A network architect is assigned the IPv4 block 172.16.32.0/20 to provision subnets for fou… — CompTIA Network+

A network architect is assigned the IPv4 block $172.16.32.0/20$ to provision subnets for four new departments in a branch office. The requirements for usable host IP addresses are as follows: Department A requires $500$ hosts, Department B requires $250$ hosts, Department C requires $100$ hosts, and Department D requires $50$ hosts. Using Variable Length Subnet Masking (VLSM) and allocating subnets contiguously starting from the lowest available network address in the block in order of largest to smallest host requirement, what is the broadcast address of Department D's subnet, and how many total IP addresses remain unallocated within the original $/20$ block?

Broadcast address: $172.16.35.191$ ; Unallocated IP addresses: $3136$ Answer
B
Broadcast address: $172.16.35.190$ ; Unallocated IP addresses: $3136$
C
Broadcast address: $172.16.35.191$ ; Unallocated IP addresses: $3128$
D
Broadcast address: $172.16.35.255$ ; Unallocated IP addresses: $2944$

Answer

Broadcast address:

172.16.35.191

; Unallocated IP addresses:

3136

Evaluating host requirements with VLSM yields block sizes of

512

(

/23

256

(

/24

128

(

/25

), and

64

(

/26

). Sequential assignment starting from

172.16.32.0

places Department D at

172.16.35.128/26

, which spans

172.16.35.128

through

172.16.35.191

, making

172.16.35.191

the broadcast address. Subtracting total allocated addresses (

960

) from the total block size of

4096

addresses leaves

3136

unallocated IP addresses.

Step-by-Step Solution

Determine total capacity of the parent block

/20

network has

32 - 20 = 12

host bits, yielding

2^{12} = 4096

total IP addresses.

Calculating total initial pool size is required to find remaining unallocated space.

Calculate required prefix lengths and block sizes using VLSM in descending order

Department A (

500

hosts): Needs

/23

(

2^9 - 2 = 510

usable, block size

512

). Department B (

250

hosts): Needs

/24

(

2^8 - 2 = 254

usable, block size

256

). Department C (

100

hosts): Needs

/25

(

2^7 - 2 = 126

usable, block size

128

). Department D (

50

hosts): Needs

/26

(

2^6 - 2 = 62

usable, block size

64

Subnet allocation requires accommodating host counts plus network and broadcast overhead (

+2

Assign contiguous subnet ranges from lowest available address

Dept A:

172.16.32.0/23

(

172.16.32.0 - 172.16.33.255

). Dept B:

172.16.34.0/24

(

172.16.34.0 - 172.16.34.255

). Dept C:

172.16.35.0/25

(

172.16.35.0 - 172.16.35.127

). Dept D:

172.16.35.128/26

(

172.16.35.128 - 172.16.35.191

Subnets must start on valid binary boundaries matching their block sizes.

Identify Department D broadcast address and calculate unallocated addresses

Broadcast address for Dept D is

172.16.35.191

. Total allocated addresses

= 512 + 256 + 128 + 64 = 960

. Remaining unallocated addresses

= 4096 - 960 = 3136

Broadcast is the last address of the subnet block, and unallocated capacity is total capacity minus used capacity.

Key Concept

Variable Length Subnet Masking (VLSM) and IPv4 Subnet Allocation

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic