefore,

$\begin{array}{c}\frac{T}{30}=4.60\\ T=4.60\times 0.30\\ =138\text{ months}\end{array}$

Hence, the length of service time for the percentage of failed units is 138 months.