Productivity improvement by reduction of idle time through application of queuing theory

Abstract

In Lean Six Sigma one of the primary objectives is to identify and eliminate the waiting or idle time for enhancing productivity and reducing production cost. However, in some situations it is better to replace ‘elimination’ by ‘optimization’ due to the presence of conflicting requirements. Optimization techniques in the form of queuing models are quite useful to resolve this conflict. In an Indian textile manufacturing plant, the waiting (idle) time in the winding process is identified as a reason for high production cost. In this situation, it is to be determined as to how many machines are to be assigned to an operator for reducing waiting or idle time of both the operators and the machines leading to reduced production cost. The situation can be compared and contrasted to a machine repairman problem aiming for optimizing the number of machines per operator. The various process parameters have been estimated and suitably modified or derived to serve as inputs to the Queuing Model. Sensitivity analysis has been carried out to identify and control the significant factors. The objective of this paper is to demonstrate through an application of queuing models the procedure of optimizing the waiting or idle time. Consequently, the focus is on the application of optimization techniques like Queuing Theory and Simulation while implementing lean six sigma.

Appendixes

1.1 Appendix 1

1.1.1 To arrive at an expression for mean patrolling time

Let one winder is assigned to N drums and ‘d’ be the center to center distance between drums. If the operator moves from the jth to ith drum then the distance walked by him is |i-j|d, where i=1,2,…,N and j = 1,2,…,N. The distance matrix with all possible distances between any two drums i and j, where i,j=1,2,..,N is given below;

Drum No	1	2		j		N-1	N
1	0	d	…….	(j-1)d	…….	(N-2)d	(N-1)d
2	d	0	…….	(j-2)d	…….	(N-3)d	(N-2)d
:	:	:		:		:	:
:	:	:		:		:	:
i	(i-1)d	(i-2)d	…….	(i-j)d	…….	(N-1-j)d	(N-i)d
:	:	:		:		:	:
:	:	:		:		:	:
N-1	(N-2)d	(N-3)d	…….	(N-1-j)d	…….	0	D
N	(N-1)d	(N-2)d	…….	(N-j)d	…….	d	0

The distance matrix shown above is symmetric. So, all possible combination are available on the other side of diagonal. Total of all possible distances covered by the operator;

$$ \begin{array}{c}\hfill =\mathrm{sum}\kern0.5em \mathrm{of}\kern0.5em \mathrm{all}\kern0.5em \mathrm{upper}\kern0.5em \mathrm{and}\kern0.5em \mathrm{lower}\kern0.5em \mathrm{diagonal}\kern0.5em \mathrm{elements}\hfill \\ {}\hfill =\left(N-1\right)N\kern0.5em d+\left(N-2\right)\left(N-1\right)d+\dots +d=d{\displaystyle \sum_{i=1}^Ni\left(i-1\right)=d\left({\displaystyle \sum_{i=1}^N{i}^2-{\displaystyle \sum_{i=1}^Ni}}\right)}\hfill \\ {}\hfill =d\times \left(\frac{N\left(N+1\right)\left(2N+1\right)}{6}-\frac{N\left(N+1\right)}{2}\right)=d\times \frac{N\left(N+1\right)\left(N-1\right)}{3}\hfill \end{array} $$

(1)

$$ \mathrm{Number}\kern0.5em \mathrm{of}\kern0.5em \mathrm{all}\kern0.5em \mathrm{possible}\kern0.5em \mathrm{distance}\kern0.5em \mathrm{the}\kern0.5em \mathrm{operator}\kern0.5em \mathrm{will}\kern0.5em \mathrm{cover}={N}^2 $$

(2)

$$ \mathrm{Hence}\kern0.5em \mathrm{the}\kern0.5em \mathrm{average}\kern0.5em \mathrm{distance}\kern0.5em \mathrm{moved}\kern0.5em \mathrm{by}\kern0.5em \mathrm{the}\kern0.5em \mathrm{operator}=\frac{ Eq.(1)}{ Eq.(2)}=\frac{\left({N}^2-1\right)}{3N}\times d $$

$$ \mathrm{Then}\kern0.5em \mathrm{the}\kern0.5em \mathrm{patrolling}\kern0.5em \mathrm{time},=\frac{\left({N}^2-1\right)}{3N}\times \frac{d}{S} $$

where

S:

walking speed per hour of operator.

1.2 Appendix 2

1.2.1 Derivation of the formula for P _o

The problem under study is a machine interference problem. Without loss of generality, following assumptions have been made for solving the problem based on queueing model.

• 1. There are fixed number (N) of machines.

• 2. There is only one operator.

• 3. The mean arrival time of a machine follows Exponential distribution with mean R

• 4. The service time of an operator follows Exponential distribution with mean W

• 5. There is only one queue

• 6. The service discipline is FCFS (First Come First Serve)

• 7. The operator will not remain idle when there is a drum waiting for his service.

So if λ be the arrival rate, then λ = 1/R and if μ be the service rate, then μ = 1/W. The most suitable queueing model is (M/M/1) with finite population.

Let λ _i is the mean arrival rate for i ^th drum and μ _i is the mean service rate for i ^th drum.

So,

$$ \begin{array}{c}\hfill {\lambda}_i=\left\{\begin{array}{lll}\left(N-i\right)\lambda \hfill & for\hfill & 0\le i\le N\hfill \\ {}0\hfill & for\hfill & i>N\hfill \end{array}\right.\hfill \\ {}\hfill {\mu}_i=\left\{\begin{array}{lll}\mu \hfill & for\hfill & i\le N\hfill \\ {}0\hfill & for\hfill & i>N\hfill \end{array}\right.\hfill \end{array} $$

When the system reaches the steady state, let P _j be the probability that there are j drums in the system (i.e., drums are not working) at any point of time, i.e., one drum is being serviced and the remaining (j-1) are in the queue, waiting for service.

At steady state, the recurrence relation between P_j, λ _j and μ _j are

• (1) $\begin{array}{ll}{\lambda}_0{P}_0={\mu}_1{P}_1\hfill & \left(\mathrm{initial}\kern0.5em \mathrm{condition}\right)\hfill \end{array}$

• (2) $\begin{array}{lll}{\mu}_j{P}_j={\lambda}_{j-1}{P}_{j-1}\hfill & for\hfill & 1\le j\le N\hfill \end{array}$

From the above relation, in steady state condition

But we know, $\begin{array}{ll}{P}_j=\frac{N!}{\left(N-j\right)!}\times {\left(\frac{W}{R}\right)}^j\times {P}_0\hfill & \mathrm{For}\kern0.5em \mathrm{j}=0,1,2,3,\dots .,\mathrm{N}\hfill \\ {}{\displaystyle \sum_{j=0}^N{P}_j=1}\hfill & \hfill \end{array}$

This implies; $\begin{array}{ll}{\displaystyle \sum_{j=0}^N\frac{N!}{\left(N-j\right)!}\times {\left(\frac{W}{R}\right)}^j\times}\hfill & {P}_0=1\kern0.7em \mathrm{or}\kern0.7em {P}_0=\frac{1}{{\displaystyle \sum_{j=0}^N\frac{N!}{\left(N-j\right)!}\times {\left(\frac{W}{R}\right)}^j}}\hfill \end{array}$