Answer :
Answer:
a. Utilization of machine A = 0.8
Utilization of machine B = [tex]\frac{2}{9}[/tex]
b. Throughput of the production system:
[tex]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[/tex]
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 = [tex]\frac{5}{6} mins[/tex]
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;
[tex]M_A = \frac{1}{4}/min[/tex]
[tex]M_B = \frac{1}{2} /min[/tex]
Inter arrival time is given as 5 mins
[tex]\beta _A = \frac{1}{5} = 0.2/min[/tex]
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,
[tex]5+4 = 9mins\\\beta _B = \frac{1}{9} /min[/tex]
a. Utilization can be defined as the proportion of time when a machine is in use, and is given by the formula [tex]\frac{\beta }{M}[/tex]
Therefore the utilization of machine A is,
[tex]P_A = \frac{\beta_A }{M_A}=\frac{0.2}{\frac{1}{4} }= 0.8[/tex]
And utilization of machine B is,
[tex]P_B = \frac{\beta_B }{M_B} = \frac{\frac{1}{9} }{\frac{1}{2} }= \frac{2}{9}[/tex]
b. Throughput can be defined as the number of jobs performed in a system per unit time.
Throughput of machines A and B,
[tex]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[/tex]
Throughput of the production system is therefore the mean throughput,
[tex]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[/tex]
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, [tex]L_q = \frac{P_A^2}{1-P_A} = \frac{0.8^2}{1-0.8}=\frac{0.64}{0.2}= 3.2[/tex]
Average waiting time = [tex]\frac{3.2}{\frac{1}{5} }= 3.2*5=16mins[/tex]
d. Since the average production time per job is 30 mins;
Probability when machine A completes in 30 mins,
[tex]P(A = 30)= e^{-M_A(1-P_A)30 }= e^{-\frac{1}{4}(1-0.8)30 }=0.225[/tex]
And probability when machine B completes in 30 mins,
[tex]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}[/tex]
The long run average number of jobs in the entire production line can be found thus;
[tex]P(S = 30)=(\frac{ {P_A}+{P_B}}{2})*30 = (\frac{ 0.225}+{0}}{2})*30= 0.1125*30\\=3.375jobs[/tex]
e. If the mean inter arrival time is changed to 1 minute
[tex]\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[/tex]
Utilization of machine A, [tex]P_A = \frac{\beta_A }{M_A} = 4[/tex]
Utilization of machine B, [tex]P_B = \frac{\beta_B}{M_B} = \frac{1}{3}[/tex]
Throughput;
[tex]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[/tex]