CHAPTER 10 Sup. (Acceptance Sampling) Statistical Process Control – “Sampling to determine if process is within acceptable limits” Learned previously Acceptance Sampling – “Sampling to accept or reject the immediate lot at hand” Learning now Purpose is to avoid 100% inspection (cost) Types of Sampling Plans 1) Single Sampling 2) Doubling Sampling 3) Sequential Sampling 1) Single Sampling Sample n items. If # of defects ≤ c then accept the lot. If # of defects > c then reject the lot. 2) Doubling Sampling Sample n 1 items. If # of defects < c 1 then accept the lot. If # of defects > c 2 then reject the lot. If c 1 < # of defects < c 2, then sample n 2 items and only accept the lot if the # of defects in (n 1 + n 2 ) < c 3. 3) Sequential Sampling After each ind. item is sampled, decide to: a) Accept the lot b) Reject the lot c) Continue sampling Quality and Risk Decisions Two levels of quality are considered: 1) Acceptance Quality Level 2) Lot Tolerance Proportion Defective 1) Acceptance Quality Level (AQL) “The percentage of defects the buyer is willing to accept” (lower limit) Producer’s Risk = α α = “probability that the lot containing defectives < AQL will be rejected” Is a Type I Error 2) Lot Tolerance Proportion Defective (LTPD) “The percentage of defects the buyer is not willing to accept” (upper limit) Consumer’s Risk = β β = “probability that the lot containing defectives > LTPD will be accepted” Is a Type II Error Notation for Sampling Plans N = number of items in a lot n = number of items in a sample M = number of defectives in a lot α = probability of rejecting good lots β = probability of accepting bad lots c = maximum level of defects X = number of defectives in the sample p = proportion of defectives in the lot p 0 = AQL p 1 = LTPD Average Outgoing Quality (AOQ) “Expected proportion of defects that the plan will allow to pass” Assumes rejected lots are 100% inspected Average Outgoing Quality Level (AOQL) “The maximum possible proportion of defects for the AOQ” Maximum value on the AOQ curve Replacing Defective Items for AOQ If replacing the defective items... 1) All defects in the lot are replaced with good items if the lot is rejected 2) All defects in the sample are replaced with good items if the lot is accepted If not replacing the defective items... 1) All defects in the lot are removed if the lot is rejected 2) All defects in the sample are removed if the lot is accepted CHAPTER 17 (Project Management) P1 Projects Project – “Series of related jobs usually directed toward some major outputs and requiring significant time to perform” Project Management – “Management activities including planning, directing and controlling resources to meet technical, cost and time constraints of a project” Projects are complex because: 1) Large number of activities 2) Precedent relationship: 3) Limited time to complete Work Breakdown Structure (WBS) Lvl. Lvl. ID Activity 1 1 Develop new Windows OS Write the code Ensure compatible w/ Compatible with XP Compatible with Vista Able to import files Critical Path Method (CPM) Is a network technique (nodes and paths) Critical Path = “path w. longest duration” (has 0 slack on all activities) Examples: Start – A – B – C – D – End or A – B – C – D Six Steps of the CPM 1) Define project and prepare the WBS 2) Develop relationships among activities (define precedents and dependents) 3) Draw the network connecting all act. 4) Assign time / cost est. to each act. 5) Compute the Critical Path 6) Use the network to help plan, schedule, monitor and control the project Planning vs. Scheduling Planning – “Determining what must be done and which tasks precede others” Scheduling – “Determining when tasks must be completed; when they can and when they must be started; which tasks are critical to the timely consumption of the project; which tasks have slack” Six Steps of Project Scheduling 1) List activities defined in the WBS 2) Identify logical sequence of activities 3) Identify resources needed for each act. 4) Identify duration of each activity 5) Develop schedule using PERT/CPM 6) Modify as needed Critical Path Analysis ES = Max EF of predecessors EF = ES + activity time LS = LF – activity time LF = Min LS of successors Slack = LS – ES or LF - EF ESEF LS LF 2A) Graphical Method - Network 1 st : Do the forward pass (ES and EF) 2 nd : Do the backward pass (LS and LF) We use activity-on-node (AON) network Methods for Determining the Outputs 1) Tabular Methods 2) Graphical Methods 2A) Network 2B) Gantt Chart Project Planning – Inputs and Outputs Inputs include: Activity completion times Activity precedence relationships Outputs include: Network representation of a project Project completion time Critical paths and activities Activity and path slack ES, EF, LS, LF for each activity CHAPTER 17 (Proj. Mgmt.) P2 Program Evaluation and Review Technique PERT – “Used for scheduling and control of large projects” Probabilistic Network Analysis “Determines the probability that project is completed within a specified time” Key notation: μ = project mean time (t e ) σ = project standard deviation x = proposed project time Z = x is Z st. dev.’s away from the mean Project Time “Sum of the duration of critical activities” According to the Central Limit Theorem, project time is normally distributed CHAPTER 17 (Proj. Mgmt.) P3 - Crashing Crashing “Shortening activities in order to shorten the project duration and thus cut costs” Options for crashing: 1) Raise direct costs (variable costs) 2) Lower indirect costs (fixed / OH costs) Information Needed for Crashing 1) Regular time and crash time estimates for each activity 2) Regular cost and crash cost estimates for each activity 3) List of activities on the Critical Path Approaches to Crashing 1) Heuristic Method 2) Linear Programming (not covered) Five Steps to the Heuristic Method 1) Determine the critical path, its length and the length of other paths 2) Rank critical activities in order of crashing cost per day (lowest to highest) 3) Determine the number of days for which each activity can be crashed 4) Begin to shorten the project one day at a time. Check after each reduction to see if another path becomes critical. 5) Crash until no additional crashing is cost-effective for the project. More on Crashing... Time Available For Crashing = Normal Time – Expedited Time Crash Cost (per unit of time) = (Expedited Cost – Normal Cost) / (Normal Time – Expedited Time) Operating Characteristic (OC) Curve “A probability curve showing the probability of accepting lots with various proportions of defects” Depends on four variables: n, c α, β α = P (x > c | True proportion of defectives in lot = AQL) β = P (X ≤ c | True proportion of defectives in the lot = LTPD) If p = AQL then α = 1 – prob. of accept. If p = LTPD then β = prob. of accept. If Defects are not Replaced p = true proportion of defects in the lot Pa = probability of accepting the lot N = lot size n = sample size If Defects are Replaced p = true proportion of defects in the lot Pa = probability of accepting the lot N = lot size n = sample size Beta Distribution “Probability distribution used for the duration of each activity” Three time estimates are used: 1) Most Likely Time = t m (mode) 2) Optimistic Time = t o 3) Pessimistic Time = t p Mean of Beta Distribution = t e Standard Deviation of Beta Dist. = σ Path Time Variability Variance of path duration = σ path Example if path is A – B – C: Variance of Path Duration = (Var. of Duration of A) + (Var. of Duration of B) + (Var. of Duration of C) Expected Path Time Expected path time = the path mean

Assumptions of Learning Curves 1) Time required to complete a task will be less each time the task is undertake 2) The time per unit will decrease at a decreasing rate 3) The reduction in time must follow a predictable pattern Calculating Unit Times (factor approach) Can also use the factor table 1 st – Select unit number on the rows 2 nd – Select LP on the columns Expected time for nth unit = T n T n = T 1 x “unit time factor” Expected total time for all n units = ∑T n ∑T n = T 1 x “total time factor” Average time for n units Avg. = ∑T n / n Estimating Unit Times When managers believe that the time for producing the 1 st unit is not representative of the rest of production... We must estimate T 1 : T 1 = (T n / “unit time factor” for n) Estimating LP (using y = a + bx) T n = Time for the n th unit (x-axis) n = number of units produced (y-axis) Use regression analysis on the function: ln(Y n ) = ln(a) + b[ln(b)] to find ‘a’ and ‘b’ Estimating LP (using y = ax b ) T n = Time for the n th unit (x-axis) n = number of units produced (y-axis) Use the linear function: Y n = an b to find ‘b’ Calculating Unit Times (formula approach) T n = Time for the n th unit T 1 = Time for the 1 st unit b = [ln(LP/100)]/ln2 Determining # of Repetitions To determine the minimum number of repetitions to achieve a given time standard we can... 1) Use the factor table to find the # of units that most closely matches a unit of time of T n / T 1 2) Use the formula: Calculating LP LP = Learning curve percentage / rate Y(n) = Production time for unit # “n” Example: LP = 80% Means... The time it takes to produce the (2n) th unit is 80% of the time it takes to produce the n th unit Therefore... Lower LP = better CHAPTER 10 Sup. (Learning Curves) Learning Curves “Explains why direct labour hours required to produce one unit of output decline as the cumulative number of units produced increases” AKA – Improvement Curve Experience Curve “Explains why marginal production costs decline as the cumulative number of units produced increases” AKA – Progress Function Calculating Unit Times (factor approach) Can also use the factor table 1 st – Select unit number on the rows 2 nd – Select LP on the columns Expected time for nth unit = T n T n = T 1 x “unit time factor” Expected total time for all n units = ∑T n ∑T n = T 1 x “total time factor” Average time for n units Avg. = ∑T n / n