PRODUCTVIVITY IMPROVEMENT MODEL OF AN UNRELIABLE THREE STATIONS AND A SINGLE BUFFER WITH CONTINUOUS MATERIAL FLOW

Transcription

1 Vol. 3, No. 1, Fall, 2016, pp ISSN X (print), (online), All Rights Reserved PRODUCTVIVITY IMPROVEMENT MODEL OF AN UNRELIABLE THREE STATIONS AND A SINGLE BUFFER WITH CONTINUOUS MATERIAL FLOW Mohammad A. Quasem, Department of Information Systems & Supply Chain Management, School of Business, Howard University, Washington D.C., United States of America, mquasem@howard.edu ABSTRACT: Production lines are a serial arrangement of a workstation. Items pass through successive stations as a specific operation is performed at each station. A major cause of line inefficiency is the blocking and starving of stations. In such a situation inter-station buffer storages are used to increase the efficiency of the productive manufacturing systems. In this study, three work stations and a single buffer is considered. The major significant insight obtained into the continuous flow production lines is that as long as there is only one continuous state variable, namely one buffer level variable, the resultant system may be solved, at least numerically. Key Words: Productivity Improvement, Continuous Materials Flow, Buffer, Work Stations, Steady State, Probabilities 1. INTRODUCTION AND LITERATURE REVIEW Production lines are a serial arrangement of a workstation. Items pass through successive stations as a specific operation is performed at each station. A major cause of line inefficiency is the blocking and starving of stations. In such a situation, inter-station buffer storages are used to increase the efficiency of the system. In the past, Wijngaard (1979) developed a production rate model of a two-station continuous materials flow with a finite buffer. Wijngaard did not find a closed form solution. Yeralan, Frank, and Quasem (1986) found a production line model with continuous materials flow to improve the productivity of a manufacturing system. They developed a closed form solution with two stations and a single buffer. Quasem (2008) examined the three stations continuous materials flow model with two finite capacity buffers. He did not find a closed form solution. Al Kaissi Mahdi (2000) used a buffer as a way of protecting soil and water quality. This buffer is known as a conservation buffer and it was described as areas or pieces of land where permanent vegetation is established around crops. Mahdi found that the buffer was helpful in managing the Iowa land. His study was related to discrete production models. Gershwin et al. (1980), Koeninberg E (1959), Muth Eginhardt J. et al (1987)., and Tan B. et al. (2007) extensively studied the role of buffers and general production line models and found the optimal solutions. Their studies did not include the continuous materials flow. Their studies were limited to discrete production line models. In this study, we consider L workstations in series and continuous materials flow. It is assumed that the time to breakdown and time to repair are exponentially distributed with rates λ i and µ i for station i, where i = 1, 2, 3 L. The single buffer storage with capacity M which is placed between two adjacent stations. The Figure 1 shows such a production line when L=3 and the buffer are placed between the first and second stations. It is assumed that any station provided is up and operating, and the service rates of the station are all the same. Notice that the unit of materials can be chosen such that the station service rate becomes unity. There exists an infinite amount of unprocessed materials before the first station and an infinite space following the last station. Thus, station one can never starved and station L can never be blocked.

2 24 Input Buffer Output Station 1 Station 2 Station 3 Figure 1. A Three Station Line with a Single Buffer There is an assumption that the time to breakdown and time to repair for any station are also exponentially distributed while the buffer remains empty or full. An idle station does not breakdown. The steady-state performance of the system is wanted. In particular, the production rate and the expected level of the buffer are of interest. A model is describing the production rate regarding station parameters and the capacity of the buffer, for example, can be used to establish the optimum location of the buffer. 2. METHOD OF ANALYSIS: Consider the case L=3 as given in Figure 2.1. At any given time, the buffer can be full, partially full, or empty. There are eight possible combinations of station states corresponding to a partially full buffer. These station state combinations are given below in table 2.1. TABLE2.1 THE STATION STATES WHEN THE BUFFER IS PARTIALLY FULL STATE STATION 1 STATION 2 STATION 3 1 U U U 2 U B D 3 U D S 4 U D D 5 D U U 6 D B D 7 D D S 8 D D D

3 25 Note: U: Station is up and operating D: Station is down and under repair B: Station is operational but blocked S: Station is operational but starved There are two other combinations of the system states. One of these combinations, denoted by system state index 9, corresponds to the station states while the buffer remains full. The other combination, denoted by system state 10, corresponds to the station states while the buffer remains empty. The system state at time t, Y t, is expressed by the 2-vector (i, y). The second variable y denotes a number of materials in the buffer. The first variable i, for i= 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 denotes the state of the stations. When the buffer becomes full, various station states are possible. All of these states are considered as I = 9 and their effects will be added to the model later. Likewise, when the buffer becomes empty, various station states are considered as I = 10 and their effect will be added to the model later. The time dependent system state probabilities while the buffer is partially full are: F i (t, y) = P [{YT= (i, s), where 0<s<y}] for i= 1, 2, 3, 4, 5, 6, 7, 8 and 0<y<1 (2.1) The time dependent system state density functions are also defined as: f i (t, y) = a F i (t, y)/ a y i = 1, 2, 3, 4, 5, 6, 7, 8 (2.2) For the states where the buffer is either empty or full we define, P 9 (t) = P [YT = (9, 1)] (2.3) P 10 (t) = P [Yt = (10, 0)] (2.4) Since the long-run characteristics of the system are of interest, the steady-state density functions are defined. f i (y) = lim f i (t, y) i = 1, 2, 3, 4, 5, 6, 7, 8 (2.5) The steady-state probability mass functions corresponding to the state where the buffer remains full or empty are respectively defined as and P 9 = lim P 9 (t) t P 10 = lim P 10 (t) t We assume that the process is ergodic and, thus, the steady-state density functions exist. The conditional steady-state density functions g i (y) is defined as: g i (y) = f i (y I 0<y, 1) for i= 1, 2, 3, 4, 5, 6, 7, 8 (2.6)

4 26 The conditional steady-state density functions for the extreme values of the buffer level are defined as and g i (0) = lim g i (y) for i = 1, 2, 3, 4, 5, 6, 7, 8 (2.7) y 0 g i (1) = lim g i (y) for i = 1, 2, 3, 4, 5, 6, 7, 8 (2.8) y 1 3. DEFINING EQUATIONS FOR THE CONDITIONAL STEADY STATE DENSITY FUNCTIONS: Conditioning the system state at time t+h on the system state at time t in a fashion similar to that of Quasem (2008) & Quasem at el. (1987) one can obtain the following equations. g 1 (t+h, y) = {1 (λ 1 + λ2 + λ 3) h} g 1 (t, y) + µ 3 hg 2 (t, y-h) + µ 2 hg 3 (t, y-h) + µ 1 hg s (t, y+h)+ 0(h 2 ) (3.1) g 2 (t + h, y) = {1-(λ 1 + λ 2b + µ 3) h} g 2 (t, y-h) + λ 3 hg 1 (t, y) + µ 2 hg 4 (t, y-h) + µ 1 hg 6 (t, y) + 0 (h 2 ) (3.2) g 3 (t + h, y) = {1-(λ 1 + µ 2 + λ 3 s) h} g 3 (t, y-h) + λ 2 hg 1 (t, y) + µ 1 hg 7 (t, y) + µ 3 hg 4 (t, y-h) + 0 (h 2 ) (3.3) g 4 (t + h, y) = {1- (λ 1 + µ 2 + µ 3) h} g 4 (t, y-h) + µ 1 hg 8 (t, y) + λ 2 6 hg 2 (t, y-h) + λ 3 s g 3 (t, y-h) + 0 (h 2 ) (3.4) g 5 (t + h, y) = {1- (µ 1 + λ 2 + λ 3) h} g 5 (t, y + h) + λ 1hg 1 (t, y) + µ 2hg 7 (t, y) + µ 3hg 6 (t, y) + 0(h 2 ) (3.5) g 6 (t + h, y) = {1- (µ 1 + λ 26 + µ 3) h} g 6 (t, y) + λ 1 hg 2 (t, y-h) + µ 2 hg 8 (t, y) + λ 3hg 5 (t, y-h) + 0(h 2 ) (3.6) g 7 (t + h, y) = {1- (µ 1 + µ 2 + λ 3s) h} g 7 (t, y) + λ 1hg 3 (t, y-h) + λ 2hg 6 (t, y+ h) + µ 3hg 8 (t, y) + 0 (h 2 ) (3.7) g 8 (t + h, y) = {1-(µ 1 + µ 2 + µ 3) h} g 8 (t, y) + λ 3shg 7 (t, y) + λ 26hg 6 (t, y + h) + λ 1hg 4(t, y-h) + 0(h 2 ) (3.8) The above equations can be written in the differential form where h 0. Similar to the manipulation in Yeralan, Frank, & Quasem (1986) we obtain the following differential equations. dg 1(t, y) = - (λ 1 +λ 2 + λ 3) g 1 (t, y) + µ 3g 2 (t, y) dt + µ 2g 3 (t, y) + µ 1g 5 (t, y) (3.9)

5 27 dg 2 (t, y) + dg 2 (t, y) = (λ 1 + λ 2 b + µ 3) g 2 (t, y) + λ 3 g 1 (t, y) + µ 2g 4 (t, y) + µ 1g 6 (t, y) dt dy (3.10) dg 3 (t, y) + dg 3 (t, y) = (λ 1 + µ 2 + λ 3 s) g 3 (t, y) + λ 2 g 1 (t, y) + µ 1g 7 (t, y) + µ 3g 4 (t, y) (3.11) dt dy dg 4 (t, y) + dg 4 (t, y) = (λ 1 + µ 2 + µ 3) g 4 (t, y) + µ 1g 8 (t, y) + λ 2 6 g 2 (t, y) + λ 3 s g 3 (t, y) (3.12) dt dy dg 5 (t, y) + dg 5 (t, y) = - (µ 1 + λ 2 + λ 3) g 5 (t, y) + λ 1g 1 (t, y) + µ 2 g 7 (t, y) + µ 3 g 6 (t, y) (3.13) dt dy dg 6 (t, y) = - (µ 1 + λ µ 3) g 6 (t, y) + λ 1 g 2 (t, y) + µ 2 g 8 (t, y) + λ 3 g 5 (t, y) (3.14) dt dg 7 (t, y) = - (µ 1 + µ 2 + λ 3 s) g 7 (t, y) + λ 1 g 3 (t, y) + λ 2 g 5 (t, y) + µ 3g 8 (t, y) (3.15) dt dg 8 (t, y) = -(µ 1 + µ 2 + µ 3) g 8 (t, y) + λ 3 s g 7 (t, y) + λ 2 6 g 6 (t, y) + λ 1g 4 (t, y) (3.16) dt In steady-state i.e. When t, we have (ag (t, y)/at) =0. Thus, the following set of simultaneous equations constitute the defining equations for the conditional steady-state density functions g i (y) for i = 1, 2, 3, 4, 5, 6, 7, 8. dg 2 (y) = (λ 1 + λ 2 b + µ 3) g 3 (y) + λ 3 g 1 (y) + µ 2 g 4 (y) + µ 1 g 6 (y) dy (3.17) dg 3 (y) = (λ 1 + µ 2 + λ 3 s) g 3 (y) + λ 2g 1 (y) + µ 1g 7 (y) + µ 3g 4 (y) dy (3.18) dg 4 (y) = (λ 1 + µ 2 + µ 3) g 4 (y) + µ 1 g 8 (y) + λ 2 6 g 2 (y) + λ 3 s (y) dy (3.19) - dg 5 (y) = - (µ 1 + λ 2 + λ 3) g 5 (y) + λ 1g 1 (y) + µ 2g 7 (y) + µ 3g 6 (y) Dy (3.20) - (λ 1 +λ 2 + λ 3) g 1 (y) + µ 3g 2 (y) +µ 2g 3 (t, y) + µ 1g 5 (y) = 0 (3.21) - (µ 1 + λ µ 3) g 6 (y) + λ 1g 2 (y) + µ 2g 8 (y) + λ 3g 5 (y) =0 (3.22) - (µ 1 + µ 2 + λ 3 s) g 7 (y) + λ 1g 3 (y) + λ 2g 5 (y) + µ 3g 8 (y) = 0 (3.23) - (µ 1 + µ 2 + µ 3) g 8 (y) + λ 3 sg 7 (y) + λ 2 6 g 6 (y) + λ 1g 4 (y) = 0 (3.24) The first four equations, 3.17 through 3.20, are linear differential equations while the later four, equations, 3.21 through 3.24, are linear equations. Since there are four first-order differential equations, four initial conditions are required to solve the defining equations. The solution to the defining equations is discussed next

6 28 4. SOLUTION OF THE DEFINING EQUATIONS FOR THE CONDITIONAL STEADY-STATE DENSITY FUNCTIONS Let the vectors w(y) and v(y) be defined as: g 1 (y) g 2 (y) w(y) = g 6 (y) and v(y) = g 3 (y) g 7 (y) g 4 (y) g 8 (y) g 5 (y) That is, v(y) corresponds to the differential equations and w(y) corresponds to the linear equations. The motivation for this arrangement is that using the linear defining equations (3.21) through (3.24), one can write w(y) a linear combination of the elements of v(y). Namely w(y) = Av(y). To find matrix A, consider equations 3.21 through 3.24 in the following form: Where, Ew(y) + Fv(y) = 0 (4.1) - (λ 1 + λ 2 + λ 3) E= 0 - (µ 1 +µ µ 3) 0 µ (µ 1 +µ 2 + λ 3 s) µ 3 0 λ 2 6 λ 3 s - (µ 1 +µ 2 +µ 3) and F= µ 3 µ 2 0 µ 1 λ λ 3 0 λ 1 0 λ λ 1 0 Then provided that matrix E is nonsingular, -1-1

7 29 w(y) = -E F v(y) or A = -E F. E is nonsingular by evaluating its determinant. E = µ 1 d 1 d 4 (d 2 + d 3) Where, d 1= (λ1 + λ 2 + λ 3) d 2= (µ 1 + λ 2 b + µ 3) d 3= (µ 1 + µ 2 + λ 3 s) d 4= (µ 1 + µ 2 + µ 3 + µ 4) Since d 1, d 2, d 3, d 4 are all positive, it follows that E is greater than zero. Matrix A = -E. F is now evaluated as. µ 3d 5 µ 2d 5 0 µ 1d 5 A= λ 1d 6 λ 1d 7 λ 1d 8 Λ 1d 9 λ 1d 10 λ 1d 11 λ 3d 9+λ 2d 10 Λ 1d 12 λ 1d 13 λ 1d 14 λ 3d 12+λ 2d 13 λ 3d 6+λ 2d 7 Where d 5= -d 2d 3d 4 + µ 2 λ 26d 3 + µ 3λ 3sd 3 d 6= -d 1d 3d 4 + µ 3λ 3sd 1 d 7= -µ 2λ 3sd 1 d 8= -µ 2d 1d 3 d 9= -µ 3λ 26d 1 d 10= -d 1d 2d 3 + µ 2λ 26d 1

8 30 d 11= -µ 3d 1d 2 d 12= -λ 26d 1d 3 d 13= λ 3sd 1d 2 d 14= -d 1d 2d 3 or equivalent d 5= (µ 1 + µ 2 + µ 3s) [-(λ 1 + λ 2 + λ 3) (µ 1 + µ 2 + µ 3) + µ 2λ 6 + µ 3λ 3s] d 6= (λ 1 + λ 2 + λ 3) [-(µ 1 + µ 2) (µ 1 + µ 2 + µ 3 + λ 3s)] d 7= µ 2λ 3s (λ 1 + λ 2 + λ 3) d 8= µ 2 (λ 1 + λ 2 + λ 3) (µ 1 + µ 2 + λ 3s) d 9= λ 2 6µ 3 (λ 1 + λ 2 + λ 3) d 10= (λ 1 + λ 2 + λ 3) [-(µ 1+ µ 3) (µ 1 + µ 2 + µ 3 + λ 2 6)] d 11= -µ 3 (λ 1 + λ 2 + λ 3) (µ 1 + µ µ 3) d 12= -λ 2 6 (λ 1 + λ 2 + λ 3) (µ 1 + µ 2 + µ 3 s) d 13= -λ 3 s (µ 1 + µ µ 3) (µ 1 + µ 2 + µ 3 s) d 14= - (λ 1 + λ 2 + λ 3) (µ 1 + λ µ 3) (µ 1 + µ 2 + λ 3 s) Similar to equation (4.1), equation (3.17) through (3.20) can be written as dv (y) = Hv (y) + Kw (y) (4.3) dy Where, -(λ 1 + λ µ 3) 0 µ (λ 1 + µ 2 + λ 3 s) µ 3 0 H= λ 2 6 λ 3 s - (λ 1 + µ 2 + µ 3) (µ 1 + λ 2 + λ 3) and λ 3 µ λ 2 0 µ 1 0 K= µ 1 λ 1 µ 3 µ 2 0

9 31 Combining equations, (3.2) and (3.3) gives: dv(y) = Wv(y) dy Where W= H+KA The matrix W can be written in terms of the system parameters in a closed-form. Namely, a 11 a 12 a 13 a 14 W= a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 Where: a 11= - (λ 1 + λ µ 3) + λ 3 µ 3d 5 + λ 1µ 1d 6 a 12= λ 3µ 2d 5 +λ 1µ 1d 7 a 13= µ 2 + λ 1µ 1d 8 a 14= λ 3µ 1d 5 + λ 3µ 1d 6 + λ 2µ 1d 7 a 21= λ 2µ 3d 5 + λ 1µ 1d 9 a 22= - (λ 2 + µ 2 + λ 3 s) + λ 2µ 2d 5 + λ 1µ 1d 10 a 23= µ 3 + λ 1µ 1d 11 a 24= λ 2µ 1d 5 + λ 3µ 1d 9 +λ 2µ 1d 10 a 31= λ 26 + λ 1µ 1d 12 a 32= λ 3 s + λ 1µ 1d 13 a 33= - (λ 1 + µ 2 + µ 3) + λ 1µ 1d 14 a 34= λ 3µ 1d 12 + λ 2µ 1d 13 a 41= λ 1µ 3d 5 + λ 1µ 3d 6 + λ 1µ 2d 9 a 42=λ 1µ 2d 5 + λ 1µ 3d 7 + λ 1µ 2d 10 a 43= λ 1µ 3d 8 + λ 1µ 2d 11 a 44= λ 1µ 1d 5λ 3µ 3d 6 + λ 2µ 3d 7 + λ 3µ 2d 9 + λ 2µ 2d 10

10 32 The solutions to equation (4.4) are readily evaluated as: Wy V(y) = e.v(0) (4.5) Notice that the vector v(0) provides the four initial values required to solve the differential equations. It can be shown that the rows of W sum to zero, i.e. (1, 1, 1, 1) W= (0, 0, 0, 0). This implies that or and, hence, (1, 1, 1, 1) dv(y) = 0 (4.6) dy dg 5(y) = dg 2(y) + dg 3(y) + dg 4(y) (4.7) dy dy dy dy g 5(y) = g 2(y) + g 3(y) + g 4(y) + K (4.8) where K is a scalar constant. 5. FIRST BOUNDARY CONDITION THE CONSERVATION OF MATERIALS To solve the linear differential equations, there is one required boundary condition. One such boundary condition follows from the conservation of the materials flow at the stations. In the long-run, the amount of materials passing through the first station is equal to the amount of materials passing through the second station. The same argument can be extended to all three stations, i.e. in the long-run the amount of materials passing through all three stations are equal. While the buffer remains empty or full, the line is productive only if all three stations are up and operating. That is, the amount of materials passing through each station is the same. Therefore, given that the buffer is partially full, the steady-state probability that the first station is up and operating is equal to the steady-state probability that the second station is up and operating which is equal to the steady-state probability that the third station is up and operating. The first station is up and operating in station states 1, 2, 3 and 4, second station is up and operating in station states 1 and 5, while the third station is up and operating in station states 1 and 5. Therefore, the following equations can be written: M 0 [g 1 (y) + g 2 (y) + g 3 (y) + g 4 (y)] dy M M = [g 1 (y) + g 5 (y)] dy = [g 1 (y) + g 5 (y)] dy (5.1) 0 0 or equivalently, M [g 2 (y) + g 3 (y) + g 4 (y)] dy

11 33 M [g 5 (y)] dy (5.2) 0 The above equation shows the constant K in equation (4.4.8) is zero. Thus g 5 (y) =g 2 (y) + g 3 (y) + g 4 (y) 6. OTHER BOUNDARY CONDITIONS The other three boundary conditions required to solve equation (4.4) are obtained from the process describing the system while the buffer remains empty. The possible station states corresponding to an empty buffer are given below. All of these states were formerly combined and given the index 10. TABLE 6.1 Possible Station States Corresponding to an Empty Buffer Station 1 Station 2 Station 3 D S S D D S D D D D SB D U U U Where U denotes up and operating, D denotes down, and under repair, S denotes operational but starved, and SB denotes operational but both starved and blocked. The only possible way the buffer becomes empty is if the system is in state 5 (DUU) of table 2.1 and buffer is partially full and no change occurs in the station states. The process may leave the empty buffer state when the first station is operational and either the second, or the third or both the second and the third stations are down. While the buffer remains empty the possible states of the system is shown below figure 1.

12 34 DSS λ 1 λ 1 λ 2 UUU UDU λ 2s µ 1 UUB µ 2 µ 3 λ 3s DDS λ 3s D, SB, D µ 3 λ 2sb DDD µ 1 UBD µ 2 Figure 1: Transition Diagram involving Three States where the Butter Remains Empty Thus after entering the buffer-empty state through system state 5, the process eventually leaves the buffer-empty state in one of the three states 2(UBD), 3(UDS), or 4(UDD). The probabilities that the empty buffer state are left in state 2, 3, and 4 are proportional to g 2 (0), g 3 (0), and g 4 (0), respectively. Since the empty-buffer process, as shown in Figure 1, is Markovian the probabilities P 5,2, P 5, 3, and P 5,4 that the buffer-empty process is left by the states 2, 3, or 4 can be evaluated. It can be shown that: P 5, 2 = -λ 2 µ 1 [µ 2 µ 3 (µ 1 + µ 2 + µ 3 + λ 3s + λ 2sb ) + µ 1 {λ 2sb (λ 3s + µ 2 ) + λ 3s µ 3 ) ] (6.1) det (S) P 5, 3 = -λ 3 µ 1 [µ 2 µ 3 (µ 1 + µ 2 + µ 3 + λ 3s + λ 2sb) +µ 1 {λ 2sb (λ 3s + µ 2) + λ 3s µ 3)] (6.2) det (S) P 5, 4 = λ 1 + λ 2 + λ 3 [λ 3s (λ 2sb + µ 3 ) + λ 3s * λ 2sb ( λ 3s + µ 2 )] (6.3) det (S) Detailed development for equations 6.1 through 6.3, as well as the structure of matrix S and its determinant are given in Quasem (1992). Thus using equations (5.3) we have

13 35 P 5, 2 v(0) = C. P 5, 3 P 5, 4 P 5, 2 + P 5, 3 + P 5, 4 Where C is a constant. To find C, one uses the normalizing conditions. 8 M Σ g i (y) dy = 1 i=1 0 This completes the evaluation of the boundary conditions and the solution of the conditional steady-state density functions. 7. EXPONENTIAL OF MATRIX W The solution of (4.4) requires the exponentiation of matrix W. Provided that matrix W has distinct eigenvectors, exp (Wy) can be evaluated as: V(y) = M e Dy M -1 V(0) Where M is the model matrix and D is the diagonal matrix containing the Eigen values of W. In this case: If exp (Wy) is evaluated numerically, a more efficient method is the Taylor Series expansion. In this case: Wy n n 1 e = Σ y W n = 0 (n + 1)! Wy Moreover, the integral of the exponential e can be evaluated as M n n 1 e WY dy = Σ W M 0 n = 0 (n+1)! 8. THE LONG-RUN CHARACTERISTICS The main interest of this study is to obtain the production rate. The amount of materials produced by a unit time when the buffer is partially full is M 0 [g 1 (y) + g 5 (y)] dy

14 36 The probability that the buffer is partially full is {1-(P 9+P 10)}. Where P 9 and P 10 are the probabilities that the buffer remains full or empty respectively. Notice that P 9 and P 10 can be computed in a manner similar to that developed by Yeralan, Frank, & Quasem (1986), given by P 9 = E [T f] g 5 (1)] 1+E [T f] g 5 (1) + E [T e] g 5 (0) (8.1) P 10 = E [T e ] g 5 (0) 1+ E [T f] g 5 (1) + E [T e] g 5 (0) (8.2) Where E[T f] and E[T e] are the expected amount of time the buffer remains full and the expected amount of time the buffer remains empty respectively. The quantity g 5 (1) is the expected number of times the buffer remains full for every time unit, and g 5 (0) is the expected number of times the buffer is empty for every time unit. The quantity g 5(1) is expressed as g 5 (1) = g 2 (1) + g 3 (1) + g 4 (1) Now, let P e and P f be the fraction of the time that the system is productive while the buffer remains empty or full respectively. Then, the production rate p can be written as follows: p = p f P 9 + p e + P 10 + [1- (P 9 + P 10)] g 1 (y) + g 5 (y)dy The evaluation of p e, P f, E[T e], and E[T f], are discussed next. M 0 9. EXPECTED OCCUPATION TIMES WHEN THE BUFFER REMAINS FULL OR EMPTY The assumption was made that the time to breakdown and time to repair are still exponentially distributed when the buffer remains empty or full and idle stations do not breakdown. To derive p f and E[T f], consider the case when the buffer is full. The buffer can become full from states 2(UBD), 3(UBD), or 4(UBD) in Table 2.1. Let N be the random variable denoting from which state the process enters the full-buffer process, (N= 2, 3, 4). The expected time the buffer remains full can now be conditioned on 4 E[T f] = Σ E[T f [N=J]. P [N=J] J=2 By considering each of the cases N=2, N=3, and N=4, we find E[T f N]. The fraction of the time the line is productive given N=J is denoted by p f, j for J= 2, 3, 4. The detailed description of the process while the buffer remains full for each possible value for E[T f N] and p f, J can be computed. The results are summarized below. E[T f N=2] = (λ 1 + λ 3) [λ 1 + µ 1 + µ 3] + µ 3 (µ 1 + µ 3) p f, 2 = µ 3 (µ 1 + µ 3) λ 1 λ 3 (λ 1 + λ 3 +µ 1 + µ 3) (9.1) (λ 1 + λ 3) [λ 1 + µ 1 + µ 3] + µ 3 (µ 1 + µ 3) (9.2)

15 37 E[T f N=3] = λ 1 + µ 1 + µ 2 (λ 1. µ 2) (9.3) and p f, 3 = µ 2 (µ 1 + µ 2) (λ 1 + λ 2 + µ 2) [λ 1 + µ 1 + µ 2] (9.4) where E [T f N=4] D/ S (9.5) D = µ 2 [λ 2 λ 3 s + λ 3µ 3 + µ 3 (λ 2 + λ 3 s +µ 2 + µ 3)] + λ 1 µ 3 (µ 2 + λ 3 s + (λ 1 + λ 3) (λ 2 + µ 2) (λ 3 s+ µ 2) + (λ 2 + µ 3) [λ 2λ 3 s + µ 3 (λ 1 + λ 2 + λ 3)] S = µ 3 [(λ 2 + µ 2 + µ 3) (λ 1 λ 3 S + λ 2 λ 3 + λ 1 µ 2 + λ 2 λ 3 S) + λ 3 S (λ 2 λ 3 + λ 2 µ 2 + λ 1 λ 2 + λ 1 µ 3 + λ 2 λ 3 + λ 3 µ 3)] and p f, 4 = µ 2 µ 3 (λ 2 + λ 3 s +µ 2 + µ 3) D (9.6) The probabilities P [N = J] for J = 2, 3, 4 are proportional to the steady-state conditional functions g J (M). Thus, P[N= 2] = g 2 (M) g 2 (M) + g 3 (M) + g 4 (M) (9.7) P [N = 3] = g 3 (M) g 2 (M) + g 3 (M) + g 4 (M) (9.8) P [N=3] = g 4 (M) g 2 (M) + g 3 (M) + g 4 (M) (9.9) The unconditional expected time the buffer remains full, and the percent of time the line is productive while the buffer remains full can be evaluated as: E [T f] = {(λ 1 + λ 3) [λ 1 + µ 1 + µ 3] + µ 3 (µ 1 + µ 3)}. g 2 (M) {λ 1 λ 3 (λ 1 + λ 3 + µ 1 + µ 3)}. {g 2 (M) + g 3 (M) + g 4 (M)} + (λ 1 + µ 1 + µ 2). g 3 (M)

16 38 (λ 1. µ 2). {g 2 (M) + g 3 (M) + g 4 (M)} (9.10) + D. g 4 (M) S. {g 2 (M) + g 3 (M) + g 4 (M)} P f = {µ 3 (µ 1 + µ 3)}. g 2 (M) {(λ1 + λ 3) [λ1 + µ 1 + µ 3] + µ 3 (µ 1 + µ 3)}. {g 2 (M) + g 3 (M) + g 4 (M)} + {µ 2 (µ 1 + µ 2)}. g 3 (M) {(λ 1 + λ 2 + µ 2) [λ 1 + µ 1 + µ 2]}. {g 2 (M) + g 3 (M) + g 4 (M)} (9.11) + {µ 2 µ 3 (λ 2 + λ 3 s + µ 2 + µ 3)}. g 4 (M) D. {g 2 (M) + g 3 (M) + g 4 (M)} The expressions for E[T e] and p e can be derived similarly and are given in equations (9.12) and (9.13) respectively. Detailed discussion is given in Quasem (1992). and E[T e] = (λ 1 + λ 2 + λ 3) (µ 1 + µ 2 + λ 3 s) + µ 1 (µ 1 + µ 3) µ 1 [λ 3 s (λ 1 + λ 2 + λ 3) + (µ 1 + µ 3) (λ 2 + λ 3)] P e= µ 2 (µ 1 + µ 3) (λ 1 + λ 2 + λ 3) (µ 1 + µ 2 + λ 3 s) + µ 1 (µ 1+ µ 3) 10. NUMERICAL EXAMPLE In real life situations, only often the parameters known are the expected time to repair. For example, given that the expected time to breakdown for station 1, station 2, and station 3 are 0.01, 0.011, and respectively and the expected time to repair for station 1, station 2, and station 3 are 0.09, and respectively, i.e. λ 1 = 0.01, λ 2 = 0.011, λ 3 = while µ 1 = 0.09, µ 2 = 0.087, and µ 3 = 0.085, then the production rate is found to be EXTENSIONS TO K STATIONS WITH A SINGLE BUFFER In principle, the above model can be extended to an L station production line with a single buffer. The buffer may be placed between any of the two adjacent lines, where the buffer is placed between the first two stations.

17 39 Input Sink Figure 10. An L-Stage Transfer Line Model In this case, the number of system states will be 2 k. Namely, UUUUU..UU, UUUUU UD, UUUUU UDU,..DDDDDD.D. The system state at time t can still be described by the 2-vector (i, y) where i is the station state combinations, i= 1, 2, 3, 2 and y is the buffer level. A simple model is where it s assumed that blocked and starved stations may not breakdown. To further investigate this case, let the downstream stations be the ones after the buffer and the upstream stations are the ones before the buffer. Let the buffer be placed after station K, K= 1, 2, 3,.L-1. Thus station 1, 2, 3 K are the upstream stations and K+1, K+2, L is the downstream stations. The upstream stations are productive only when all upstream stations are operational, and the K-th station is not blocked. Similarly, the downstream stations are productive only when all downstream stations are operational, and station K+1 is not starved. Time to the first breakdown among upstream stations, given that all are productive, is exponentially distributed with rate λ = Σ λ i where λ i is the breakdown rate of station i. The probability that jth station breaks down is λ j.. Thus time to the breakdown for the upstream stations is exponentially distributed. If µ i = µ for all i= 1,2,3,.K, where µ i is the repair rate for station i, then time to repair the upstream stations becomes exponentially distributed with rate µ. Otherwise, time to repair the upstream stations is hyper-exponentially distributed. Similarly, let λ = Σ λ i and µ = µ for i= K+1, K+2,.L. The line can equivalently be modeled as having two stations with a buffer between the stations. The first station equivalent to the set of upstream stations has parameters λ and µ. Similarly, station 2 equivalents to the downstream stations, has parameters λ and µ. The line can then be analyzed using the result found in Yeralan, Frank, and Quasem (1986) as well as Quasem (2008). 12. CONCLUSIONS AND REMARKS The most significant insight obtained into the continuous flow production lines is that as long as there is only one continuous state variable, namely one buffer level variable, the resultant system may be solved, at least numerically. The introduction of various station states makes the model more complicated but not entirely non-traceable. Quasem (2008) has discussed in detail how the complexity of the model is increased when there are two continuous state variables, i.e., two buffer levels. Note: The introduction and literature review of this paper is similar to that in the paper The Role of Buffers for Productivity Improvement in an Unreliable Stochastic Production Line Model because both papers objectives are similar, except the number of stations and a buffer taken into consideration.

18 40 REFERENCES Al- Kaisi, Mahdi (June 2000), Use Conservation Buffers to make Dollars and Sense," Integrated Crop M Management, Iowa State University Gershwin, S. B., and Schick, I. C. (1980), Continuous Model of an Unreliable Two-Stage Material Flow System with Finite Interstage Buffers, MIT Laboratory of Information and Design Systems Report, LIDS 979 Koeninsberg, E (1959), Production Lines and Internal Storage A Review, Management Science Vol. 5, Pp Muth Eginhard J. and Abkaff Abdullah. (1987), The Throughput Rali of Three-station Production Line a Unifying Solution. International Journal of Production Research, Vol. 25, No. 10 Quasem, Mohammad (December 2008), The Role of Buffers for Productivity Improvement in an Unreliable Stochastic Production Line Model" Journal of Global Information Technology Vol 6 N0 1 & 2 Pp Quasem, Mohammad A., Yeralan, S. and Franck, Wallace, Jr., (1985) The Effect of Station Repair Policies on the Productivity of Production Lines, American Statistical Association 1985 proceedings of the Section on statistical Education (Pp ) Tan, B & Gershwin, Stanley B., (2007), Modeling and Analysis of Markovian Continuous Flow Production Systems with a Finite Buffer: A General Methodology and Applications Operations Research Center working Paper, Massachusetts Institute of Technology Wijngaard, J. (1979), The Effect of Interstage Buffer on the Two Unreliable Production Units in Series with Different Production Rates, AIIE Transactions, Vol. 11, No. 1, Pp Yeralan, S. Franck, Wallace, Jr., Quasem, Mohammad A. (1986), A Continuous Materials Flow Production Line Model with Station Breakdown has been published in the European Journal of Operational Research in December, (Pp )