To calculate the elastic response of a rubber sample, the three chain force models (regimes Ia, Ib and II) and the network morphology must be combined in a micro-mechanical network model. Using the joint probability distribution in equation (4) and the force extension models, it is possible to devise numerical algorithms to both construct a faithful representative volume element of a network and to simulate the resulting mechanical stress as it is subjected to strain. An iterative relaxation algorithm is used to maintain approximate force equilibrium at each network node as strain is imposed. When the force constant obtained for kinks having 2 or 3 isoprene units (approximately one Kuhn length) is used in numerical simulations, the predicted stress is found to be consistent with experiments. The results of such a calculation are shown in Fig. 1 (dashed red line) for sulfur cross-linked natural rubber and compared with experimental data (solid blue line). These simulations also predict a steep upturn in the stress as network chains become taut and, ultimately, material failure due to bond rupture. In the case of sulfur cross-linked natural rubber, the S-S bonds in the cross-link are much weaker than the C-C bonds on the chain backbone and are the network failure points. The plateau in the simulated stress, starting at a strain of about 7, is the limiting value for the network. Stresses greater than about 7 MPa cannot be supported and the network fails. Near this stress limit, the simulations predict that less than 10% of the chains are taut, i.e. in the high chain extension regime and less than 0.1% of the chains have ruptured. While the very low rupture fraction may seem surprising, it is not inconsistent with the common experience of stretching a rubber band until it breaks. The elastic response of the rubber after breaking is not noticeably different from the original.