Mathematics: A production line has two machines, Machine A and Machine B, that are arranged in series. Each jol needs to processed by Machine A first. Once it finishes the processing by Machine A, it moves to the next station, to be processed by Machine B. Once it finishes the processing by Machine B, it leaves the production line. Each machine can process one job at a time. An arriving job that finds the machine busy waits in a buffer. (The buffer sizes are assumed to be infinite.) The processing times for

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A production line has two machines, Machine A and

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veaseyjermarius1234

18.01.2022

Mathematics

a. Utilization of machine A = 0.8

Utilization of machine B = $\frac{2}{9}$

b. Throughput of the production system:

$E_S = \frac{E_A+E_B}{2} = \frac{20+\frac{18}{7} }{2}=(\frac{1}{2}*20 )+ (\frac{1}{2}*\frac{18}{7} )= 10+\frac{9}{7}= \frac{79}{7} mins$

c. Average waiting time at machine A = 16 mins

d. Long run average number of jobs for the entire production line = 3.375 jobs

e. Throughput of the production system when inter arrival time is 1 = $\frac{5}{6} mins$

Step-by-step explanation:

Machines A and B in the production line are arranged in series

Processing times for machines A and B are calculated thus;

$M_A = \frac{1}{4}/min$

$M_B = \frac{1}{2} /min$

Inter arrival time is given as 5 mins

$\beta _A = \frac{1}{5} = 0.2/min$

since the processing time for machine B adds up the processing time for machine A and the inter arrival time,

Inter arrival time for machine B,

$5+4 = 9mins\\\beta _B = \frac{1}{9} /min$

a. Utilization can be defined as the proportion of time when a machine is in use, and is given by the formula $\frac{\beta }{M}$

Therefore the utilization of machine A is,

$P_A = \frac{\beta_A }{M_A}=\frac{0.2}{\frac{1}{4} }= 0.8$

And utilization of machine B is,

$P_B = \frac{\beta_B }{M_B} = \frac{\frac{1}{9} }{\frac{1}{2} }= \frac{2}{9}$

b. Throughput can be defined as the number of jobs performed in a system per unit time.

Throughput of machines A and B,

$E_A = \frac{\frac{1}{M_A} }{1-P_A}= \frac{4}{1-0.8} = \frac{4}{0.2}= 20 mins\\ E_B = \frac{\frac{1}{M_B} }{1-P_B}= \frac{2}{1-\frac{2}{9} } = \frac{18}{7}mins$

Throughput of the production system is therefore the mean throughput,

$E_S = \frac{E_A+E_B}{2} = \frac{20+\frac{18}{7} }{2}=(\frac{1}{2}*20 )+ (\frac{1}{2}*\frac{18}{7} )= 10+\frac{9}{7}= \frac{79}{7} mins$

c. Average waiting time according to Little's law is defined as the average queue length divided by the average arrival rate

Average queue length, $L_q = \frac{P_A^2}{1-P_A} = \frac{0.8^2}{1-0.8}=\frac{0.64}{0.2}= 3.2$

Average waiting time = $\frac{3.2}{\frac{1}{5} }= 3.2*5=16mins$

d. Since the average production time per job is 30 mins;

Probability when machine A completes in 30 mins,

$P(A = 30)= e^{-M_A(1-P_A)30 }= e^{-\frac{1}{4}(1-0.8)30 }=0.225$

And probability when machine B completes in 30 mins,

$P(B = 30)= e^{-M_B(1-P_B)30 }= e^{-0.5(1-\frac{2}{9} )30 }=e^{-\frac{15*7}{9} }=e^{-11.6}$

The long run average number of jobs in the entire production line can be found thus;

$P(S = 30)=(\frac{ {P_A}+{P_B}}{2})*30 = (\frac{ 0.225}+{0}}{2})*30= 0.1125*30\\=3.375jobs$

e. If the mean inter arrival time is changed to 1 minute

$\beta _A= \frac{1}{1}= 1/min\\\beta _B= \frac{1}{6}/min\\ M_A = \frac{1}{4}min\\ M_B = \frac{1}{2} min$

Utilization of machine A, $P_A = \frac{\beta_A }{M_A} = 4$

Utilization of machine B, $P_B = \frac{\beta_B}{M_B} = \frac{1}{3}$

Throughput;

$E_A = \frac{\frac{1}{M_A} }{1-P_A} = \frac{4}{1-4} = \frac{4}{3} \\E_B= \frac{\frac{1}{M_B} }{1-P_B} = \frac{2}{1-\frac{1}{3} } = 3\\\\E_S= \frac{E_A+E_B}{2} = \frac{\frac{4}{3}+3 }{2}=(\frac{4}{3} *\frac{1}{2} )+(3*\frac{1}{2} ) =\frac{2}{3} + \frac{3}{2} \\= \frac{5}{6} min$

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