301b Exam Question 16
Which command should the LTM Specialist use to determine the current system time?
301b Exam Question 17
An LTM Specialist is receiving reports from customers about multiple applications failing to work properly. The LTM Specialist looks at the services running and notices that the bigd process has NOT started.
How are monitored LTM device objects marked when the bigd process is stopped?
How are monitored LTM device objects marked when the bigd process is stopped?
301b Exam Question 18
The LTM device is configured to provide load balancing to a set of web servers that implement access control lists (ACL) based on the source IP address of the client. The ACL is at the network level and the web server is configured to send a TCP reset back to the client if it is NOT permitted to connect.
The virtual server is configured with the default OneConnect profile.
The ACL is defined on the web server as:
Permit: 192.168.136.0/24
Deny: 192.168.116.0/24
The packet capture is taken of two individual client flows to a virtual server with IP address 192.168.136.100.
Client A - Src IP 192.168.136.1 - Virtual Server 192.168.136.100:
Clientside:
09:35:11.073623 IP 192.168.136.1.55684 > 192.168.136.100.80: S 869998901:869998901(0) win 8192 <mss
1460,nop,wscale 2,nop,nop,sackOK>
09:35:11.073931 IP 192.168.136.100.80 > 192.168.136.1.55684: S 2273668949:2273668949(0) ack
869998902 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:35:11.074928 IP 192.168.136.1.55684 > 192.168.136.100.80: . ack 1 win 16425
09:35:11.080936 IP 192.168.136.1.55684 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:35:11.081029 IP 192.168.136.100.80 > 192.168.136.1.55684: . ack 299 win 4678 Serverside:
09:35:11.081022 IP 192.168.136.1.55684 > 192.168.116.128.80: S 685865802:685865802(0) win 4380 <mss
1460,nop,wscale 0,sackOK,eol>
09:35:11.081928 IP 192.168.116.128.80 > 192.168.136.1.55684: S 4193259095:4193259095(0) ack
685865803 win 5840 <mss 1460,nop,nop,sackOK,nop,wscale 6>
09:35:11.081943 IP 192.168.136.1.55684 > 192.168.116.128.80: . ack 1 win 4380
09:35:11.081955 IP 192.168.136.1.55684 > 192.168.116.128.80: P 1:299(298) ack 1 win 4380
09:35:11.083765 IP 192.168.116.128.80 > 192.168.136.1.55684: . ack 299 win 108 Client B - Src IP 192.168.116.1 - Virtual Server 192.168.136.100:
Clientside:
09:36:11.244040 IP 192.168.116.1.55769 > 192.168.136.100.80: S 3320618938:3320618938(0) win 8192
<mss 1460,nop,wscale 2,nop,nop,sackOK>
09:36:11.244152 IP 192.168.136.100.80 > 192.168.116.1.55769: S 3878120666:3878120666(0) ack
3320618939 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:36:11.244839 IP 192.168.116.1.55769 > 192.168.136.100.80: . ack 1 win 16425
09:36:11.245830 IP 192.168.116.1.55769 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:36:11.245922 IP 192.168.136.100.80 > 192.168.116.1.55769: . ack 299 win 4678 Serverside:
09:36:11.245940 IP 192.168.136.1.55684 > 192.168.116.128.80: P 599:897(298) ack 4525 win 8904
09:36:11.247847 IP 192.168.116.128.80 > 192.168.136.1.55684: P 4525:5001(476) ack 897 win 142 Why was the second client flow permitted by the web server?
The virtual server is configured with the default OneConnect profile.
The ACL is defined on the web server as:
Permit: 192.168.136.0/24
Deny: 192.168.116.0/24
The packet capture is taken of two individual client flows to a virtual server with IP address 192.168.136.100.
Client A - Src IP 192.168.136.1 - Virtual Server 192.168.136.100:
Clientside:
09:35:11.073623 IP 192.168.136.1.55684 > 192.168.136.100.80: S 869998901:869998901(0) win 8192 <mss
1460,nop,wscale 2,nop,nop,sackOK>
09:35:11.073931 IP 192.168.136.100.80 > 192.168.136.1.55684: S 2273668949:2273668949(0) ack
869998902 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:35:11.074928 IP 192.168.136.1.55684 > 192.168.136.100.80: . ack 1 win 16425
09:35:11.080936 IP 192.168.136.1.55684 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:35:11.081029 IP 192.168.136.100.80 > 192.168.136.1.55684: . ack 299 win 4678 Serverside:
09:35:11.081022 IP 192.168.136.1.55684 > 192.168.116.128.80: S 685865802:685865802(0) win 4380 <mss
1460,nop,wscale 0,sackOK,eol>
09:35:11.081928 IP 192.168.116.128.80 > 192.168.136.1.55684: S 4193259095:4193259095(0) ack
685865803 win 5840 <mss 1460,nop,nop,sackOK,nop,wscale 6>
09:35:11.081943 IP 192.168.136.1.55684 > 192.168.116.128.80: . ack 1 win 4380
09:35:11.081955 IP 192.168.136.1.55684 > 192.168.116.128.80: P 1:299(298) ack 1 win 4380
09:35:11.083765 IP 192.168.116.128.80 > 192.168.136.1.55684: . ack 299 win 108 Client B - Src IP 192.168.116.1 - Virtual Server 192.168.136.100:
Clientside:
09:36:11.244040 IP 192.168.116.1.55769 > 192.168.136.100.80: S 3320618938:3320618938(0) win 8192
<mss 1460,nop,wscale 2,nop,nop,sackOK>
09:36:11.244152 IP 192.168.136.100.80 > 192.168.116.1.55769: S 3878120666:3878120666(0) ack
3320618939 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:36:11.244839 IP 192.168.116.1.55769 > 192.168.136.100.80: . ack 1 win 16425
09:36:11.245830 IP 192.168.116.1.55769 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:36:11.245922 IP 192.168.136.100.80 > 192.168.116.1.55769: . ack 299 win 4678 Serverside:
09:36:11.245940 IP 192.168.136.1.55684 > 192.168.116.128.80: P 599:897(298) ack 4525 win 8904
09:36:11.247847 IP 192.168.116.128.80 > 192.168.136.1.55684: P 4525:5001(476) ack 897 win 142 Why was the second client flow permitted by the web server?
301b Exam Question 19
A web application sends information about message integrity and content life time to the client.
Which two HTTP headers should be used in sending the client information? (Choose two.)
Which two HTTP headers should be used in sending the client information? (Choose two.)
301b Exam Question 20
While investigating the cause of a device failover, an LTM Specialist discovers the following events in /var/ log/ltm:
01010029:5: Clock advanced by 518 ticks
01010029:5: Clock advanced by 505 ticks
01010029:5: Clock advanced by 590 ticks
01010029:5: Clock advanced by 568 ticks
01010029:5: Clock advanced by 1681 ticks
01010029:5: Clock advanced by 6584 ticks
01140029:5: HA daemon_heartbeat tmm fails action is failover and restart.
010c0026:5: Failover condition, active attempting to go standby.
Which issue caused the failover?
01010029:5: Clock advanced by 518 ticks
01010029:5: Clock advanced by 505 ticks
01010029:5: Clock advanced by 590 ticks
01010029:5: Clock advanced by 568 ticks
01010029:5: Clock advanced by 1681 ticks
01010029:5: Clock advanced by 6584 ticks
01140029:5: HA daemon_heartbeat tmm fails action is failover and restart.
010c0026:5: Failover condition, active attempting to go standby.
Which issue caused the failover?