Solve Control for Target

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Justinn
Posts: 2
Joined: Wed May 11, 2022 7:04 pm

Solve Control for Target

Postby Justinn » Tue Jul 12, 2022 8:25 pm

I am interested in the Solve Control for Target procedure. However, I've never had luck navigating the error messages and can tell I'm doing something wrong. Is there an example workfile anywhere that I can use just so I can better understand and implement the procedure? Such as the workfile for this- https://eviews.com/EViews12/ev12models_n.html

If not, would someone be able to simulate some data with a working model that successfully demonstrates the procedure? I would be really grateful if so.

Regards,
Justin

EViews Matt
EViews Developer
Posts: 528
Joined: Thu Apr 25, 2013 7:48 pm

Re: Solve Control for Target

Postby EViews Matt » Wed Jul 13, 2022 10:34 am

Hello,

Here's a trivial artificial model with an example of using the control procedure over a single variable.

Code: Select all

wfcreate u 100
series inA = 1 + @abs(@sin(@obsid / 25)) / 2 + rnd / 8
series inB = rnd * @sin(@obsid / 15)
series out
model m
m.append out = inA + inB
m.solve

series out_target = out_0
smpl 81 100
out_target = out_0 * (1 + (@obsid - 80) / 40)
smpl @all
show m
m.control(create) inA out out_target

If you share the error message you're encountering I may be able shed some light on what can be done to resolve them. There's quite a variety of errors that could occur, from structural mistakes (e.g. trying to use an exogenous variable to control an endogenous variable that it doesn't influence) to numerical issues (e.g. failing to find control values that satisfy the target series).

Justinn
Posts: 2
Joined: Wed May 11, 2022 7:04 pm

Re: Solve Control for Target

Postby Justinn » Wed Jul 13, 2022 5:14 pm

Matt, thank you very much for the response! This illuminates the procedure for me now that I can follow a worked example.

I tend to encounter this error, "An equal number of control, target, and trajectory variables must be specified." I think part of that is I wasn't understanding the concept of the trajectory variable.

Thanks again,
Justin


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