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hargunk329
21.12.2019 •
Mathematics
Ajob shop consists of three machines and two repairmen. the amount of time a machine works before breaking down is exponentially distributed with mean 10. if the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) what is the average number of machines not in use? (b) what proportion of time are both repairmen busy?
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Ответ:
Step-by-step explanation:
Let X(t) denote the number of machines breakdown at time t.
The givenn problem follows birth-death process with finite space
S={0, 1, 2, 3} with
The birth-death process having balance equations![\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2](/tpl/images/0428/6710/c9747.png)
since, state rate at which leave = rate at which enter
0![\lambda_0P_0=\mu_1P_1](/tpl/images/0428/6710/15dc3.png)
1![(\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0](/tpl/images/0428/6710/52f6d.png)
2![(\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1](/tpl/images/0428/6710/4f2b9.png)
Since![\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}](/tpl/images/0428/6710/4aeab.png)
a)
Average number not in use equals the mean of the stationary distribution![P_1+2P_2+3P_3=\frac{2136}{751}](/tpl/images/0428/6710/6c2e9.png)
b)
Proportion of time both repairmen are busy![P_2+P_3=\frac{672}{1522}=\frac{336}{761}](/tpl/images/0428/6710/086c3.png)
Ответ:
2 * $3.50 = 7
4 * 3.25 = 13
2 * 1.90 = 3.80
7 + 13 + 3.8 = 23.8
0.05 * 23.8 = 1.19
23.8 - 1.19 = 22.61
Nathaniel spent $22.61.
PLEASE CHECK MY WORK!