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31 0 974 26 16 Sesame Chicken 6 Oz 670 40 0 9 49 1 540 29 19 Shrimp & Vegetable SF 6 Oz 130 5 0 4 15 6 890 6 7 Sweet & Sour Chicken 6 Oz 280 11 0 3 30 1 290 17 16 Thai Coconut Curry Chicken SF 6 Oz 230 12 0 6 10 7 504 20 8 Thai Sweet Chili Chicken SF 6 Oz 200 6 0 1 22 5 1008 13 17 Rice And Noodles Serving Size Calories Per Serving Fat (g) Trans Fat (g) Saturated Fat (g) Carbs (g) Fiber (g) Sodium (mg) Protein (g) Sugar (g) Asian Noodles 4 Oz 160 3 0 1 26 6 594 5 5 Steamed Brown Rice 4 Oz 130 1 0 0 27 3 11 4 0.2 Vegetable Fried Rice 4 Oz 170 5 0 1.5 27 1 1460 4 5 Steamed White Rice 4 Oz 150 0 0 0 32 1 0 3 0 Desserts Serving Size Calories Per Serving Fat (g) Trans Fat (g) Saturated Fat (g) Carbs (g) Fiber (g) Sodium (mg) Protein (g) Sugar (g) Chocolate Dipped Fortune Cookie 1 100 4 0 4 15 1 25 2 11 Fortune Cookie 1 30 0 0 0 5 0 6 0 2 Good Fortune Bar 3 Oz Bar 370 17 0 12 48 3 160 3 40 Lemon Yogurt 8 Oz 280 0.5 0 0 60 0 110 9 54 Raspberry Yogurt 8 Oz 280 0.5 0 0 60 0 110 9 54 Beverages Serving Size Calories Per Serving Fat (g) Trans Fat (g) Saturated Fat (g) Carbs (g) Fiber (g) Sodium (mg) Protein (g) Sugar (g) Jasmine Blackberry Ice Tea 20 Oz 10 0 0 0 0 0 0 0 0 Strawberry Lemonade 20 Oz 250 0 0 0 65 0 0 0 62 Traditional Ice Tea 20 Oz 10 0 0 0 1 0 4 0 0 Vanilla Red Ice Tea 20 Oz 110 0 0 0 28 0 0 0 28 Side Sauces Serving Size Calories Per Serving Fat (g) Trans Fat (g) Saturated Fat (g) Carbs (g)

B GateGateway of last resort is not set 1.0.0.0/32 is subnetted, 1 subnetsS 1.1.1.1 [1/0] via 10.12.12.1 2.0.0.0/32 is subnetted, 1 subnetsC 2.2.2.2 is directly connected, Loopback0 3.0.0.0/32 is subnetted, 1 subnetsO 3.3.3.3 [110/2] via 10.23.23.3, 00:05:12, FastEthernet0/1 4.0.0.0/32 is subnetted, 1 subnetsS 4.4.4.4 [1/0] via 10.24.24.4 10.0.0.0/8 is variably subnetted, 6 subnets, 2 masksC 10.12.12.0/24 is directly connected, FastEthernet0/0L 10.12.12.2/32 is directly connected, FastEthernet0/0C 10.23.23.0/24 is directly connected, FastEthernet0/1L 10.23.23.2/32 is directly connected, FastEthernet0/1C 10.24.24.0/24 is directly connected, FastEthernet1/0L 10.24.24.2/32 is directly connected, FastEthernet1/0However, at this point of configuration R3 doesn’t know how to reach R1 and R4 loopback because R2 doesn’t forward this information to R3 due to different routing protocol between R1/R4 and R3. Below is the routing table on R3:R3#sh ip route | b GateGateway of last resort is not set 2.0.0.0/32 is subnetted, 1 subnetsO 2.2.2.2 [110/2] via 10.23.23.2, 00:06:56, FastEthernet0/0 3.0.0.0/32 is subnetted, 1 subnetsC 3.3.3.3 is directly connected, Loopback0 10.0.0.0/8 is variably subnetted, 2 subnets, 2 masksC 10.23.23.0/24 is directly connected, FastEthernet0/0L 10.23.23.3/32 is directly connected, FastEthernet0/0This is the root cause of why R3 is unable to communicate with neither R1 nor R4 and vice versa. To fix this, R3 needs to be informed from R2 about the routes to R1 and R4 loopback address, and that means R2 needs to distribute its static routes to R3. The command to distribute static route via OSPF in Cisco IOS Router is “redistribute static subnets” and the way we applied this in the example is:R2(config)#router ospf 1R2(config-router)#redistribute static subnetsNow after the redistribution command is applied, R3 will have the knowledge of how to reach R1 and R4 loopback address, as shown in the updated routing table below:R3#sh ip route | b GateGateway of last resort is not set 1.0.0.0/32 is subnetted, 1 subnetsO E2 1.1.1.1 [110/20] via 10.23.23.2, 00:00:27, FastEthernet0/0 2.0.0.0/32 is subnetted, 1 subnetsO 2.2.2.2 [110/2] via 10.23.23.2, 00:00:51, FastEthernet0/0 3.0.0.0/32 is subnetted, 1 subnetsC 3.3.3.3 is directly connected, Loopback0 4.0.0.0/32 is subnetted, 1 subnetsO E2 4.4.4.4 [110/20] via 10.23.23.2, 00:00:32, FastEthernet0/0 10.0.0.0/8 is variably subnetted, 2 subnets, 2 masksC 10.23.23.0/24 is directly connected, FastEthernet0/0L 10.23.23.3/32 is directly connected, FastEthernet0/0R3#ping 1.1.1.1 source 3.3.3.3Type escape sequence to abort.Sending 5, 100-byte ICMP Echos to 1.1.1.1, timeout is 2 seconds:Packet sent with a source address of 3.3.3.3!!!!!Success rate is 100 percent (5/5), round-trip min/avg/max = 36/41/48 msR3#ping 4.4.4.4 source 3.3.3.3Type escape sequence to abort.Sending 5, 100-byte ICMP Echos to 4.4.4.4, timeout is 2 seconds:Packet sent with a source address of 3.3.3.3!!!!!Success rate is 100 percent (5/5), round-trip min/avg/max = 16/40/60 msThe routes to R1 and R4 are shown as an external OSPF routes and it also can be seen that ping from R3 to both R1 and

132∴ Exponent = (132)10 = (1000 0100)2Mantissa (23 bits ) = 0100 0000 1100 1100 1100 110 Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) 1 1000 0100 0100 0000 1100 1100 1100 110 (1100 0010 0010 0000 0110 0110 0110 0110)2 = (C2206666)16(C2206666)16 = 0xC2206666 The format of the single-precision floating-point representation of a real number as per the IEEE 754 standard is as follows:Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard? exponent = 00000000 and mantissa = 00000000000000000000001exponent = 00000001 and mantissa = 00000000000000000000000exponent = 00000001 and mantissa = 00000000000000000000001​exponent = 00000001 and mantissa = 000000000000000000000001Answer (Detailed Solution Below) Option 2 : exponent = 00000001 and mantissa = 00000000000000000000000 Option 2) is correct answer.Concept:In IEEE- 754 single precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Normalized Mantissa (M’) (23 bits) Sign bit value 0 means a positive number, and 1 means a negative number.The floating-point number can be obtained by formula: (-1)s × 1.M × 2E – 127Explanation: Smallest normalized positive number Sign bit Biased Exponent (E’) Normalized Mantissa (M’) 0 0000 0001 00000000000000000000000 Smallest normalized positive = (-1)0 × 1.00...0 × 21 – 127 = 2-126 ≈1.1755 × 10–38 Consider the following representation of a number in IEEE 754 single-precision floating point format with a bias of 127.S: 1 E: 10000001 F : 11110000000000000000000Here S, E and F denote the sign, exponent and fraction components of the floating point representation.The decimal value corresponding to the above representation (rounded to 2 decimal places) is ______Answer (Detailed Solution Below) -7.75 Concept:In IEEE- 754 single-precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Mantissa (M’) (23 bits) Sign bit value 0 means a positive number, and 1 means a negative number.The floating-point number can be obtained by formula: (-1)s × 1.M × 2E – 127Explanation: Sign bit Biased Exponent (E’) Mantissa (M’) 1 10000001 11110000000000000000000 Exponent = 10000001 = 129decimal value = (-1)1 × 1.1111...0 × 2129 – 127 = (-1)1 ×1.1111 × 22 The decimal value corresponding to the above representation (rounded to 2 decimal places) is -7.75 The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n - f. The range of decimal values for X in this representation is. 2-f to 2i2-f to (2i - 2-f) 0 to 2i0 to (2i - 2-f) Answer (Detailed Solution Below) Option 4 : 0 to (2i - 2-f) Diagram:i represents an integral part of the and f represents the fractional part of the number.Since, the n

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