lingo例题讲解1

MODEL: ! Assembly line balancing model; ! This model involves assigning tasks to stations
in an assembly line so bottlenecks are avoided. Ideally, each station would be assigned an equal amount of work.; SETS: ! The set of tasks to be assigned are A through K, and each task has a time to complete, T; TASK/ A B C D E F G H I J K/: T; ! Some predecessor,successor pairings must be observed(e.g. A must be done before B, B before C, etc.); PRED( TASK, TASK)/ A,B B,C C,F C,G F,J G,J J,K D,E E,H E,I H,J I,J /;
! There are 4 workstations; STATION/1..4/; TXS( TASK, STATION): X; ! X is the attribute from the derived set TXS that represents the assignment. X(I,K) = 1 if task I is assigned to station K; ENDSETS DATA: ! Data taken from Chase and Aquilano, POM; ! There is an estimated time required for each task: A B C D E F G H I J K; T = 45 11 9 50 15 12 12 12 12 8 9; ENDDATA ! The model; ! *Warning* may be slow for more than 15 tasks; ! For each task, there must be one assigned station;
The times to complete the various tasks are given in the table below: Task: A B C D E F G H I J K Minutes:45 11 9 50 15 12 12 12 12 8 9 We need to find an assignment of tasks to workstations that minimize the assembly line抯 cycle time.
@FOR( TASK( I): @SUM( STATION( K): X( I, K)) = 1); ! Precedence constraints; ! For each precedence pair, the predecessor task I cannot be assigned to a later station than its successor task J; @FOR( PRED( I, J): @SUM( STATION( K): K * X( J, K) - K * X( I, K)) >= 0); ! For each station, the total time for the assigned tasks must be less than the maximum cycle time, CYCTIME; @FOR( STATION( K): @SUM( TXS( I, K): T( I) * X( I, K)) <= CYCTIME); ! Minimize the maximum cycle time; MIN = CYCTIME; ! The X(I,J) assignment variables are binary integers; @FOR( TXS: @BIN( X)); END
data: capacity=1,1,1,1,1; demend=1,1,1,1,1; cost=8 6 10 9 12
9 12 7 11 9 7 435 8 9 5 8 11 8 4 6 7 5 11; Enddata end
练习
For our example, we have eleven tasks (A through K) to assign to four workstations (1 through 4). The task precedence diagram looks like this:
例题讲解
Model: title Assignment Problem; sets: worker/w1,w2,w3,w4,w5/:capacity; job/j1,j2,j3,j4,j5/:demend; routes(worker,job):cost,volume; endsets min=@sum(routes:cost*volume); @for(job(j):@sum(worker(i):volume(i,j)) =demend(j)); @for(worker(i):@sum(job(j):volume(i,j)) =capacity(i));
Fra Baidu bibliotek
The Sets
We have two primitive sets in this model総he tasks (TASK) and the workstations (STATION).
From these two primitive sets, we create two derived sets.