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a Date: ... 4.3 Ci) yg if jettl 7. % ree Pale) pop ch Pg lt) =)" for ji Pa exp (a) Ci) | States = 0 - being repaired Po =| (por day } [= working or = 4 (per day ) g ' t oc ty aia ©. 0 Git) EBDE = How (t) = Plo Pwlt) r Peo “ Poolt) = 4 Prot) -H Doo (t) = KFDE = Gpool*) 5 Pa (t) * Uo + Peoolt) - Hoo = Pult)+ 1-4 Doolt) Civ) |_Since there've only 2 states, Por (t) = |= Poo(t) Sub into the KFOE equation : de Poot) = (1-poolt)) -4 Poot) = l- D Poo(t) & Poot) + 5 foo (t) =| Integrating _e1° s oF peolt) + 50% poolt) = 0% factor ~ =0* at A(U-poolt)) at = (oat OF Poole) = see | Scanned with | CamScanner’: No.: Date: Poolt) = 5 * co 5t we know that poo (0) =1 , so poo(0) = 3 +=] ea “ Paolt)= Ys + 3 0% ew Scanned with | CamScanner’: No.: Date: (iii) We want +0 find 4 Pant) - Ck: Panltth) = 5, Pavlt) - Peath) = Pant) + Pan Ch) Note that ths is the only path possible Paa(h) = 1+ hYaa + o0(h) = |-0-5h +0(h) Sub into CK equation © pan (tth) = paalt) - [ 1-0.5h+ 0h) ] Dan (tth) - Paalt) 9-6. -te) > 004) +— h = 0-5 Pantt) + n ») Take the limit h20, fe Pant) = -0-5 pane) Epance) +0. 5 pan (t) = ee) Integrating factor = efoS4t = 9?St MULiply both sides of O by IF: 00-5t. FZ aa(t) + 0.50°5* pratt) =0 ale. “Pan t)] at = fo dt e°7* Daa lt) =¢ ») Since pan) =|) 1+ paa(o) = ¢ c =| Therefore, Pan(t) = o-o # \ amr? Scanned with | CamScanner’: 
