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- Title: MM2
- Type: Notes
- School: UF
- Course: EIN 6392
- Term: Spring
Chain Supply Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Manufacturing Management Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling Factory Flow Basics s Definitions: x We should already have a sense for commonly used terminology such as: 3 Workstations, parts, raw materials, components, subassemblies, final assemblies, end items, BOM, routings, orders, jobs, raw materials inventory, finished goods inventory Throughput (TH): Average output per unit time (throughput rate, e.g., parts per hour) WIP: Inventory between start and end points of a routing Cycle Time (CT): Average time from job release to end of routing x We need to formalize some key definitions 3 3 3 1 Factory Flow Basics x Important parameters: 3 Bottleneck rate (rb): processing rate of the workstation having the least long-term capacity Considering breakdowns, setups, rework Bottleneck rate gives maximum throughput rate for line 3 3 Raw process time (T0): sum of average processing times at each station on a line; average time it takes a single job to traverse an empty line Critical WIP (W0): WIP level that gives maximum throughput and minimum cycle time on a line with no variability in process times W0 = rbT0 Penny Fabrication Examples s Penny Fab One (PF1) x Four machines in series that make giant souvenir one-cent pieces 3 3 Line runs 24/7 Each machine takes exactly 2 hours (rb = 0.5/hr, T0 = 8 hrs, W0 = 4) W0 = number of machines for balanced serial lines s Penny Fab Two (PF2) x Similar to PF1, except 3 3 3 Station 2 has 2 machines, each taking 5 hours Station 3 has 6 machines, each taking 10 hours Station 4 has 2 machines, each taking 3 hours 2 Penny Fab Two Example 3 3 3 3 3 3 3 3 Station 1 has processing rate 0.5/hr Station 2 has processing rate 2*(1/5) = 0.4/hr Station 3 has processing rate 6*(1/10) = 0.6/hr Station 4 has processing rate 2*(1/3) = 0.67/hr rb = 0.4/hr, Station 2 T0 = 20 hours W0 = (0.4)(20) = 8 pennies W0 < number of machines (11) Some machines will not be fully utilized 3 If W0 is not a whole number, no constant WIP level will allow a throughput rate equal to the bottleneck rate Measuring Performance s Best Case (PF1) x WIP = 1 3 3 TH = (1/8)/hr CT = 8 hrs WS 1 t=0 t=2 t=4 t=6 WS 2 WS 3 WS 4 3 Measuring Performance s Best Case (PF1) x WIP = 2 3 3 TH = (1/4)/hr CT = 8 hrs WS 1 t=0 t=2 t=4 t=6 WS 2 WS 3 WS 4 Measuring Performance s Best Case (PF1) x WIP = 3 3 3 TH = (3/8)/hr CT = 8 hrs WS 1 t=0 t=2 t=4 t=6 WS 2 WS 3 WS 4 4 Measuring Performance s Best Case (PF1) x WIP = 4 3 3 TH = (1/2)/hr (= rb) CT = 8 hrs WS 1 t=0 t=2 t=4 t=6 WS 2 WS 3 WS 4 Measuring Performance s Best Case (PF1) x WIP = 5 3 3 TH = (1/2)/hr CT = 10 hrs WS 1 t=0 t=2 t=4 t=6 WS 2 WS 3 WS 4 5 Measuring Performance s What can we learn from this simple example? x There is a relationship between WIP, CT, and TH 3 3 TH increases as WIP increases, up to a point When we exceed W0, TH cannot be improved, and CT increases s Little s Law (Queuing Theory) x WIP = (TH)(CT) 3 3 3 For a constant WIP level, increasing throughput reduces cycle time (and vice versa) At a constant throughput rate, increasing WIP reduces cycle time (and vice versa) For a constant cycle time, Increasing throughput rate decreases WIP (and vice versa) Best Case Performance s For a fixed WIP level, we can characterize the best possible CT and TH: Avg. CT w W0 T , CTbest = 0 w / rb , otherwise w / T0 , w W0 = otherwise rb , T0 W0 Avg. TH rb WIP, w TH best W0 WIP, w 6 Worst Case Performance s We next consider the absolute worst case for a line with bottleneck rate rb and raw process time T0. x x x x Suppose that we have 4 jobs in PF1, and job 1 requires 8 hours on each machine, while jobs 2, 3, and 4 require zero hours on each machine. All centers have the same long-term capacity of producing 4 jobs every 2 hours, or rb = 0.5/hr. The long-term average raw processing time per job gives T0 = 8 hours These are the same values of rb and T0 as in the best case performance, however, there is significant variability in the processing requirements of the individual jobs Worst Case Performance s s s s WIP = 4 jobs CT = 32 hours TH = (1/8) job/hour WIP = CT*TH x Little s Law holds WS 1 t=0 t=8 t = 16 t = 24 WS 2 WS 3 WS 4 7 Worst Case Performance s Of course this is an extreme and impractical example, but it gives a bound on the worst performance possible and actually can happen in practice x x If the processing time at each station for every job was 2 hours as before, but we moved items from station to station in batches of size four, the same performance would result Note, however, that this worst case performance requires that the entire WIP in the system is transported in a single batch move; it is more likely that we would see several batches moving between different stations at the same time CTworst = wT0 THworst = 1/T0 s We can characterize these relationships in general as x x Practical Worst Case s s s s Maximum randomness case State of the system describes the location of jobs at stations Maximum randomness is the case in which all possible states of the system are equally likely Maximum randomness is guaranteed under the following conditions: x x x The line is balanced (all stations have same average processing times) All stations consist of a single machine Processing times are exponentially distributed 8 Exponential Distribution s The exponential distribution is a single-parameter distribution x x The normal distribution, for example, is a two parameter distribution; we must specify a mean, and a standard deviation, to characterize it The probability density function (pdf) of the exponential distribution has the form f(x) = exp(- x) 3 The mean and standard deviation of this distribution both equal 1/ x The cumulative distribution function (cdf) of the exponential distribution is given by: t 3 Prob{X t} = exp( x ) dx = 1 exp(- t) 3 Prob{X t} = exp(- t) 0 Exponential Distribution x The exponential distribution is memoryless 3 3 The probability that X t1 + t2 given that X t1 equals the probability X t2. If X is a random variable for the time something lasts, then the probability something lasts t2 more time units given that it has lasted until t1 has nothing to do with the value of t1. There is no memory of how long it has lasted Prob(X t1 + t2 | X t1) = Prob(X t1 + t2 AND X t1)/Prob(X t1) This is Bayes Rule for probability For the exponential distribution using Bayes rule we have Prob(X t1 + t2 | X t1) = Prob(X t1 + t2)/ Prob(X t1) = exp(- [t1 + t2])/exp (- t1) = exp (- t1)exp(- t2)/exp(- t1) = exp (- t2) = Prob(X t2) 9 Practical Worst Case x Suppose we have N single-machine stations, each with an avg. processing time of t, and we have a constant WIP = w in the PWC system 3 3 3 3 3 rb = 1/tT0 = Nt As a job arrives at the first station, it will, on average, see the other w 1 jobs evenly distributed at the N stations The number of jobs ahead of the arrival at machine 1 will then equal (w 1)/N So the average time spent at this station will equal t(w 1)/N + t = [1 + (w 1)/N]t Since this holds for arrivals at each station, the average cycle time equals CT = N[1 + (w 1)/N]t = Nt + (w 1)t = T0 + (w 1)/rb Practical Worst Case x We can use Little s Law to calculate the average throughput: 3 3 TH = WIP/CT = w/[T0 + (w 1)/rb] = w/[W0/rb + (w 1)/rb] = [w/(W0 + w 1)]rb To summarize these results, for a constant WIP level, w, we have: CTPWC = T0 + (w 1)/rb THPWC = [w/(W0 + w 1)]rb 3 3 When w = 1, CTPWC = T0, THPWC = rb/W0 = 1/T0 When w , CTPWC , THPWC rb Obtaining throughput that is close to bottleneck avg. capacity requires high WIP levels 3 PWC is between Best and Worst case performance 10 Practical Worst Case s s Recall that the assumptions of the PWC required a balanced line, single-machine stations, and exponential processing times How can we improve upon a PWC system? x x x Unbalance lines (improves level of randomness, even we add capacity at non-bottleneck stations) Parallel machines (even if we exchange two machines with the capacity of a single machine, parallel processing reduces waiting times) Reduce variability below that implied by the exponential distribution Variability s s s Variability is the root of all evil in operations (?) The higher the variability of a process, in general, the worse the predictability and performance of the process Variability is relative x If my standard deviation, , of weekly demand is 1,000 units, x x is this high or low? It depends on the average weekly demand level, . If average demand is 500, this is relatively high variability; if average demand is 100,000, this is relatively low variability We need a relative measure of variability 3 3 Coefficient of variation, c = / (Note that c = 1 for exponential distribution) Squared Coefficient of variation (SCV), c2 = 2/ 2 (sometimes more convenient mathematically) 11 Sources of Variability s Variability in manufacturing processes comes from many sources: x x x x x Machine breakdowns Setup times Operator variability Rework Natural variation due to many factors that change with time, temperature, materials Process Time Variability s Natural variability effects x Let t0 and 0 denote mean and standard deviation of natural process time (natural variability is inherent in the process and is difficult to effect) Let mf denote the mean time to failure for a machine (the avg. time it takes until the next breakdown once it is up and running) Let mr denote the mean time to repair for a machine (the avg. time it takes to repair the machine once it breaks down) s Variability from breakdowns x x 12 Variability from Breakdowns s The availability, A, of a machine is the proportion of time the machine is working x x A = mf/(mf + mr) We must adjust the natural processing time to account for the effects of breakdowns 3 3 We define the effective processing time by the equation te = t0/A The effective capacity (rate) of a machine with natural processing time t0 then is re = A/t0 = Ar0 Breakdown Variability Illustration s Consider two machines, one with short but frequent outages, and another with infrequent but long outages. x Both machines have t0 = 10 min and 0 = 2 min x x x x x Machine 1: mf = 5,400 min and mr = 600 min Machine 2: mf = 540 min and mr = 60 min A = 0.9 for both machines Which would you rather have? Which one would require more WIP to ensure that it does not starve during a breakdown? 3 It can be shown that machine 1 ce2 = 10.84 and machine 2 ce2 = 1.12 assuming exponentially distributed failure and repair times 13 Nonpreemptive Outages s s Variability due to random failures is preemptive, since it preempts processing without regard for the state of the system Variability due to planned events, such as setups or rework, is called nonpreemptive x x Setups and rework reduce the capacity of machines, and must be factored into effective machine capacity and variability Table 8.1 summarizes equations for te, e2, and ce2 as a result of preemptive and nonpreemptive outages Flow Variability s s The types of variability we considered so far deal with processing time variability We are often concerned with the variability of the time it takes jobs to flow through a machine or a series of machines x x This depends not only on the processing time variability, but on the nature of arrivals to the system Let ta denote average time between arrivals to a station, and let ra denote the arrival rate, where ra = 1/ta 14 Flow Variability s s s Note that the only way a workstation with variable processing times can keep up with its work is if re > ra (why?) Letting a2 denote the variance of the time between arrivals we define the SCV of arrivals as ca2 = a2/ta2 In addition to arrival times, we need to characterize departure times. x x Let td denote the average time between departures, and we define rd and cd2 as above Note that in a serial production line, the departures at station i become the arrivals at station i + 1 Utilization and Departure Rates s s How do the arrival rate and processing rate interact to determine the departure rate, and therefore the arrival rate, at the next station? Utilization, u, is the average fraction of time a machine is processing jobs x x x x For a single machine station, u = te/ta = ra/re = rate For an m-machines station, u = rate/m In a serial system with no yield loss or rework, the (avg.) arrival rate to every workstation equals the (avg.) throughput rate, where the throughput rate is the departure rate (we would accumulate inventory in front of the station indefinitely if ra > TH and it is impossible to have ra < TH) How do we characterize the variability of departures? 15 Inter-Departure Time Variability s s s If utilization, u, is near 1, the station is virtually always working and we expect that the variability of inter-departure times and processing times are nearly equal, i.e., cd2 = ce2 If u is close to zero, then the station is often idle and we would expect that the variability of inter-departure times nearly equals that of inter-arrival times, i.e., cd2 = ca2 We can interpolate between these two extremes to approximate cd2 for different levels of u: x x cd2 = u2ce2 + (1 u2)ca2 m-machine workstation approximation: 3 cd2 = 1 + (1 u2)(ca2 1) + u2(ce2 1)/m1/2 Queuing Fundamentals s Before we move from our one-machine flow results, we need to understand basics of queuing systems x x x Queuing theory is the study of performance of systems involving servers and waiting lines Typical examples include bank lines, retail store lines, amusement park lines A workstation is a queuing system in which the machine is the server, and WIP waits in a line for the server 16 Queuing Fundamentals s Kendall s Notation: A/B/m x x x x A describes inter-arrival distribution B describes (server) processing time distribution m is the number of servers Distribution Notation: 3 3 3 D = deterministic M = Markovian (exponential distribution) G = general distribution (normal, uniform, etc.) x M/G/3, for example describes a system with exponentially distributed inter-arrival times, generally distributed service times, and 3 servers Queuing Fundamentals s We typically differentiate the time spent in queue from the time in the system, since this impacts service levels in service systems, and WIP inventory costs in manufacturing systems x x x CT = CTq + te WIPq = ra*CTq WIP = ra*CT (note that ra replaces TH since they are equivalent) 17 M/M/1 Queue s s This is the simplest queuing system (other than deterministic systems) to analyze We are first interested in characterizing the WIP of an M/M/1 queue as a function of arrival rate and processing times x x The expected WIP in the system equals n n =0 where pn denotes the long-run probability that we find n jobs in the system at a random point in time Our job then is to determine the values of the pn s np , M/M/1 Queue s When the system is in state n (i.e., n units are in the system), it can move to either state n + 1 or state n 1 x x x x x Rate of moving from n 1 to n when in state n 1 equals the arrival rate, ra Rate of moving from n to n 1 when in state n equals the processing rate, re Overall rate of moving from n 1 to n equals pn-1ra Overall rate of moving from n to n 1 equals pnre For the system to be stable (not accumulate inventory indefinitely or run out of work) we require that pn-1ra = pnre, or pn = (ra/re)pn-1 = upn-1 18 M/M/1 Queue s We need some starting pn value x What is p0? 3 This is the proportion of time the system is idle; therefore, p0 = 1 u x We therefore have the following: 3 3 3 3 3 p1 = up0 = u(1 u) p2 = up1 = u2(1 u) In general, pn = un(1 u) We also require p0 + p1 + p2 + = 1, or (1 u)(1 + u + u2 + ) = 1 Note that the above equation implies we must have u < 1 M/M/1 Queue s Going back to our WIP equation: x x x x x x x n =0 n =0 n n 1 WIP = npn = (1 u) nu = u(1 u) nu It is possible to show that nu n =1 n 1 = (1 u ) 2 n =1 This gives WIP(M/M/1) = u/(1 u) By Little s Law, CT(M/M/1) = WIP(M/M/1)/ra = te/(1 u), and CTq(M/M/1) = CT(M/M/1) te = teu/(1 u) WIPq(M/M/1) = ra*CTq(M/M/1) = u2/(1 u) We now have a simple set of equations to characterize the performance of an M/M/1 queue 19 More General Queuing Systems s s s Queuing systems with general arrival and processing time distributions are much more difficult to analyze The table below summarizes best-known approximations for characterizing more general queuing systems Note that given CTq we can determine WIPq, WIP, and CT using CT = CTq + te and Little s Law: WIPq = ra*CTq, WIP = ra*CT M/M/m G/G/1 G/G/m System CTq teu ( 2( m +1) 1) m(1 u ) 2 2 2 2 ca + ce ca + ce CTq (M/M/1) CTq (M/M/m) 2 2 Supply Chain Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Manufacturing Management Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling 20 Push versus Pull Systems s s s s s Push system schedules work releases according to forecasted demand Pull system authorizes work releases based on system (production system and demand) status Most all real world systems use some combination of the two We will tend to take MRP/MRP II as synonymous with Push and CONWIP/Kanban as synonymous with pull Push systems control throughput and observe WIP, Pull systems control WIP and observe throughput x WIP is more observable than throughput What is so good about pull? s s s The Japanese success was largely attributed to the implementation of JIT systems through pull systems using Kanban The authors argue that the reason pull systems perform so well is because they put a cap on WIP Why is a WIP cap good? x x Shop floor disruptions (quality problems, machine breakdowns do not result in a WIP explosion as in push) Holding back work releases during disruptions gives greater flexibility 3 Physical items are not committed to orders 21 Pull Systems Reduce Variability s s Good customer service requires low cycle time variability Kanban cycle times are less variable than under push systems x Eliminating WIP explosions eliminates cycle time explosions s Pull systems reduce the water level to expose the rocks by keeping WIP low x x Highlights problems in the process Promotes continuous improvement Pull Systems Promote High Quality s JIT and associated pull systems require and promote high levels of quality x By capping WIP, we cannot afford to have poor quality 3 If we try to have a WIP cap with poor quality levels, throughput will suffer through downstream starvation x x Inspection is made easier under low WIP Workers must go upstream to get materials 3 3 They can directly inspect input to their stations and identify problems before processing Downstream stations will put pressure on upstream stations to provide good materials so they can keep working 22 Pull Systems are More Flexible s Once a job is released to the floor it takes an identity x x x s By starting work on raw materials we differentiate them so they can serve in less end products Work on the floor is often assigned to a particular order Pull systems keep orders on paper longer Pull systems allow the facility to work ahead when WIP falls below the cap x Push systems do not permit work ahead due to scheduled releases CONstant Work In Process s A CONWIP system is equivalent to running the entire line as a single Kanban station x x Instead of circulating cards through each station, CONWIP circulates cards through the entire production line Output pulled from the last station sends a card to the beginning of the line All parts follow the same routing through the line WIP can be easily measured We can still implement CONWIP if these assumptions are violated, but we must establish different CONWIP levels for each routing s Pure CONWIP assumes: x x x 23 CONWIP s CONWIP is a closed queuing network x x x x Items that finish at the last station always return to the beginning of the line The number of items in the queue is always the same MRP is an open queue in which items leave the system, and the number of items in the system can vary Kanban is a closed queue with blocking 3 When an item is finished at a station, it can be blocked at the next station, unlike CONWIP CONWIP versus MRP s Observability x x x Recall that pull systems control WIP and push systems control throughput Since WIP is more easily observable, it is easier to control Throughput is a function of capacity, which may be difficult to estimate dynamically 24 CONWIP versus MRP s How do we compare the efficiency of these systems? x x x Compare the WIP required to achieve a given throughput Consider a 5-machine serial line, where each machine processes at a average rate of 1 job per hour, and processing times are exponentially distributed A balanced CONWIP line with exponential processing times behaves according to the practical worst case performance we discussed CONWIP versus MRP s The PWC analysis led us to the following relationship between WIP, w, and throughput, TH: x x TH(w) = rbw/(w + W0 - 1) = w/(w + 4) for a 5-machine balanced line If the time between releases is exponentially distributed, the system behaves as 5 independent M/M/1 queues WIP(M/M/1) = u/(1 - u) WIP of the line = 5u/(1 - u) s In a push system the release rate determines TH x x x 25 CONWIP versus MRP x x Since u = rate, and te = 1 at each station, u = ra at each station. By conservation of flow, ra = TH. This implies that for the MRP system: 3 w(TH) = 5TH/(1 - TH) s Suppose w = 6 in the CONWIP system: x TH = w/(w + 4) = 0.6 jobs per hour s Now let TH = 0.6 in the MRP system: x w = 5TH/(1 - TH) = 7.5 s s The CONWIP system can achieve the same throughput with less WIP By Little s Law, for a fixed TH, a push system will have longer cycle times than a pull system CONWIP versus MRP s Variability of cycle times is also greater for a push system when compared with an equivalent pull system x x x Higher cycle time variability requires quoting longer lead times to meet due dates at the same rate (with the same service level in terms of on-time delivery) Push systems also have higher average cycle times for a given throughput rate Longer average production lead times (cycle times) coupled with more variable lead times results in much higher cost to provide equivalent service 26 CONWIP versus MRP s CONWIP is also much more robust to deviations in WIP than MRP is to deviations in throughput x Consider a simple profit function: 3 3 Profit = pTH - hw, p = marginal profit per job and h is WIP holding cost x x Recall that in a CONWIP system we choose WIP, w, and in MRP we choose throughput, TH For a balanced serial line we had: 3 3 CONWIP: TH(w) = rbw/(w + W0 - 1) MRP: w(TH) = mu/(1 - u) (m is # of machines) CONWIP versus MRP x Profit function for CONWIP: 3 ProfitCW = prbw/(w + W0 - 1) - hw Profit = pTH - hmu/(1 - u), where u = TH*te, which gives ProfitMRP = pTH - hmteTH/(1 - teTH) x Profit function for MRP: 3 3 x x For given system parameters, we differentiate ProfitCW wrt w and set the result equal to 0 to find the optimal w* and profit; we differentiate ProfitMRP wrt TH and set the result equal to 0 to find the optimal TH* and profit In our prior 5 station system, we find that w* = 16 and ProfitCW (w*) = $63.30; TH* = 0.776 jobs per hour, and ProfitMRP(TH*) = $60.30 27 CONWIP versus MRP s The following graph shows the robustness of CONWIP as a function of deviations from the optimal WIP level and MRP as a function of deviations from the optimal throughput, TH Profit 0 100 Parameter value as a percentage of optimal value (w for CONWIP; TH for MRP) CONWIP versus Kanban s Both CONWIP and Kanban are pull systems that use a WIP cap, so what s the difference? x x x x x Recall that CONWIP is equivalent to running the entire line as a Kanban station Kanban requires setting WIP level at every station Cards are part-specific in Kanban CONWIP uses a backlog list instead of part-specific cards, so the same card can authorize different work each time through the loop Part-specific cards generate work for the part each time the part is consumed downstream 3 Slow moving items are always produced and held somewhere in WIP, regardless of the fact that the next demand may not occur for a long time 28 CONWIP versus Kanban s Kanban systems can have trouble dealing with product mixes on the same line x x x x x When the product mix changes, the bottleneck can change, depending on which products are produced Kanban always requires more cards at the bottleneck, so it does not starve Setting card counts at each station dynamically as a function of product mix can be extremely complex CONWIP requires one single card count for the entire line Since CONWIP allows work ahead , WIP naturally accumulates in front of the bottleneck, where it is always needed CONWIP versus Kanban s People issues x x x Kanban can induce friction between stations, since downstream stations pressure upstream stations for good quality and continuous work flow This has its good points in terms of quality and communication, but can induce stressful relationships CONWIP essentially pushes at every station except the first, relieving this source of friction 29 Supply Chain Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Manufacturing Management Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling Shop Floor Control s s s Where planning meets parts Real time tracking of plant status Material Flow Control is the heart of SFC x x x Which jobs to release? What parts to work on at stations WIP Tracking Material movement between stations Status Monitoring Throughput Tracking Material Flow Control Work Forecasting Capacity Feedback Quality Control 30 Shop Floor Control s WIP tracking x Identifies current location of parts on the line Machine, staffing status WIP Throughput Measures output against Tracking Tracking quotas and due dates Anticipates overtime and Material Flow Status Work Control Monitoring staff shifting needs Forecasting Statistical throughput control Capacity Feedback s Status monitoring x s Throughput tracking x x x Quality Control Shop Floor Control s Capacity Feedback x Ensures whether current capacity status is consistent with high-level planning Adjusts material flow control, WIP tracking to account for reworked items Status Monitoring s Quality Control x WIP Tracking Throughput Tracking Material Flow Control Work Forecasting Capacity Feedback Quality Control 31 Gross Capacity Control s s Attempts to control drastic swings in line utilization by ensuring lines are optimally loaded Options for gross capacity control (measures to prevent wild swings in line loading) x x x x x Varying the number of shifts Varying the number of work days per week 3 Planned overtime Varying the number of hours per day Varying staffing levels Using outside vendors Bottleneck Planning s The bottleneck determines the pace of the line x x x x Throughput is directly related to bottleneck utilization In multi-product systems with multiple routings, bottlenecks can float, making bottleneck control difficult Stable bottlenecks are easier to manage/focus on Bottlenecks can be designed 3 Add inexpensive capacity at certain stations 32 Basic CONWIP s Easy to control basic CONWIP line that has x x x x Constant routings Similar processing times at different stations No significant setups No assemblies (linear flows) s s WIP is controlled through cards or containers We still need to control other aspects of the CONWIP system CONWIP Control Issues s Work backlog x Rules must be established to set the work order of the CONWIP backlog (we will cover more of this in scheduling) Typically FIFO, but must allow for exceptions 3 s Line discipline x Establish passing points at which hot jobs can pass others in line 33 CONWIP Control Issues s Card counts x x x Not an exact science in practice Establish a reasonable cycle time and let WIP = CT*rp, where rp is a reasonable estimate of the line rate If a station s queue virtually never empties, the count can be reduced without affecting throughput If a machine fails downstream from the bottleneck, the bottleneck will starve We can release extra jobs without cards to keep the bottleneck running When the non-bottleneck machine goes up it will catch up and the jobs without cards will clear from the system s Card deficits x x x CONWIP Control Issues s Work ahead rules x x Although CONWIP allows stations to work ahead when the plant is running well, we don t want to work too far ahead, or we will create too much inventory in finished goods We can set a limit on how far ahead the system will work 3 Release jobs into the system, provided that their due date is less than n weeks away 34 Tandem CONWIP s Even when we have all of the conditions for CONWIP to work well (linear flow, similar processing times), we might want to break it up into two or more CONWIP lines x x x If the line is particularly long, it may be difficult to manage all of the stations as one big line It may be easier to separate the line into different loops for better managerial control Different CONWIP loops are separated by a WIP buffer that creates independence between the loops CONWIP Assembly Lines s s s The line below requires assembly of one item from line A and one from line B Each time an assembly is completed, a signal is sent to the beginning of both lines for a new release If lines operate at different speeds, the two releases sent following an assembly need not be paired at assembly Line A 2 1 4 1 Line B 3 3 2 3 35 CONWIP Assembly s s s In the system shown, the bottleneck is machine 3 of line A, where rb = 0.25 jobs/hr W0A = rbT0A = 2 W0B = rbT0B = 3 Under PWC, suppose we want a throughput equal to 90% of the bottleneck rate, i.e., TH = 0.9rb s s TH(w) = rbw(W0 + w 1) = 0.9rb Implies w = 9(W0 1), or wA = 9 and wB = 18 Line A 2 1 4 1 Line B 3 3 2 3 Kanban Control Issues s Kanban is just tandem CONWIP at its extreme x x Kanban is more robust to multiple routings Control disadvantages: 3 3 3 3 3 Complexity of setting card counts at all stations Less work ahead flexibility Requires a container for each product type Not flexible to changing relative product mix at a station since it has fixed product-specific cards Not good for slow moving items x Route-specific cards and implementing a work backlog at each station can help alleviate these problems 36 Pull from the Bottleneck s Both Kanban and CONWIP can result in the following: x x Bottleneck starvation under downstream machine failures Premature work releases that do not consider due date or next time part time is demanded s Pull from bottleneck methods can alleviate these problems (suggested in The Goal) x x WIP is constant in all machines up to and including the bottleneck, but is pushed out freely from the bottleneck Breakdowns downstream from the bottleneck will not cause bottleneck starvation; these machines can always catch up Supply Chain Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Management Manufacturing Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling 37 Production Scheduling s Given a set of orders waiting to be processed, each with a due date and processing time requirements, how do we decide the order in which they will be processed? x The factors that drive these decisions are 3 3 3 3 desire for high rate of on-time delivery desire for high machine utilization desire for short customer lead times desire to keep WIP levels low Production Scheduling s Measures for performance in meeting due dates x x x x Service level in Make-to-Order systems percentage of orders not delivered late Fill Rate in Make-to-Stock systems percentage of orders filled directly from stock Lateness if cj is completion time of job j and dj is the due date, lateness is given by Lj = cj dj (note that negative lateness is earliness) Tardiness, Tj = Lj+ = max{Lj, 0} 38 Production Scheduling s Maximizing utilization x x x x Recall that maximizing utilization is a traditional objective based on accounting principles We typically want bottleneck resources to have high utilization, but high utilization combined with variability can lead to high cycle times and congestion One way to ensure high utilization is through minimizing makespan of a set of orders Makespan is the total time it takes to process all jobs from start to finish Production Scheduling s Reducing WIP and cycle times x Why do we want low cycle times? 3 3 3 Increases flexibility by leaving orders on paper longer Supports higher quality through faster detection Reduces reliance on forecasts further in future x Little s law: CT = WIP/TH implies reducing CT and WIP are equivalent if throughput remains the same 3 However, reducing WIP without reducing variability will cause throughput to decrease and cycle time may be unchanged 39 Classic Scheduling Research s Because of the complexity of scheduling problems, past research has made many simplifying assumptions that don t hold well in practice. These include: x x x x x x One, two, or at most three machine problems All jobs are available at time zero Deterministic processing times No machine breakdowns No preemption of jobs No cancellation of orders Classic Scheduling Problems s Minimize average cycle time on a single machine x Shortest Processing Time (SPT) rule Earliest Due Date (EDD) rule No known efficient algorithm for optimal solution Johnson s Rule (1954) 3 3 3 s Minimize maximum lateness on a single machine x s Minimize average tardiness on a single machine x s Minimizing makespan on two machines in sequence x List all processing times on machines A (1st machine) and B Choose the smallest among all processing times If the smallest number is in column A, schedule the job in the first available position, otherwise schedule it last. Cross the corresponding job off the list and repeat 40 Classic Scheduling Problems s Minimize makespan in a job shop x No known efficient algorithm for optimal solution 3 10 job, 10 machine problem has nearly 4 x 1065 possible schedules (more than there are atoms on the earth?) s Why don t we have efficient algorithms for such seemingly simple problems? x x x x A problem of sequencing 25 jobs has 1.55 x 1025 possible schedules Combinatorial problems like this fall into a class of problems known as NP-hard which means that means that their solution time grows exponentially with the problem size Advances in computing time grow at a much slower pace A computer that can analyze 1 Million schedules per second would take 77,147 years to analyze all schedules for a 20 job problem Production Scheduling s s s The good news is that many algorithms exist for these complex problems that provide near-optimal solutions in short computing times we don t need an optimal schedule, just a good one Note that the firms has discretion over setting due dates, so it has some control over the difficulty of the problems Single machine results give us insights on handling harder problems x SPT and Johnson s work because they take advantage of the flexibility given by jobs with short processing times Limiting batch sizes creates more flexibility Scheduling a single machine is not as difficult as an entire shop s We should focus on scheduling bottleneck machines anyway x 41 Supply Chain Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Manufacturing Management Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling Aggregate Planning s Long-range plans for resources x Staffing x Procurement: supplier contracts x Subcontracting component manufacture x Marketing: promotional decisions based on capacity and new product plans 42 Aggregate Planning Model s Simple single-product model with capacity constraints x x x x x x Neglects randomness and yield losses t = index for time periods, t = 1, , T dt = demand in period t ct = capacity in period t 3 measured in same units as demand r = profit per unit sold (before subtracting inventory costs) h = cost of holding one unit for one period Aggregate Planning Model x x x Xt = production quantity in period t St = quantity sold in period t It = inventory at the end of period t; I0 is given max t =1 ( rSt hI t ) T subject to: St d t , X t ct , I t = I t 1 + X t St , X t , St , I t 0, t = 1,..., T , t = 1,..., T , t = 1,..., T , t = 1,..., T . 43 Aggregate Planning Model s Note that this is a simplified model x Does not account for lot sizing decisions 3 3 3 Assumes profit per unit can be easily calculated to obtain r Assumes production in period t can be sold in period t However our goal is to determine capacity feasibility of a long-range plan, not an economical production plan x x A simple linear program Similar to Wagner-Whitin model, but no fixed cost for production and includes capacity constraint Product Mix Model s s Extends simple model to multiple products and multiple workstations Notation x x x x x i = index for products, i = 1, , m j = index for workstations, j = 1, , n ditu = maximum demand for product i in period t ditl = minimum sales required for product i, period t aij = time on workstation j to produce a unit of product i 44 Product Mix Model x x x x x x cjt = capacity of workstation j in period t (consistent with units of aij) ri = net profit from one unit of product i hi = holding cost for one unit of product i for one period Xit = production of product i in period t Sit = sales of product i in period t Iit = inventory of product i at end of period t Product Mix Model max t =1 i =1 ( ri Sit hi I it ) T m subject to: l u d it Sit d it , for all i, t , for all i , t , for all i, t , for all i, t. m i =1 ij a X it c jt , I it = I it 1 + X it Sit , X it , Sit , I it 0, 45 Product Mix Model s This formulation, although still somewhat simplified allows us to answer some important questions: x x x Are our projected demands capacity feasible? Which workstation constraints are binding? 3 Which are the bottleneck stations? Proportions of capacity allocated to each product What is the best relative product mix? 3 Model Extensions s Including additional resource constraints x Labor, materials, material transport capacities 3 3 3 Let bij denote the usage of resource j per unit of product i Let kjt denote the capacity of resource j in period t m Add constraints: i =1 bij X it k jt s Planning for utilization < 100%: x x Let qj denote the percentage utilization of resource j. To account for this factor, multiply cjt by qj in the capacity constraints 46 Model Extensions s Backorders x x Suppose we allow for backordering items, and it costs i to place a unit of product i on backorder for one period Add the constraint Iit = Iit+ - Iit-, where Iit+ is the inventory held and Iit- is the inventory on backorder in period t for product i, and Iit- is non-negative 3 Iit now denotes the inventory position, and Iit+ is the onhand inventory x Subtract the term i Iit- in the objective function for each product, i Model Extensions s Overtime availability x x x Let lj denote the cost per hour of overtime at workstation j. Let Ojt denote the overtime usage at workstation j in period t, a non-negative decision variable The capacity constraint for workstation j becomes: m i =1 aij X it c jt + O jt And we subtract the term lj Ojt in the objective function for all workstations and time periods x 47 Model Extensions s Yield loss x We account for yield loss similar to the way we accounted for machine utilization 3 3 3 Let yij denote the cumulative yield from station j onward (including station j) for product i If we want to get Xit final units produced, we must release Xit/yij into production at workstation j The resource capacity constraints become: m i =1 aij X it yij c jt Workforce Planning s To illustrate this model, we take a single-product approach x x Valid for situations in which product routings are identical or similar so we can treat them as a single product, or when routings are entirely separate (have no common workstations) 48 Workforce Planning Model s Notation changes/additions x x x x x x x x x aj = time one workstation j to produce one unit b = number of labor hours to produce one unit l = regular labor cost, $/labor-hour l = overtime labor cost, $/labor-hour e = cost to increase by one labor-hour per period e = cost to decrease by one labor-hour per period Wt = workforce in period t, in labor-hours of regular time (W0 is given) Ht = Increase in workforce from period t 1 to period t (laborhours) Ft = Decrease in workforce from period t 1 to period t (laborhours) Workforce Planning Model max t =1{rSt hI t lWt l Ot eH t e Ft } T subject to: d tl St d tu , a j X t c jt , I t = I t 1 + X t St , Wt = Wt 1 + H t Ft , bX t Wt + Ot , X t , St , I t , Ot ,Wt , H t , Ft 0, for all t , for all j , t , for all t , for all t , for all t , for all t. 49 Assembly Line Balancing s Classical IE problem x x x n distinct tasks must be completed on each item ti = time required for task i Goal: divide tasks into groups, where each group represents a workstation 3 Let W denote the number of workstations x Time allocated to each group = C, where C is the cycle time 3 3 Each workstation will complete its tasks in C time units and pass their work to the next station A job comes off the line every C time units Assembly Line Balancing x x Let T denote the total work content of each item, n where T = t i =1 i If we want a cycle time C, what is the number of workstations required? 3 3 3 3 Note that this implies a production rate of 1/C We need to divide the total work content by the cycle time to get the minimum number of stations, i.e., Wmin = T/C T/C is likely to be non-integer, and we require W to be an integer number of stations A stronger condition is Wmin = T/C where x is the smallest integer greater than or equal to x 50 Assembly Line Balancing x For a given number of workstations, W, what is the minimum cycle time, Cmin? 3 3 3 We know that if we can split tasks between workstations, we could achieve C = T/W. However, this typically cannot happen, so T/W just gives a lower bound on the cycle time. A better lower bound then is Cmin = T/W x Our goal should be to obtain the values of W and C that give the best line balance 3 However, this is an NP-hard problem, even to determine the best W for a fixed C Analogous to the classical Bin-packing problem We must pack processing times into bins of size C and determine the minimum number of bins (workstations) Assembly Line Balancing x For a desired cycle time C, we use the ranked positional weight heuristic (Helgeson and Birnie [1961]) 3 3 Assigns a weight to each task based on the time required for all successor tasks Sequentially assign tasks to stations based on ranked positional weights x The following example from Nahmias [1997] illustrates the ranked positional weight technique 51 Assembly Line Balancing Example Task 1 2 3 4 5 6 7 8 9 10 11 12 Immediate Predecessor(s) 1 2 2 2 2 3,4 7 5 9,6 8,10 11 Time 12 6 6 2 2 12 7 5 1 4 6 7 3 7 4 1 2 5 6 9 10 11 12 8 Assembly Line Balancing Example Task 1 2 3 4 5 6 7 8 9 10 11 12 Positional Weight 70 58 31 27 20 29 25 18 18 17 13 7 Task Time 6 3 2 12 6 7 5 7 1 8 6 7 4 2 2 1 11 9 4 12 5 12 6 10 52 Assembly Line Balancing Example s We Make assignments to workstations in decreasing order of ranked positional weight x x x x Assignments violating precedence constraints or desired cycle time are not made Suppose we desire a cycle time of 15 minutes Note that Wmin = 70/15 = 4.67 = 5 If we can find a solution with W = 5, it is optimal (why?) Assembly Line Balancing Example 3 s Ranking: x x 7 4 1 2 5 9 8 11 12 (1,2,3,6,4,7,5,8,9,10,11,12) Follow the following steps 3 3 3 3 Open WS1 Assign task 1 to WS1, t1= 12, slack1 = 3 6 10 Task 2 must be assigned before any other task or precedence constraints will be violated t2 =6, since t1 + t2 > 15, we must open WS2 and assign task 2 to WS2, slack2 = 9 WS1 must be closed since task 2 is in WS2 and assigning any other task to WS1 would violate precedence constraints 3 3 Next assign task 3 to WS2, t3 = 6, slack2 = 3 Next try to assign task 6 to WS2. t6 = 12, Insufficient slack exists, so we open WS3 and assign task 6, slack3 = 3 53 Assembly Line Balancing Example s Ranking: x 3 7 WS 2 8 11 12 (1,2,3,6,4,7,5,8,9,10,11,12) 3 4 5 6 3 Next assign task 4 1 1 2 2 9 10 t4 = 2; We can assign it to WS2 without violating precedence constraints slack2 = 1 3 Next assign task 7 t7 = 7, so it won t fit in WS2 or WS3 and we must open WS4 and assign task 7, slack4 = 8 3 Next assign task 5 t5 = 2; since slack3 = 3, we can assign it to WS3; slack3 = 1 Assembly Line Balancing Example WS 2 3 2 4 s Ranking: x 7 8 11 9 10 12 1 (1,2,3,6,4,7,5,8,9,10,11,12) 3 2 4 3 1 2 Next assign task 8 5 3 We cannot assign task 8 before WS4 or we violate precedence constraint 6 t8 = 5 so we assign it to WS4; slack4 = 3 3 Next assign task 9 t9 = 1 and we cannot assign it before WS3 or we violate precedence constraint; since slack3 = 1, we assign it to WS3; slack3 = 0 54 Assembly Line Balancing Example WS s 2 3 2 4 4 Ranking: x 7 3 8 11 12 1 (1,2,3,6,4,7,5,8,9,10,11,12) 3 2 4 3 1 2 Next assign task 10 5 9 10 t10 = 4 and we cannot assign it before 3 WS3; Neither WS3 nor WS4 has 6 enough slack, so we must open WS5 and assign it there; slack5 = 11 3 Next assign task 11 t11 = 6 and we cannot assign it before WS5, which has enough slack so we assign it there; slack5 = 5 3 Next assign task 12 t12 = 7 and we cannot assign it before WS5, which does not have enough slack; we must open WS6 and assign it there; slack6 = 5 Assembly Line Balancing Example s Final Solution: x Six workstations Workstation Tasks Idle Time WS 1 1 3 2 2 2,3,4 1 3 5,6,9 0 4 4 4 7,8 3 5 10,11 5 6 12 8 3 2 7 8 5 3 1 2 4 2 3 6 1 11 12 5 3 9 5 6 10 55 Supply Chain Management Manufacturing Strategy Inventory Control/Mgmt Requirements Planning Plant Layout Push/Pull Systems Manufacturing Management Manufacturing Flexibility Factory Flow Dynamics Quality Control Shop Floor Control Aggregate Planning Production Scheduling Inventory Management s We ve looked at many aspects of Inventory Control x x x Single-stage models MRP systems JIT systems 3 3 Kanban CONWIP s So what is left to cover in inventory management? x x We need to look at overall principles for managing inventory Not all real world situations will fit nicely into our model assumptions 56 Inventory Management s Back to the beginning x Why 3 do firms hold inventory? Batching/Economies of Scale Cycle Stock 3 Randomness Safety stocks, safety lead times 3 Speculation on future values Forward buying 3 Quantity discounts induce forward buys Principal Inventory Types s Raw Materials x Decisions affected by batching, randomness of supply/consumption, obsolescence WIP can be found in 3 3 3 3 3 s Work-in-progress (WIP) x Queue waiting for processing Processing at a station Waiting for batch move Movement between stations Waiting to match waiting for another part for assembly 57 Principal Inventory Types s s Finished Goods Inventory (FGI) Reasons for holding FGI x x x x x Customer service Batch production/economies of scale Forecast errors (items that were not sold as expected) Production variability (higher yield or earlier completion than expected) Demand seasonality/Production smoothing Principal Inventory Types s Spare parts x x x x Reasons for holding spare parts Customer service Production lead times for creating parts needed for replacement (hand-in-hand with customer service) Batching/economies of scale 58 Managing Raw Materials s Lack of raw materials can and often does result in extremely costly production stoppage x It is typically far less expensive to hold excess raw materials than to stop production 80/20 rule 20% of items typically account for 80% of value 3 3 3 3 s ABC classification x A items account for 75 80% of inventory investment (5 20%) B items account for 10 15% of inventory investment (10 15%) C items account for 5 10% of inventory investment (80%) Devote most of inventory management efforts to A items, some to B items, and very little to C items Managing Raw Materials s A items are good candidates for JIT supplier shipments x Frequent deliveries in small batches 3 3 3 May impose heavy burden on suppliers JIT suppliers should have visibility of production schedules Vendor certification is important 59 Managing Raw Materials s Purchased components x A items Lot-for-lot (assuming order cost is small) 3 3 Reduces safety stock of expensive items Different from JIT, since deliveries are in concert with production schedule Purchasing lead times should guarantee a high probability of ontime delivery Supplier delivery performance is critical Safety lead times for independent items that must be assembled need to be inflated further to ensure a high probability that all items are on time If we want a 0.95 of on-time delivery for two independent items, each should have a 0.975 probability of being on time x Safety lead times 3 Managing Raw Materials s Our approach for determining order quantities in Chapter 2 assumed that items were independent from each other (EOQ) x x Items from the same supplier often share truck capacity One simple method is to compute the EOQ for each item and then round the corresponding Ti* to the nearest power-of-two 3 Items will be delivered every week, every 2 weeks, every 4 weeks, etc. 60 Managing Raw Materials s Joint order costs x x Items that can share resources also share the associated fixed costs of the resources Suppose we want to order a set of n items, indexed by i, that share an order frequency, and therefore order cost. 3 3 3 3 Item i has constant demand rate i and annual holding cost hi Let AJ denote the joint order cost for ordering the items together If we order items together, they will have the same delivery frequency, i.e., T = Qi/Di for all i. The total annual average cost equation for the items is G (T ) = AJ n h DT + i =1 ci Di + i i T 2 Managing Raw Materials s Joint order costs (cont d) x Differentiating G(T) and setting the result equal to zero gives: T* = x G(T*) x x = i =1 c i D i + 2 A J i =1 h i D i Note that Qi* = DiT* n Let G(Ti*) = i = 1 c i D i + 2 Ai D i h i n n 2 AJ n i =1 i h Di ( ) 3 G(Ti*) is the average annual cost if all items are ordered separately with item i s order cost = Ai 2 AJ i = 1 h i D i < n x We should order the items together only if n i =1 2 Ai h i D i 61 Setting Order Frequencies s s s Consider the assumptions of the EOQ model (i.e., constant, deterministic demand rate) and suppose we have N items that share a delivery system or purchasing department capacity Suppose these are not A items, so we don t necessarily want to have JIT deliveries We would like to minimize inventory investment and would like average order frequency to be less than or equal to some value F (we don t want them delivered too often so we don t demand too much of our purchasing department or Ndelivery system). Annual average inventory investment equals ciQi / 2 i =1 s The delivery frequency of item i equals Di/Qi, so the average N delivery frequency equals 1 Di / Qi N i =1 Setting Order Frequencies s The problem we want to solve is: minimize cQ / 2 i =1 i i N s 1N subject to: Di / Qi F N i =1 One way to solve this is by dualizing the constraint x This means including it in the objective function with a penalty term for violating the constraint 62 Setting Order Frequencies s s s s s N 1 N L = ciQi / 2 + Di / Qi F N i =1 i =1 If we differentiate L wrt Qi, and set the result equal to zero we obtain Qi ( ) = 2 Di / ci 1N For a given we have F( ) = N Di / Qi ( ) i =1 Let be a penalty multiplier for violating the constraint We minimize the Lagrangian function: We follow the following procedure: x x x Choose some and calculate Qi( ) for all i and F( ) If |F( ) F | < stop ( is some tolerance level); otherwise: If F( ) < F, decrease ; If F( ) > F, increase , and repeat the procedure for the new value of . s This can be easily implemented in Excel Managing WIP s WIP management should focus on x x Finding the right utilization levels for workstations Reducing variability 3 Variability of arrivals, processing times, and utilization x x Reducing batch moves Synchronizing flows 63 Managing WIP s Recall that for a G/G/1 queue we had: 2 2 c 2 + c 2 u ca + ce u CT a e te + te WIP = CT ra u + u 2 1 u 2 1 u x x Reducing WIP then focuses on reducing either the arrival variability, the effective processing variability, or the utilization We next consider ways to achieve these objectives Reducing WIP in Queue s Potential options for reducing WIP include: x Equipment changes/additions 3 Adding capacity reduces utilization Pull system obtains same throughput with lower WIP than push system Schedule should always be capacity feasible or congestion occurs, increasing WIP (MRP systems don t ensure this) Increases capacity and, therefore, reduces utilization Reduces effective processing variability at the workstation x Pull systems 3 x Finite-capacity scheduling 3 x Setup reduction 3 3 64 Reducing WIP in Queue s Options for reducing WIP (cont d) x Improving machine reliability/maintenance 3 3 Decreasing MTTR decreases machine effective variability Also increases capacity, reducing utilization Reduces consumption of capacity used by rework, increasing effective capacity Increases utilization Cross-trained workers can move to bottleneck stations under systems that often experience a floating bottleneck, increasing effective capacity and reducing utilization x Improving quality 3 3 x Flexible workers 3 x Classify WIP as ABC and focus on A classified WIP reduction Reducing Wait-for-Batch WIP s High WIP often results from batch processes and moves x Methods for reducing wait-for-batch WIP 3 3 3 Lot splitting Batches processed together do not have to move together Flow-oriented layout create a layout with small move distances and use automated material handling equipment Cart sharing carts used for material transport should take output from parallel production lines, and not be dedicated to a single line 65 Reducing Wait-to-Match WIP s Assembly systems often result in higher WIP because component wait for partners used in assembly x Improving synchronization of matching parts 3 3 3 Pull system synchronizes work releases to fabrication lines with final assembly consumption Common work backlog If multiple fabrication lines have different backlog sequences in CONWIP, production releases are scrambled and don t match between lines. Matching backlog sequences helps synchronize production Balanced batching If certain fabrication lines have high batching requirements, they will have to be started earlier. Reducing setups can also allow for smaller batches. Finished Goods Inventory s Focal points for improving FGI levels: x x x x x x Improve forecasting methods reduces safety stock needs Dynamic lead time quoting quote lead times based on factory conditions, don t use fixed lead time quotes for all orders Cycle time reduction Naturally decreases forecast errors because better information is available as time goes on Cycle time variability reduction reduces required safety lead times Delayed differentiation assemble to order Chase strategy vs. Production smoothing 66 Stocking Spare Parts for Emergencies s Suppose we stock N spare parts for serving emergency repairs, whose demand is stochastic x Assuming continuous review system we will use (Qi, ri) policy s We are concerned with the overall perception of service by our customers x x Let Si denote the service level for part i 1 The average service level is given by S = DTOT where DTOT is the sum over all Di s D S i =1 N ii s Recall that the average inventory in a (Q, r) system equals Q/2 + s = Q/2 + r - , where is the mean demand during lead time Stocking Spare Parts for Emergencies s s We wish to minimize average inventory investment while meeting an overall service level, S, and not exceeding a specified order frequency, F Consider a Type I service level: Si = Gi(ri) = probability demand during lead time is less than or equal to ri (i.e., the probability we do not stock out in a cycle). Our formulation N becomes: minimize ci (Qi / 2 + ri i ) i =1 subject to: 1 N D /Q i =1 i N i =1 i N i F i i 1 DTOT D G (r ) S 67 Stocking Spare Parts for Emergencies s s We dualize both constraints, again using for the first constraint and for the second Without going through the math, if demand follows a Poisson distribution (which is a reasonable assumption for spare parts), we end up with equations for Qi and ri: Qi ( ) = 2 Di / Nci 2 i DTOT ci ri = i + 2 i ln Di s The algorithm is very similar to the one we did for multiple products with a maximum average order frequency, except we now have two parameters to adjust at each step, and Stocking Spare Parts for Emergencies x x x x x x x x We begin by choosing some initial values for and (note that should be large so we don t get a negative term under the square root) Compute Qi( ) and ri( ) for all products for the chosen and N Compute F( ) as before, i.e., F( ) = 1 Di / Qi ( ) N N i =1 Compute S( ) = 1 DiGi (ri ( )) DTOT i =1 If |F( ) F | < and |S( ) S | < stop; otherwise: If F( ) < F, decrease ; If F( ) > F, increase If S( ) < S, increase ; If S( ) > S, decrease 3 Repeat the procedure with the new values of and Note that Gi(ri( )) can be computed in Excel using the function POISSON(ri, i, TRUE), so this whole procedure can be done in Excel 68 Capacity Management s Key Capacity Questions: x x How much capacity should be added and when? 3 Before or after demand exceeds capacity Flexible versus dedicated equipment Existing or new location What kind of capacity should we add? 3 x Where should we add capacity? 3 Economies of Scale s Short-term economies of scale x In the short-run, many costs are fixed (labor, equipment, etc.) 3 Unit cost = Fixed Cost/Throughput + variable unit cost s Medium-term economies of scale x Function of production run length and changeover costs between products 3 Unit cost = Changeover cost/Units per run + running cost per unit s Long-term economies of scale x x Cost per unit of equipment capacity, not related to production Cost of capacity increases at a decreasing rate 69 Modeling Capacity Decisions s s We consider the decision on how to determine production capacity of a single line that produces a single product Assume we have an M-station line x x We ll only consider a linear flow with no rework, although models exist for relaxing these assumptions Each station has a set of technology options corresponding to different equipment types Modeling Capacity Decisions s Each technology option at a station has five associated parameters: x te x x x x ce2 = mean effective processing time = effective SCV of processing times m = number of identical machines at the station k = cost per machine A = fixed cost of the machine option 3 Total cost of installing a technology option = A + km 70 Modeling Capacity Decisions s s Assume we have a minimum required throughput rate, TH, and a maximum allowable cycle time, CT for the line Given all of our technology option parameters and an arrival process for jobs, we can calculate: x x x x u(m) = utilization of a station with m machines s(m) = cycle time at a station with m machines ca2 = SCV of arrivals at the station cd2 = SCV of departures from the station Modeling Capacity Decisions s We use the following formulas we have studied: x x u(m) = [rate/m] [ 2 2 s(m) = ca + ce u 2 t +t m(1 u ) e e 2( m +1) 1] (= CT(G/G/m)) x x ca2 is given for station 1 and ca2(i) = cd2(i 1) for all other stations From Chapter 8 we have (note that Equation 18.4 in the book is slightly incorrect): 2 2 cd (i ) = 1 + [ca (i ) 1][1 u 2 ( m)] + u 2 (m ) 2 [ce (i ) 1] m 71 Modeling Capacity Decisions s s s s Given a technology option and a value of m we can calculate s(m) and check if we are meeting the maximum cycle time allowed If not, we need a different technology option or a greater value of m, or a combination of both How do we obtain a minimum cost, capacity feasible line? For a given technology option, what is the minimum number of machines we need to be capacity feasible? Modeling Capacity Decisions x Capacity feasibility implies u < 1 3 3 u(m) = rate/m < 1 m > rate calculate rate and round up to smallest integer that gives u(m) < 1 x x x x x The minimum cost capacity feasible (MCCF) configuration is the choice of capacity feasible technology options at each station that result in the lowest cost After determining the MCCF configuration, compute its cycle time, s(m) If s(m) is less than or equal to desired maximum cycle time then this is the best option Otherwise, we need to either add machines, change technology options, or modify existing machines within an option Determining the least-cost combination of technology and machines is an extremely difficult problem 72 Modeling Capacity Decisions x x x Suppose we can modify existing machines and add machines Modifying existing machines involves reducing effective SCV through additional replacement parts or more maintenance We can use a greedy heuristic approach 3 3 3 3 3 At each station, consider the cost of adding a machine and the resulting decrease in cycle time for the line At each station, consider the cost of reducing SCV and the resulting decrease in cycle time for the line For each type of change, compute the ratio of the cost increase to the cycle time decrease and choose the option that gives the minimum cost per hour of cycle time reduction Check whether the desired cycle time is met If not, repeat the greedy procedure 73
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Path: UF >> ENC >> 3310 Fall, 2008
Path: UF >> ENC >> 3310 Fall, 2008
Path: UF >> ENC >> 3310 Fall, 2008
Path: UF >> ENC >> 3310 Fall, 2008
Path: UF >> ENC >> 3310 Fall, 2008
Path: UF >> ENC >> 3310 Fall, 2008