# any body can help me to change this nl constraint to linear?

 1 $st(j+1,p,m,w,c,h) \ge x(j+1,p,m,w,c,h) \cdot ft(j,p,d,ww,z,h)$ $\begin{eqnarray} ft &\in &\;\;\mathbb{Z}_0 &\text{[= finish time]}\\ st &\in &\;\;\mathbb{Z}_0 &\text{[= start time], and}\\ x &\in &\{0,1\} &\text{[= decision variable if operation j+1}\\ & & &\quad\;\text{of part p on machine m with worker w}\\ & & &\quad\;\text{in cell c and horizon h is done or not]}\\ \end{eqnarray}\\$ $\ldots\text{all are decision variables.}$ asked 04 Jul '12, 02:56 girl 31●1●5 accept rate: 0% fbahr ♦ 4.6k●7●16 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

 4 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 281●2●13 accept rate: 0% i think it will work,thanks a lot (04 Jul '12, 06:01) girl 1 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
 toggle preview community wiki

By Email:

Markdown Basics

• *italic* or _italic_
• **bold** or __bold__
• image?![alt text](/path/img.jpg "Title")
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported

Tags: