Can you solve it quickly please!!
Random Servicing . It is not known when and for how long a facility needs to be serviced Mean times are usually known and laws of probability is used to determine number of machines to be assigned to an operator. (usually for small number of m/c (1-6 m/c)) If : n:is no of m/c to be assigned to an operator p: probability of down time of m/c q: probability of run time where q= (1-p) m: no of m/c down out of n P(m out of n) = (n!/(m!(n-m)!)) pqn-m (binomial expansion) man assume average m'c run time is 60%, therefore down time (servicing time) 40% if3 m/c are assigned to an operator the probabilities of m (out of n) are: m/c down Probability of being down 0 3!(0!(3-0)!(0.4. 0.63)-0.216 1 3!(1:(3-1)!(0.41 . 0.62 )-0.432 2 3!/(2(3-2)!( 0.42 +0.6) = 0.288 3 3!(3!(3-3)!(0.43.0.6) = 0.064 Resulting lost time for one operator/3 machines No of m/c down probability m/c hrs lost per 8 hrs day 0 0 1 0 0.216 0.432 0.288 0.064 2 No m'c is down so time lost 1m/c is down so opr. can attand this 1*(0.288)*8 = 2.304 2*(0.064)*8 = 1.024 3.328 3 Total 1.000 Proportion of m/c time lost= 100*(3.328)/3*8 = 13.9% f each machine can produce 60 units/hr Hourly rate of operator: K = 10 TL Hourly rate of machine: K = 60 TL Number of m'c assigned: n=3 Rate of production from n mc: R=(60pieces per m/c)*3 Total expected cost piece: TEC Then: TEC = (K+n*K/R Knowing that 3.328 production hours are lost during & hr day total real production time =24-3.328=20.672 hrs R=((60*3)units/hr * (20.672) hrs/day) (24hrs/day allowed time) =155.04 units/hr TEC = ( 10+3*60,155.04 = 190/155.04 =1.23 TL/piece HOMEWORK for today untill 23:59 Every thing being identical calculate the unit cost if 2 operators are assigned Random Servicing . It is not known when and for how long a facility needs to be serviced Mean times are usually known and laws of probability is used to determine number of machines to be assigned to an operator. (usually for small number of m/c (1-6 m/c)) If : n:is no of m/c to be assigned to an operator p: probability of down time of m/c q: probability of run time where q= (1-p) m: no of m/c down out of n P(m out of n) = (n!/(m!(n-m)!)) pqn-m (binomial expansion) man assume average m'c run time is 60%, therefore down time (servicing time) 40% if3 m/c are assigned to an operator the probabilities of m (out of n) are: m/c down Probability of being down 0 3!(0!(3-0)!(0.4. 0.63)-0.216 1 3!(1:(3-1)!(0.41 . 0.62 )-0.432 2 3!/(2(3-2)!( 0.42 +0.6) = 0.288 3 3!(3!(3-3)!(0.43.0.6) = 0.064 Resulting lost time for one operator/3 machines No of m/c down probability m/c hrs lost per 8 hrs day 0 0 1 0 0.216 0.432 0.288 0.064 2 No m'c is down so time lost 1m/c is down so opr. can attand this 1*(0.288)*8 = 2.304 2*(0.064)*8 = 1.024 3.328 3 Total 1.000 Proportion of m/c time lost= 100*(3.328)/3*8 = 13.9% f each machine can produce 60 units/hr Hourly rate of operator: K = 10 TL Hourly rate of machine: K = 60 TL Number of m'c assigned: n=3 Rate of production from n mc: R=(60pieces per m/c)*3 Total expected cost piece: TEC Then: TEC = (K+n*K/R Knowing that 3.328 production hours are lost during & hr day total real production time =24-3.328=20.672 hrs R=((60*3)units/hr * (20.672) hrs/day) (24hrs/day allowed time) =155.04 units/hr TEC = ( 10+3*60,155.04 = 190/155.04 =1.23 TL/piece HOMEWORK for today untill 23:59 Every thing being identical calculate the unit cost if 2 operators are assigned
