This question's explanation and your answers so far.
問91
頻出度★★☆
Mさんの部署では，図に示すような，PC2台（PC1とPC2），サーバ1台とプリンタ1台を，ハブ1台と4本のケーブルa～dでつないだLANで構成されるシステムを使っている。
このLANに接続されているプリンタは，プリントサーバ機能を有しており，PCから直接プリンタに出力できる設定になっている。ある日，MさんはPC1からプリンタに出力を行ったが，出力できなかった。前日に同じ処理を行ったときは，出力できていた。
なお，各機器と各ケーブルの接続箇所については，ソケットの不具合は機器側の障害，コネクタの不具合はケーブル側の障害として切り分けられるものとする。また，障害が複数箇所で同時に発生する可能性はなく，PC1では障害が発生していないものとする。
M さんは，障害の原因を特定するための手順を，流れ図に書いて考えてみることにした。このとき，次の図の X に入る適切な字句はどれか。
なお，次の流れ図は，作成途中のものである。
（端子）「問題の切分け」
↓
〈判断1〉「PC2 からプリンタに出力できるか」
・No → 右へ分岐（以降は表示していない）
・Yes ↓
〈判断2〉「PC1 と PC2 をハブにつないでいるケーブル c と d をハブ側で差し替え，PC2 からプリンタに出力できるか」
・No → （処理）「X」
・Yes → （処理）網掛けの箱（表示していない）
注 網掛けの部分は，表示していない。
出典 平成21年度 春期 ITパスポート試験 問91
解説
問 91
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○ア
正解The correct answer is ア. X is the box you reach when Decision 2 is “No” — that is, even after cables c and d are swapped at the hub side, PC2 too becomes unable to output to the printer. First, Decision 1 was Yes, so PC2 can output normally, which tells us that the printer itself, cable b connecting the printer to the hub, and the relay function of the hub are all working. Since the premises are that there is only one fault and that no fault has occurred in PC1, only two suspects remain: cable c, and the socket on the hub side into which c is plugged. If c and d are now swapped at the hub side only, PC2 cable d goes into the socket that c used to occupy. If PC2 then becomes unable to output (No), the symptom did not travel with the cable but stayed at the socket position, so the cause is on that socket side. Because the premises classify a defective socket as a fault on the device side, this can be judged to be a fault in the hub. Therefore X should read “Could it be a fault in the socket connecting PC1 to the hub, that is, a fault in the hub?”
×Incorrectイ
A fault in cable c is the conclusion for the case where Decision 2 is “Yes”; it belongs in the shaded box, not in X (the No side). If the cause were on the cable c side (a defective connector, for example), then after the swap PC2 would be using the normal cable d, so the output would succeed and Decision 2 would be Yes. The moment the result is No, cable c is known to be innocent, so it cannot go in X.
×Incorrectウ
Cable b, which connects the printer to the hub, was already confirmed to be normal at the point where PC2 could output in Decision 1 (Yes), because print data from PC2 must also pass through cable b. Since the premise is that only one fault occurs, re-suspecting a component that has already been cleared as normal is wrong for a flowchart. Suspecting cable b is the role of the right-hand branch taken when Decision 1 is No (the part that is not shown).
×Incorrectエ
A fault in the printer itself was likewise ruled out at the point where PC2 could output in Decision 1, because if the printer were broken, no PC at all would be able to output. This too is a candidate to be considered when Decision 1 goes down the No side, so it does not belong in X, which is reached after a Yes.
ここがポイント
The basis of fault isolation is to remove from suspicion, one by one, the components that lie on a path known to work. At the point where PC2 can print, the printer, cable b, and the relay function of the hub are confirmed to be normal; combined with the premises that there is only one fault and that PC1 is normal, the suspects narrow down to just two: cable c and the hub socket used by PC1.
How to read the swap (substitution) method: after exchanging parts, if the symptom moves together with a part, that part is the cause; if the symptom stays in the same place, the cause is on the socket (device) side. In this question a defective socket is defined as a fault on the device side, so if PC2 still cannot output after the swap (No) the cause is the hub, and if it can output (Yes) the cause is cable c.
In the Yes / No branches of a flowchart, write only what can be asserted when that decision actually comes out that way. A candidate that an earlier decision has already ruled out, or a conclusion that should be stated on the opposite branch, must not be written there.