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I have a following generalized linear model with $\log$ as a link function.

Call:
glm(formula = time ~ I(1/nprocs) + ndoms + nDOF + nDOF:ndoms + 
    I(nDOF^2):ndoms + I(1/nprocs):nDOF + I(1/nprocs):nDOF, 
    family = gaussian(link = "log"), 
    data = dataFact)

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.10398  -0.02385  -0.01383   0.01055   0.09257  

Coefficients:
                   Estimate Std. Error t value Pr(>|t|)    
(Intercept)      -4.082e+00  9.860e-02 -41.401  < 2e-16 ***
I(1/nprocs)       8.230e-01  8.693e-02   9.468  < 2e-16 ***
ndoms             4.169e-04  1.210e-04   3.447 0.000682 ***
nDOF              1.693e-04  7.458e-06  22.706  < 2e-16 ***
ndoms:nDOF        2.288e-07  1.692e-08  13.524  < 2e-16 ***
ndoms:I(nDOF^2)  -1.120e-11  6.918e-13 -16.185  < 2e-16 ***
I(1/nprocs):nDOF -4.167e-05  6.739e-06  -6.183 3.08e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 0.0009785681)

    Null deviance: 12.39092  on 223  degrees of freedom
Residual deviance:  0.21235  on 217  degrees of freedom
AIC: -907.61

But I'm not sure, what's the resulting mathematical formula now. Is it correct to just exponentiate the whole expression like this?

$$ \operatorname{time} = \\ \small\exp \left[ c_1 \!+\! \frac{c_2}{\scriptsize\operatorname{nprocs}}\! +\! c_3 {\scriptsize(\operatorname{ndoms})}\! +\! c_4 {\scriptsize(\operatorname{nDOF})}\! +\! c_5 {\scriptsize(\operatorname{nDOF})}{\scriptsize(\operatorname{ndoms})}\! +\! c_6 {\scriptsize(\operatorname{nDOF})}^2 {\scriptsize(\operatorname{ndoms})}\! +\! c_7 \small\frac{\scriptsize\operatorname{nDOF}}{\scriptsize\operatorname{nprocs}}\right] $$

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1 Answer 1

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As you haven't provided any data or told us what the variables mean, it's not entirely clear. But you are correct in inverting the link function and applying it to the linear combination of predictor variables to retrieve the expected value of 'time'. So yes, the expression seems correct. You may want to look at and modify the answer here stackoverflow.com/a/26640226/4269699 to extract the required expression efficiently.

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