Approaches to Line balancing Optimal Solutions Active Learning Module 3 Dr. César O. Malavé Texas A&M University

Background Material Modeling and Analysis of Manufacturing Systems by Ronald G. Askin, Charles R. Standridge, John Wiley & Sons, 1993, Chapter 2. Manufacturing Systems Engineering by Stanley B. Gershwin, Prentice – Hall,1994, Chapter 2. Any good manufacturing systems textbook which has detailed explanation on reliable serial systems.

Lecture Objectives At the end of this module, the students should be able to Explain the Optimal Solutions approach to line balancing Find the optimal solutions to line balancing problems using the above technique.

Time Management 3Assignment 10Team Exercise 4Practical Issues 50 MinsTotal Time 15Tree Exploration 10Tree Generation 5Readiness Assessment Test (RAT) 3Introduction

Readiness Assessment Test (RAT) TaskTime (Seconds)Predecessor A45 - B11A C9B D50A E15D F12C G C H E I E J8F, G, H, I K9J Draw the precedence diagram for the tasks shown below

RAT – Solution A B D C E F G H I JK Precedence Chart :

Optimal Solutions Imagine a decision tree containing all possible sequences of tasks that obey the precedence constraints. “0” node is called root of the tree, terminal nodes at the other end are called the leaves. A unique path leads from root to each leaf. Each leaf represents a complete sequence. Each feasible sequence is represented by exactly one leaf. Best solution is found by examining every path.

Tree Generation Backtracking – Allows both symmetric exploration of the tree and efficient storage of our location and history during exploration. General applicability is to an environment in which N sequential decisions are to be made. In generating the tree of possible sequences, the order of performing the assembly tasks is important rather than the workstations. Tree initially grows by selecting first alternative at each stage until there is a complete assembly sequence

Tree Generation – Cont… Backtrack until we reach the first node of the unexplored branch. Continue moving forward, decision by decision, as far as possible. Process continues until all possible leaves are explored. A task is fittable if it satisfies three conditions. Task must fit in the remaining idle time of the station. It must be currently unassigned. All its predecessors are assigned.

Tree Generation Algorithm 0. Input Bound and task data 1. Setup k = 1, p = 0 c k = C, B = 0 i’ exist ? 2. Select new task Find i’ = lowest i i fittable, i>B i > i* unless new station. 3. Assign task A i* = k, p = p+1 c k = c k – t i* TA p = i*, B = 0 p = N ? ck = C ? No i* = i’ Yes 6. Sequence complete Save if best solution Yes p = 0 ? Yes i* > 0 ? No Stop Enumeration Complete Yes No 7. Backtrack to B. (Remove B from Station K) If c k = C, k = k-1 B = Ta p, A B = 0 c k = c k +t B p = p-1, i* = 0 No 4. Open New Station k = k+1 c k = C Yes

Tree Exploration Creating only those new nodes required to find or disprove the existence of a better solution. In practice, we need not create each node of the tree. At any partial sequence node, if we can establish that all completions of this partial sequence are non-optimal; then we will start backtracking immediately instead of completing a sequence. Two Rules – Problem Structure Rules & Fathoming Rules.

Problem Structure Rules Basic Optimality Principle – Never close a workstation while “fittable” tasks remain. Open a new workstation once when necessitated by time. Long task becomes a station only when no other task can feasibly fit in the same station. A lower bound on the number of workstations needed is useful in determining if we have an optimal solution.

Fathoming Rules These rules help us to prune the tree rapidly. Allows us to implicitly consider all leaves without explicitly enumerating many of them. Rules only be checked when a new workstation must be opened. Rule 1 : Task Dominance Suppose Task i can be feasibly replaced by a longer task j and all successors must also follow j. If we substitute, we reduce the workload without losing any possible sequence completions.

Fathoming Rules – Cont… Rule 2 : Station Dominance Suppose we try to form a station that is identical to a “first” station that was explored earlier. As no new sequences can be considered, new node is fathomed. Rule 3 : Solution Dominance We know that K 0 gives a lower bound on optimal number of solutions. We can stop once we have got a solution as that of lower bound

Fathoming Rules – Cont… Rule 4 : Bound Violation Suppose we have K workstations. This is optimal unless a K – 1 solution is found. Let A i be the station to which task i is assigned. If we require at most K – 1 stations, the upper bound is given by Nodes containing at least one of A i outside these bounds can be pruned from the tree.

Fathoming Rules – Cont… Rule 5 : Excessive Idle Time Whenever cumulative idle time exceeds (K – 1)C –  i t i,we may not fathom the partial sequence All the above mentioned fathoming rules are adapted from FABLE – Fast Algorithm for Balancing Lines Effectively described by Johnson

Optimal Solutions - Example TaskActivityAssembly Time Immediate Predecessor aInsert Front Axle / Wheels 20- bInsert Fan Rod6a cInsert Fan Rod Cover5b dInsert Rear Axle / Wheels 21- eInsert Hood to Wheel Frame 8- fGlue Windows to top35- gInsert Gear Assembly15c, d hInsert Gear Spacers10g iSecure Front Wheel Frame 15e, h jInsert Engine5c kAttach Top46f, i, j lAdd Decals16k

Example Solution Product Structure Rules : 1. Task Time Augmentation Longest Station = t k = 46 Shortest Station = t c = 5 t k + t c  C. So task would require own station. 2. Solution Lower Bound Lower Bound K 0 = 3 Only one task exceeds C/2 and two tasks exceeding C/3. No better solution exists now..

Example Solution – Cont… Fathoming Rules : 1.Task Dominance Several Tasks can be replaced. Task Pairs satisfying the relationship are (j, d), (j, e), (j, f), (j, g), (j, h), (j, i), (i, f), (e, a), (e, h), (e, g), (e, d) 2.Task Bound Violation Assume current best solution has four stations. The natural ordering being (a, b, c, d, e), (f, g, h), (i, j, k), (l) Still need to search for three-station solution. Bound violations will be useful.

Example Solution – Tree Generation 0 1 a 2 b 3 c 4 d 5 e 13 g 6 j 7 f 8 g 9 h 10 g 11 h 1212 i 14 e 15 f 16 h 17 i 18 j 19 k 20 l

Team Exercise Consider the five assembly tasks given below Evaluate each sequence in the tree to determine the number of workstation required. List all optimal sequences TaskTimeImmediate Predecessor a40- b75a c50a d35c e80d

Exercise Solution 0a bcde cbde dbe eb WorkstationOptimal Yes

Practical Issues C has been fixed so far but in reality we only forecast demand. C should be set such that  i t i / C is an integer or jus less than an integer. Randomness in performance times exists. Let  i 2  Variance of performance times.  ij  Correlation between tasks i and j. s k  Random variable for the time required by station k for a cycle Sk  Set of tasks assigned to station k.

Practical Issues – Cont… From basic statistics we know When added to the station, the task must satisfy

Assignment Solve by implicit enumeration technique and find the optimal solution g15h e24g d26f c24e c11d a, b10c -5b -3a PredecessorsTimeOperation