Hi, In another post I suggested this equation

e_T_pi(i+1) .. T_pi(i+1)*(m_rc_pb(i+1)+m_ts_pb(i+1)) =e= (m_rc_pb(i+1)*T_ro + m_ts_pb(i+1)*T_to(i+1));

## Search found 84 matches

- 1 year ago
- Forum: Syntax
- Topic: logical equations & binary variables
- Replies:
**18** - Views:
**2924**

- 1 year ago
- Forum: Syntax
- Topic: logical equations & binary variables
- Replies:
**18** - Views:
**2924**

### Re: logical equations & binary variables

Hi, the equation listing and column Listing shows the value on the initial point, not in the solution. The equations are fine.

You should review the SolEQU and SolVAR section.

Bye!

You should review the SolEQU and SolVAR section.

Bye!

- 1 year ago
- Forum: Syntax
- Topic: logical equations & binary variables
- Replies:
**18** - Views:
**2924**

### Re: logical equations & binary variables

Hi, It's similar

eq1.. m =g= m.lo*(1-x);

eq2.. m =l= m.up*(1-x)

Bye!

eq1.. m =g= m.lo*(1-x);

eq2.. m =l= m.up*(1-x)

Bye!

- 1 year ago
- Forum: Syntax
- Topic: logical equations & binary variables
- Replies:
**18** - Views:
**2924**

### Re: logical equations & binary variables

Hi, try these equations

eq1.. m =g= m.lo*x;

eq2.. m =l= m.up*x;

m.lo (lower bound)

m.up (upper bound)

if x=1

m greater than m.lo*1;

m lower than m.up*1

if x=0

m greater than m.lo*0;

m lower than m.up*0;

so, m = 0

Bye!

eq1.. m =g= m.lo*x;

eq2.. m =l= m.up*x;

m.lo (lower bound)

m.up (upper bound)

if x=1

m greater than m.lo*1;

m lower than m.up*1

if x=0

m greater than m.lo*0;

m lower than m.up*0;

so, m = 0

Bye!

- 1 year ago
- Forum: Syntax
- Topic: Synthax of binary variables
- Replies:
**3** - Views:
**744**

### Re: Synthax of binary variables

I think this can work. I do not understand why you use (i + 1) instead of (i) e_T_pi(i+1) .. T_pi(i+1)*(m_rc_pb(i+1)+m_ts_pb(i+1)) =e= (m_rc_pb(i+1)*T_ro + m_ts_pb(i+1)*T_to(i+1)); If you want to avoid the first value of i, you can use this equation. e_T_pi(i)$(ord(i) ne 1).. T_pi(i)*(m_rc_pb(i)+m_t...

- 1 year ago
- Forum: Modeling
- Topic: Ensuring realations between decision variables
- Replies:
**6** - Views:
**1785**

### Re: Ensuring realations between decision variables

Hi Peter, I think that with the auxiliary variable "on(t)" can work. If on(t)=1 x(t) or y(t) = 1 If on(t)=0 x(t) and y(t) = 0 eq1(t).. x(t) + y(t) =l= 1; eq2(t).. 1-on(t) + x(t) + y(t) =g= 1; eq3(t).. on(t) + 1-x(t) =g= 1; eq4(t).. on(t) + 1-y(t) =g= 1; eq5(t,tp)$(ord(t) gt 1 and ord(tp) gt ord(t) a...

- 1 year ago
- Forum: Syntax
- Topic: Synthax of binary variables
- Replies:
**3** - Views:
**744**

### Re: Synthax of binary variables

e1_npb(i) .. m_pb_rc(i) =l= (1-npb(i) )*M;

e2_npb(i) .. m_pb_ts(i) =l= (1-npb(i) )*M;

With these equations when npb(i) take the value of 1, m_pb_rc (i) and m_pb_ts (i) are 0.

I can not understand the idea of the problem but this may help you.

Best

e2_npb(i) .. m_pb_ts(i) =l= (1-npb(i) )*M;

With these equations when npb(i) take the value of 1, m_pb_rc (i) and m_pb_ts (i) are 0.

I can not understand the idea of the problem but this may help you.

Best

- 1 year ago
- Forum: Syntax
- Topic: logical condition with equation definitions and variables
- Replies:
**4** - Views:
**966**

### Re: logical condition with equation definitions and variables

Hi, the equations can not vary during the resolution. To relate variables or take logical decisions you will need to use binary variables.

Best!

Best!

- 1 year ago
- Forum: Syntax
- Topic: logical condition with equation definitions and variables
- Replies:
**4** - Views:
**966**

### Re: logical condition with equation definitions and variables

Hi, logical conditions apply to the equations.

e_charge(i+1)$(m_rc_ts(i+1) > 0).. dec(i+1) =e= 1;

Best

e_charge(i+1)$(m_rc_ts(i+1) > 0).. dec(i+1) =e= 1;

Best

- 1 year ago
- Forum: Syntax
- Topic: Questions about Summation
- Replies:
**1** - Views:
**834**

### Re: Questions about Summation

Hello, you have many alternatives: sets t /t1*t10/ ; (1) sum(t$(ord(t) ge 2),x(t)) (2) sum(t$(ord(t) ne 9 and ord(t) ne 10),x(t)) =e= 1 sum(t$(ord(t) ge 2 and ord(t) le 7),x(t)) =e= 1 using subsets: sets t /t1*t10/ sum1(t) /t2*t10/ sum2(t) /t1*t8/ sum3(t) /t2*t7/ ; (1) sum(t$sum1(t),x(t)) (2) sum(t$...