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NEW QUESTION: 1
プロジェクトマネージャーは、リスクの再評価をツールおよび手法としてどのプロセスで使用できますか？
A. 定性的リスク分析を実行する
B. リスクの監視と管理
C. プロジェクト作業の監視と制御
D. リスク対応を計画する
Answer: B

NEW QUESTION: 2
Your network consists of a large hub-and-spoke Frame Relay network with a CIR of 56 kb/s for each spoke. Which statement about the selection of a dynamic protocol is true?
A. EIGRP would be appropriate, because the Frame Relay spokes could be segmented into their own areas.
B. EIGRP would be appropriate, because you can manage how much bandwidth is consumed over the Frame Relay interface.
C. EIGRP would be appropriate, because by default, queries are not propagated across the slow speed Frame Relay links.
D. EIGRP would be appropriate if LMI type ANSI is NOT used.
Answer: B
Explanation:
Explanation By default, EIGRP will limit itself to using no more than 50% of the interface bandwidth. The primary benefit of controlling EIGRP's bandwidth usage is to avoid losing EIGRP packets, which could occur when EIGRP generates data faster than the interface line can absorb it. This is of particular benefit on Frame Relay networks, where the access interface bandwidth and the PVC capacity may be very different. For example, in our Frame Relay topology a Hub is connected with 4 Spoke routers. The main Frame Relay interface on Hub router is 512Kpbs which is not enough to use for 6 links of 128 Kbps ( = 768 Kbps).

The solution here is we can use 512 / 6 = 85 Kbps on each subinterface of Hub by using "bandwidth 85 command. For example: Hub(config)#interface Serial0/0.1 point-to-pointHub(config-subif)#bandwidth 85 Also on Spoke routers we need to set this value. For example on Spoke1: Spoke1(config)#interface Serial0/1.0 point-to-pointSpoke1(config-subif)#bandwidth 85 Notice that by default, EIGRP limits itself to use no more than 50% of the configured interface bandwidth. In this case EIGRP will not use more than 42.5 Kbps (50% of 85 Kbps). (For more information about implementing EIGRP over Frame Relay, please read http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a0080094063.shtml)

NEW QUESTION: 3
What happens when you attempt to compile and run the following code?
#include <vector>
#include <iostream>
# include <algorithm>
# include <functional>
using namespace std;
class B { int val;
public:
B(int v=0):val(v){}
int getV() const {return val;}
B operator +(const B &b )const { return B(val + b.val);} };
ostream & operator <<(ostream & out, const B & v) { out<<v.getV(); return out;} template<class T>struct Out {
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) { out<<val<<" "; } };
B Add(B a, B b) { return a+b; }
int main() {
int t[]={1,2,3,4,5,6,7,8,9,10};
vector<B> v1(t, t+10);
vector<B> v2(10);
transform(v1.begin(), v1.end(), v2.begin(), bind2nd(ptr_fun(Add),1));
for_each(v2.rbegin(), v2.rend(), Out<B>(cout));cout<<endl;
return 0;
}
Program outputs:
A. 11 10 9 8 7 6 5 4 3 2
B. 2 3 4 5 6 7 8 9 10 11
C. compilation error
D. 1 2 3 4 5 6 7 8 9 10
E. 10 9 8 7 6 5 4 3 2 1
Answer: A