Industrial Engineering Test 4

Concept:

The PERT (Project Evaluation and Review Technique) technique is used, when activity time estimates are stochastic in nature. For each activity, three values of time (optimistic, most likely, pessimistic) are estimated.

• Optimistic time (to) estimate is the shortest possible time required for the completion of the activity
• Most likely time (tm) estimate is the time required for the completion of activity under normal circumstances
• Pessimistic time (tp) estimate is the longest possible time required for the completion of the activity

In PERT expected time of an activity is determined by using the below-given formula:

\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)

Variance, σ2 = \({\left[ {\frac{{{t_P} - {t_0}}}{6}} \right]^2}\)

Calculation:

Given, t_o = 3 days, t_m = 10 days and t_p = 8 days

\({t_e} = \frac{{{t_o} + 4{t_m} + {t_p}}}{6} = \frac{{3 + 4\left( 10 \right) + 8}}{6} = \frac{{51}}{6} = 8.5 \ days\)

Concept:

Mean waiting time of customer in queue,

\({W_q} = \frac{\lambda }{{\mu \left( {\mu - \lambda } \right)}}\)

Calculation:

λ = 30 per hour

\(\mu = \frac{{60\; \times \;60}}{{100}} = 36\;{\rm{per\;hour}}\)

\(\therefore {W_q} = \frac{{30}}{{36\left( {36 - 30} \right)}}\)

W_q = 0.138 hour

∴ W_q = 8.33 minute

Explanation:

Given:

λ = 20/hour, μ = 30/hour

Mean waiting time in queue \(= \frac{\lambda }{{\mu \left( {\mu - \lambda } \right)}}\)

Mean waiting time in queue \(= \frac{{20}}{{30\left( {30 - 20} \right)}}\)

Mean waiting time in queue \(= \frac{{20}}{{300}} = \frac{1}{{15}}{\rm{\;hour}}\)

∴ Mean waiting time in queue \(= \frac{1}{{15}} \times 60\)

∴ Mean waiting time in queue = 4 minutes

Concept:

No. of jobs in system, \({L_S} = \frac{\lambda }{{\mu - \lambda }}\)

No. of jobs in queue, \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

Calculation:

Arrival rate λ = 6 jobs/minute

Service rate, μ = 10 jobs/minute

∴ No. of jobs in system, \({L_S} = \frac{\lambda }{{\mu - \lambda }}\)

\(\Rightarrow {L_S} = \frac{6}{{10 - 6}} = 1.5\)

Points to remember

• Average arrival time and the time spent in the system (Waiting time in system) = \({W_s} = \frac{1}{{\mu - \lambda }}\)
• Average arrival time and the time spent in the queue (before being served) (Waiting time in queue) = \({W_q} = \frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the system = \({L_s} = \lambda {W_s} = \lambda .\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the queue = Average queue length = \({L_q} = \lambda {W_q} = \lambda .\frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }} = \frac{{{\lambda ^2}}}{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)
• Probability that there are k customers in the system = (ρ)^k(1 - ρ) = \(\frac{\lambda }{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)

• Probabili ty that there are more than k customers = \({\left( {\frac{\lambda }{\mu }} \right)^{k + 1}}\)

Case I: Repairman ‘A’

λ = 3, μ = 4

Average downtime of machine\({\rm{\;}} = \frac{1}{{\mu - \lambda }} = \frac{1}{{4 - 3}}\)

Average downtime of machine = 1 hour

Thus,

Downtime of 3 machines that arrive in an hour = 3 × 1 = 3 hours

Now,

Total cost = Downtime cost + Repairman charges

Total cost = (60 × 3) + (20 × 3)

∴ Total cost = Rs. 240

Case II: Repairman B

λ = 3, λ = 6

Average downtime\(= \frac{1}{{\mu - \lambda }} = \frac{1}{{6 - 3}} = \frac{1}{3}\)

Downtime of 3 machines that arrive in an hour\(= \frac{1}{3} \times 3 = 1\;hour\)

Total cost = (60 × 1) + (30 × 1)

∴ Total cost = Rs. 90

So, Repairman ‘B’ should be hired and total cost will be Rs. 90.

Concept:

The idle time will be the product of the probability of no customers in the system and the shift duration.

Probability of no customer is \({P_0} = 1 - \frac{\lambda }{\mu }\)

Where λ and μ are the arrival and service rates

Calculation:

The arrival rate is:

\(\lambda = \frac{4}{{12}} = \frac{1}{3}\;h{r^{ - 1}}\)

Service rate is:

\(\mu = 1\;h{r^{ - 1}}\)

Thus:

\({P_0} = 1 - \frac{1}{3} = \frac{2}{3}\)

Idle time in hours:

\(T = \frac{2}{3} \times \;12 = 8\;hours\)