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diff --git a/tests/testthat/test_mkinpredict_SFO_SFO.R b/tests/testthat/test_mkinpredict_SFO_SFO.R
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+# $Id: jranke $
+
+# Copyright (C) 2012-2015 Johannes Ranke
+# Contact: jranke@uni-bremen.de
+
+# This file is part of the R package mkin
+
+# mkin is free software: you can redistribute it and/or modify it under the
+# terms of the GNU General Public License as published by the Free Software
+# Foundation, either version 3 of the License, or (at your option) any later
+# version.
+
+# This program is distributed in the hope that it will be useful, but WITHOUT
+# ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS
+# FOR A PARTICULAR PURPOSE.  See the GNU General Public License for more
+# details.
+
+# You should have received a copy of the GNU General Public License along with
+# this program. If not, see <http://www.gnu.org/licenses/>
+
+# This was migrated from an RUnit test in inst/unitTests/runit.mkinpredict.R
+
+context("Model predictions with mkinpredict")
+test_that("Variants of model predictions for SFO_SFO model give equivalent results", {
+  # Check model specification and solution types for SFO_SFO
+  # Relative Tolerance is 0.01%
+  # Do not use time 0, as eigenvalue based solution does not give 0 at time 0 for metabolites
+  # and relative tolerance is thus not met
+  tol = 0.01
+  SFO_SFO.1 <- mkinmod(parent = list(type = "SFO", to = "m1"),
+         m1 = list(type = "SFO"), use_of_ff = "min")
+  SFO_SFO.2 <- mkinmod(parent = list(type = "SFO", to = "m1"),
+         m1 = list(type = "SFO"), use_of_ff = "max")
+
+  ot = seq(0, 100, by = 1)
+  r.1.e <- subset(mkinpredict(SFO_SFO.1, 
+             c(k_parent_m1 = 0.1, k_parent_sink = 0.1, k_m1_sink = 0.1), 
+             c(parent = 100, m1 = 0), ot, solution_type = "eigen"), 
+                 time %in% c(1, 10, 50, 100))
+  r.1.d <- subset(mkinpredict(SFO_SFO.1, 
+             c(k_parent_m1 = 0.1, k_parent_sink = 0.1, k_m1_sink = 0.1), 
+             c(parent = 100, m1 = 0), ot, solution_type = "deSolve"), 
+                 time %in% c(1, 10, 50, 100))
+
+  r.2.e <- subset(mkinpredict(SFO_SFO.2, c(k_parent = 0.2, f_parent_to_m1 = 0.5, k_m1 = 0.1), 
+	    c(parent = 100, m1 = 0), ot, solution_type = "eigen"),
+                  time %in% c(1, 10, 50, 100))
+  r.2.d <- subset(mkinpredict(SFO_SFO.2, c(k_parent = 0.2, f_parent_to_m1 = 0.5, k_m1 = 0.1), 
+	    c(parent = 100, m1 = 0), ot, solution_type = "deSolve"),
+                  time %in% c(1, 10, 50, 100))
+
+  # Compare eigen and deSolve for minimum use of formation fractions
+  dev.1.e_d.percent = 100 * (r.1.e[-1] - r.1.d[-1])/r.1.e[-1]
+  dev.1.e_d.percent = as.numeric(unlist((dev.1.e_d.percent)))
+  dev.1.e_d.percent = ifelse(is.na(dev.1.e_d.percent), 0, dev.1.e_d.percent)
+  expect_equivalent(dev.1.e_d.percent < tol, rep(TRUE, length(dev.1.e_d.percent)))
+
+  # Compare eigen and deSolve for maximum use of formation fractions
+  dev.2.e_d.percent = 100 * (r.1.e[-1] - r.1.d[-1])/r.1.e[-1]
+  dev.2.e_d.percent = as.numeric(unlist((dev.2.e_d.percent)))
+  dev.2.e_d.percent = ifelse(is.na(dev.2.e_d.percent), 0, dev.2.e_d.percent)
+  expect_equivalent(dev.2.e_d.percent < tol, rep(TRUE, length(dev.2.e_d.percent)))
+
+  # Compare minimum and maximum use of formation fractions
+  dev.1_2.e.percent = 100 * (r.1.e[-1] - r.2.e[-1])/r.1.e[-1]
+  dev.1_2.e.percent = as.numeric(unlist((dev.1_2.e.percent)))
+  dev.1_2.e.percent = ifelse(is.na(dev.1_2.e.percent), 0, dev.1_2.e.percent)
+  expect_equivalent(dev.1_2.e.percent < tol, rep(TRUE, length(dev.1_2.e.percent)))
+
+})