R help - Aug 2012 - Solving binary integer optimization problem

Hi,

I am new to R for solving optimization problems, I have set of communication
channels with limited capacity with two types of costs, fixed and variable
cost. Each channel has expected gain for a single communication.
I want to determine optimal number of communications for each channel
maximizing ROI)return on investment) with overall budget as constraint.60000
is the budget allocated.

Channel	Fixed_Cost	Variable_Cost	Capacity	Expected_Gain
C1	400	2.5	5000	0.25
C2 	10000	0	30000	0.3
C3	4000	0.15	20000	0.15
C4	2000	2	10000	0.36
C5	100	3	4000	0.09

Channel_Select
<-data.frame(Channel=c('c1','c2','c3','c4','c5'),
 	                      Fixed_Cost=c(400,5000,4000,2000,100), 
				    Variable_Cost=c(2.5,0,0.15,2,3),
				    Capacity=c(5000,30000,20000,10000,4000),
				    Expected_gain=c(0.25,0.3,0.15,0.36,0.09))


Let  x1,x2,x3,x4,x5 are the decision variables  for c1,c2,c3,c4,c5 channel
and z1,z2,z3,z4,z5 are the indicator binary variables if channel has
allocated communication if any.

max((0.25*x1+0.30*x2+0.15*x3+0.36*x4+0.09*x5)-(2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5)/(
2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5))

Constraints:
(2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5))
<60000 ##Budget Constraint
x1-5000*z1<=0
x2-30000*z2<=0
x3-20000*z3<=0
x4-10000*z4<=0
x5-4000*z5<=0
x1 >= 200
x2 >= 100
x3>=100
x4>=500
x5>=0

I had tried lp function from lpsolve but not able to set objective.in for
objective function. Any help or hint is welcomed!




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On Fri, Aug 10, 2012 at 05:41:07AM -0700, aajit75 wrote:
> Hi,
> 
> I am new to R for solving optimization problems, I have set of
communication
> channels with limited capacity with two types of costs, fixed and variable
> cost. Each channel has expected gain for a single communication.
> I want to determine optimal number of communications for each channel
> maximizing ROI)return on investment) with overall budget as
constraint.60000
> is the budget allocated.
> 
> Channel	Fixed_Cost	Variable_Cost	Capacity	Expected_Gain
> C1	400	2.5	5000	0.25
> C2 	10000	0	30000	0.3
> C3	4000	0.15	20000	0.15
> C4	2000	2	10000	0.36
> C5	100	3	4000	0.09
> 
> Channel_Select
<-data.frame(Channel=c('c1','c2','c3','c4','c5'),
>  	                      Fixed_Cost=c(400,5000,4000,2000,100), 
> 				    Variable_Cost=c(2.5,0,0.15,2,3),
> 				    Capacity=c(5000,30000,20000,10000,4000),
> 				    Expected_gain=c(0.25,0.3,0.15,0.36,0.09))
> 
> 
> Let  x1,x2,x3,x4,x5 are the decision variables  for c1,c2,c3,c4,c5 channel
> and z1,z2,z3,z4,z5 are the indicator binary variables if channel has
> allocated communication if any.
> 
>
max((0.25*x1+0.30*x2+0.15*x3+0.36*x4+0.09*x5)-(2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5)/(
> 2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5))
> 
> Constraints:
> (2.5*x1+0*x2+0.15*x3+2*x4+3*x5+400*z1+10000*z2+4000*z3+2000*z4+100*z5))
<> 60000 ##Budget Constraint
> x1-5000*z1<=0
> x2-30000*z2<=0
> x3-20000*z3<=0
> x4-10000*z4<=0
> x5-4000*z5<=0
> x1 >= 200
> x2 >= 100
> x3>=100
> x4>=500
> x5>=0
> 
> I had tried lp function from lpsolve but not able to set objective.in for
> objective function. Any help or hint is welcomed!
Hi.

Your objective function is not a linear function, since it contains
a ratio of two linear functions. So, the current form of the problem
cannot be formulated as an input to lpSolve.

Petr Savicky.