\[st(j+1,p,m,w,c,h) \ge x(j+1,p,m,w,c,h) \cdot ft(j,p,d,ww,z,h)\]

\[ \begin{eqnarray} ft &\in &\;\;\mathbb{Z}_0 &\text{[= finish time]}\\ st &\in &\;\;\mathbb{Z}_0 &\text{[= start time], and}\\ x &\in &\{0,1\} &\text{[= decision variable if operation j+1}\\ & & &\quad\;\text{of part p on machine m with worker w}\\ & & &\quad\;\text{in cell c and horizon h is done or not]}\\ \end{eqnarray}\\ \]

\[\ldots\text{all are decision variables.}\]

asked 04 Jul '12, 02:56

girl's gravatar image

accept rate: 0%

edited 06 Jul '12, 06:06

fbahr's gravatar image

fbahr ♦

very efficient, @girl, but too efficient for me. could you give a little more context, tell us what kind your variables are (continuous, integer, binary?), where does the constraint come from, etc.? Thanks.

(04 Jul '12, 03:53) Marco Luebbecke ♦

oh yes,x is binary and st and ft are integer.ft=finish time.st=start time.x is decision variable if operation j+1 of part p on machine m with worker w in cell c and horizon h is done or not

(04 Jul '12, 04:42) girl

@fbahr thanks for editting...

(04 Jul '12, 05:58) girl

st(j+1,p,m,w,c,h)≥⋅ft(j,p,d,ww,z,h) - M(1-x(j+1,p,m,w,c,h)) does this solve the problem? usually this arises in Job Shop Scheduling models if I am not mistaken. But that causes numerical instabilities,


answered 04 Jul '12, 05:51

ShahinG's gravatar image

accept rate: 0%

i think it will work,thanks a lot

(04 Jul '12, 06:01) girl

For performance (and stability) reasons, try to pick M as small as is safely possible in each constraint (something like worst-case makespan).

(04 Jul '12, 17:21) Paul Rubin ♦♦

yea im working on somehow scheduling model,thanks for ur help

(05 Jul '12, 16:08) girl
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Asked: 04 Jul '12, 02:56

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Last updated: 06 Jul '12, 06:06

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