Balancing the flow. Part 3. Two-Machine Flowshop SCHEDULING SCHEDULING MODELS. Two-Machine Flowshop Two-Machine Job Shop Extensions

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1 PRODUCTION PLANNING AND SCHEDULING Part 3 Andrew Kusiak 3 Seamans Center Iowa City, Iowa Tel: Fax: andrew-kusiak@uiowa.edu SCHEDULING Assignment of operations to s in time PS RP Balancing Scheduling SCHEDULING ODELS Two-achine Flowshop Two-achine Job Shop Extensions Two-achine Flowshop Flow of parts odified Johnson s Algorithm inimization of ax Flow (in Fax) What does inimization of the ax Flow (in Fax) mean Balancing the flow Two-achine Flowshop odel odified Johnson s Algorithm Step. Set k =, l = n. Step. For each part, store the shortest processing time and the corresponding number. Step 3. Sort the resulting list, including the triplets "part number/time/ number" in increasing value of processing time. Step 4. For each entry in the sorted list: IF number is, then (i) set the corresponding part number in position k, (ii) set k = k +. ELSE (i) set the corresponding part number in position l, (ii) set l = l -. END-IF. Step 5. Stop if the entire list of parts has been exhausted.

2 Example: Two-achine Flowshop odel Schedule 7 parts Two operations per part, each performed on a different Part Number Processing t ij of Part i on achine j i j= j = in For each part calculate in {ti, ti} in processing time calculated Part Number min {t i, t i } achine Number Triplets ordered on the processing time (4,, ) (5,, ) (,, ) (3, 3, ) (, 3, ) (6, 4, ) (7, 6, ) (4,, ) (5,, ) (,, ) (3, 3, ) (, 3, ) (6, 4, ) (7, 6, ) Optimal Schedule: (4,, 6, 7,, 3, 5) achine achine Think of incorporating in Johnson s Algorithm: Due dates Precedence Constraints Two-achine Job Shop Use of Johnson s algorithm by dividing parts into four types Two-achine Job Shop Type A: parts to be processed only on. D A B C Type B: parts to be processed only on. Type C: parts to be processed on both s in the order,. Type D: parts to be processed on both s in the order,.

3 Step. Step. Step 3. Step 4. Step 5. Schedule the parts of type A in any order to obtain the sequence SA. Schedule the parts of type B in any order to obtain the sequence SB. Scheduling the parts of type C according to Johnson s algorithm produces the sequence SC. Scheduling the parts of type D according to Johnson s algorithm produces the sequence SD (Note that is the first, whereas is the second one). Construct an optimal schedule as follows: The Optimal Schedule (SC, SA, SD ) (SD, SB, SC ) Example: Two-achine Job Shop First Second Processing Order and First achine Second achine Type A parts: Parts 7 and 8 are to be processed on alone. An arbitrary order SA = (7, 8) is selected. Type B parts: Parts and 0 require alone. Select an arbitrary order SB = (, 0). Type A: parts to be processed only on. Type B: parts to be processed only on. Processing Order and First achine Second achine Type C parts: Parts,, 3, and 4 require first and then. SC =(4, 3,, ) Type C Parts Number achine achine in {ti, ti} Processing Order and First achine Second achine Type D parts: Parts 5 and 6 require first and then. SD = (5, 6) Type D Parts Number achine achine in {ti, ti} 5 6 st() 6 5 st() 6 5 st () 5 6 st () Optimal Schedule Partial Schedules Optimal Schedule : (SC, SA, SD) : (SD, SB, SC) SA = (7, 8) SB = (, 0) SC = (4, 3,, ) SD = (5, 6) : (4, 3,,, 7, 8, 5, 6) : (5, 6,, 0,4, 3,, )

4 Optimal Schedule : (4, 3,,, 7, 8, 5, 6) : (5, 6,,0,4, 3,, ) Gantt Chart of the Optimal Schedule in F max = 46 for the optimal schedule What is the main difference between Two flow shop schedule and Two job shop schedule both solved with Johnson s algorithm Answer: The Sequence of Operations Two flow shop schedule Two job shop schedule SAE on the two s DIFFERENT on each Special Case of Three-achine Flow Shop odel n n Either in {ti} ax {ti} i= i= n n or in {ti3} ax{ti} i= i= ai = ti + ti bi = ti + ti3 Example: Special Case of Three-achine Flow Shop odel Scheduling Data Actual Processing s t i t i t i3 Part Constructed Processing s Actual Processing s a i b i First achine Second achine ti ti ti3 Part in {ti } = 3; ax {ti} = 3; and in {ti3} = i= i= i= The first condition is met 6 6 in {ti} = 3 3 = ax {ti } i = i=

5 Number achine achine in {ti, ti} (, 4, 5,, 3, 6) Gantt Chart of the Optimal Solution Optimal Solution (, 4, 5,, 3, 6)