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eimsori [14]
3 years ago
9

Is this correct....help me with the 2 questions

Mathematics
1 answer:
torisob [31]3 years ago
6 0
Yes I believe that these are correct
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A production line has two machines, Machine A and Machine B, that are arranged in series. Each jol needs to processed by Machine
marysya [2.9K]

Answer:

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

3 0
3 years ago
A waitress earned $7 per hour at her job plus an additional $50 in tips on Friday. She earned more than $99 total. Which inequal
aivan3 [116]

Answer:

Step-by-step explanation:

7h + 50= 99

99-50 =49

7h=49

49/7

h=7

here's the steps i don't know which part you need

5 0
3 years ago
Andre performs a 90-degree counterclockwise rotation of Polygon P and gets Polygon P’, but he does not say what the center of th
Alik [6]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Need help pls someone
dimaraw [331]

\sqrt[3]{250a^{11}}=\sqrt[3]{125\cdot2a^{2+9}}=\sqrt[3]{125}\cdot\sqrt[3]{2a^2\cdot a^9}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^9}\\\\=5\sqrt[3]{2a^2}\cdot\sqrt[3]{a^{3\cdot3}}=5\sqrt[3]{2a^2}\cdot\sqrt[3]{(a^3)^3}=5\sqrt[3]{2^2}\cdot a^3\\\\=\boxed{5a^3\sqrt[3]{2a^2}}

7 0
3 years ago
Please help. I don't understand this. Thanks.
neonofarm [45]
Yes;k=-3 and y=-3x is the answer because 2 x -3 = -6 -3 x -3 = -9 and etc..
4 0
3 years ago
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