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The target collision energy requires $175$ GeV per lepton beam. Its initial energy after the half-linac is $62.5$ GeV, so each lepton must gain $112.5$ GeV. Define the transformer ratio as accelerating field experienced by the witness divided by decelerating field experienced by the driver. If the useful stage length exhausts the driver's $62.5$ GeV while delivering the required gain, thenThis matches depletion to acceleration and avoids leaving unused driver energy at the witness target. It is an ideal energy-transfer estimate, assuming suitable loading and longitudinal field profiles; a field ratio alone does not specify a practical wake or its efficiency.

For ideal transfer of the proton bunch's kinetic energy to the witness bunch,ThusThis is a minimum driver population for unit efficiency and full depletion under the stated energy budget. If only a fraction $\eta$ is usefully transferred, the required ratio becomes $1.8/\eta$. The equality to the transformer ratio uses the same ideal matching assumption as part b.

If the driver can be fully depleted with transformer ratio six, the largest ideal witness gain from a $62.5$ GeV proton is $375$ GeV. Each lepton then hasFor equal opposing electron and Positron beam energies,The associated ideal population ratio would be $N_p/N_e=6$. This energy-budget ceiling ignores wake-generation and transfer losses, staging limits, emittance preservation and the difficulty of accelerating both charge signs with suitable focusing. It is the consequence of the stipulated transformer-ratio bound, not a prediction of an achievable upgrade.