--- title: "BOQ for DOECE four computer labs" subject: "Internet and Intranet" chapter: 5 key: "semester_8/internet_and_intranet:boq-doece-four-labs" source: https://www.bctengineer.com/q/semester_8/internet_and_intranet/boq-doece-four-labs site: BCT Engineering --- # BOQ for DOECE four computer labs Internet and Intranet ยท Chapter 5: Designing Internet Systems and Servers - Asked in: 2074 Magh, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/internet_and_intranet/boq-doece-four-labs - Whole chapter: https://www.bctengineer.com/md/semester_8/internet_and_intranet/chapter-5 ## Question DOECE has four computer labs each having 24 computers. One lab is located as ground floor of library building, one is at ground floor of CIT building, one as first floor of DOECE building and one research lab is at third floor of ICT building. Each computer has redundant network connection via dual ports CAT6 information outlet connected directly from Network Server room which is at first floor of CIT building. Prepare a Bill of Quantity (BOQ) with necessary network resources required in quantity for the complete networking. Your BOQ should include estimation of information outlet (faceplate), wireless AP, CAT6 cables, Patch cords, 24 ports patch panel and 24 port switches. Include logical IP addressing in each computers in each labs with address assigned by DHCP Server (in Server room) which has one dedicated IP address for DOECE building. Assume IP address as your own. ## Short answer For 96 dual-homed PCs plus 4 APs, plan 196 CAT6 runs totaling 13,206 m, so 44 boxes. Use 100 faceplates, 196 jacks, 9 patch panels, 8 access switches, 1 PoE switch, and 2 core switches. Use 10.10.0.0/16 with each lab in its own /27 subnet via DHCP. ## Answer #### Assumptions - 4 labs x 24 computers = 96 computers. Each computer has a dual-port CAT6 outlet, so 2 cables per computer = 192 cables. - One wireless AP per lab (4 APs), each with one CAT6 drop and PoE power from the switch. - All runs are under the 90 m CAT6 permanent-link limit. Average run lengths (horizontal plus vertical, with slack) from the Server room (CIT building, 1st floor): | Lab | Location | Average run | |---|---|---| | Lab 1 | Library building, ground floor | 70 m | | Lab 2 | CIT building, ground floor | 30 m | | Lab 3 | DOECE building, 1st floor | 60 m | | Lab 4 (research) | ICT building, 3rd floor | 85 m | - Wastage and slack: 10%. Cable comes in 305 m boxes. Switches have 24 x GE ports plus 4 SFP uplinks. The two ports of a computer go to two different switch groups (A and B) so that one switch failure does not isolate it. #### Topology ```text DHCP server (one IP for DOECE) | [Core L3 switch pair] <- Server room, CIT 1st floor | | Switch group A Switch group B (4 x 24-port) (4 x 24-port) | patch panels (9) | +------+------+--------+ | CAT6 (196 runs) Lab1 Lab2 Lab3 Lab4 (Lib) (CIT) (DOECE) (ICT) 24 PC 24 PC 24 PC 24 PC dual dual dual dual outlet outlet outlet outlet + 1 AP each ``` #### Quantity estimation 1. Dual-port outlets: 24 x 4 = 96. Single-port outlets for APs: 4. Total faceplates = **100**. Jacks = 96 x 2 + 4 = **196**. 2. Cables = 192 + 4 = 196 runs. Each lab has 49 runs (48 PC cables + 1 AP cable) of its own average length, so total length = 49 x (70 + 30 + 60 + 85) = 49 x 245 = 12,005 m. 3. With 10% wastage: 12,005 x 1.1 = 13,206 m. Boxes = 13,206 / 305 = 43.3, so **44 boxes**. 4. Patch panel ports needed = 196. 196 / 24 = 8.2, so **9 patch panels** (24-port). 5. Switch ports needed = 196 (192 PC + 4 AP). 8 x 24 = 192 ports for PCs (4 in group A, 4 in group B, each PC port 1 on A, port 2 on B), and one extra 24-port PoE switch for the 4 APs and servers = **9 access switches**, plus **2 core 24-port switches** for the uplink and DHCP server = **11 switches** in all. 6. Patch cords: user side (outlet to PC) 2 m: 192 + 4 (APs) = 196. Panel to switch 1 m: 196. Core to access uplinks 1 m: 12 (CAT6 or fibre). Total CAT6 patch cords = **392 + 12**. #### Bill of Quantity | S.N. | Item | Specification | Qty | |---|---|---|---| | 1 | CAT6 UTP cable | 305 m box, LSZH | 44 boxes | | 2 | Dual-port faceplate | With back box | 96 | | 3 | Single-port faceplate | For APs | 4 | | 4 | RJ45 CAT6 jack | Keystone | 196 | | 5 | Patch panel | 24-port CAT6, 1U | 9 | | 6 | Patch cord 2 m | CAT6 | 196 | | 7 | Patch cord 1 m | CAT6 | 208 | | 8 | Access switch | 24-port GE, 4 SFP | 8 | | 9 | PoE switch | 24-port PoE+ | 1 | | 10 | Core L3 switch | 24-port, SFP+ | 2 | | 11 | Wireless AP | Dual-band, PoE | 4 | | 12 | DHCP server | Rack server | 1 | | 13 | Rack | 42U with cable managers | 1 | | 14 | UPS | 3 kVA | 1 | | 15 | Conduit / trunking, labels | | lot | | 16 | Cable tester | Certification | 1 | #### Logical IP addressing Private range 10.10.0.0/16 is used for DOECE. The DHCP server in the Server room has one dedicated static address, 10.10.0.10, for the whole DOECE building. Each lab has its own VLAN and subnet. A /27 (30 usable hosts) holds 24 PCs, the gateway and the AP. The second NIC port is for redundancy (NIC teaming) and shares the same IP. | Network | VLAN | Subnet | Gateway | Static | DHCP pool | |---|---|---|---|---|---| | Server | 5 | 10.10.0.0/24 | 10.10.0.1 | DHCP server 10.10.0.10 | none | | Lab 1 Library | 11 | 10.10.10.0/27 | 10.10.10.1 | AP 10.10.10.2 | .3 - .30 (28) | | Lab 2 CIT | 12 | 10.10.10.32/27 | 10.10.10.33 | AP .34 | .35 - .62 | | Lab 3 DOECE | 13 | 10.10.10.64/27 | 10.10.10.65 | AP .66 | .67 - .94 | | Lab 4 ICT | 14 | 10.10.10.96/27 | 10.10.10.97 | AP .98 | .99 - .126 | | Management | 99 | 10.10.99.0/24 | 10.10.99.1 | switches | static | The core switch holds the gateway of each VLAN and forwards DHCP requests to the server with `ip helper-address 10.10.0.10`. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the number of CAT6 boxes found? A. Total length 12,005 m plus 10% wastage = about 13,206 m, divided by 305 m, rounded up to 44 B. Number of computers divided by 24 C. 12,005 m divided by 90 m D. Number of labs times 10 Answer: A. 12,005 x 1.1 = 13,206 m; /305 = 43.3 so 44 boxes. (Key point: Cable total) 2. Why are 9 patch panels needed? A. One per lab plus spare B. 196 ports needed / 24 = 8.2, rounded up to 9 C. 192 divided by 24 is exactly 8 and one for the server D. Because there are 9 buildings Answer: B. 196 runs need 196 ports; 24-port panels give 9. (Key point: Patch panels) 3. Why are a computer's two ports connected to switch groups A and B? A. To double the speed only B. To share one IP between users C. So one switch failure does not isolate the computer D. To reduce cable usage Answer: C. Dual connection to different switch groups gives redundancy. (Key point: Redundancy) 4. How do lab VLANs reach the single DHCP server? A. Each lab has its own DHCP server B. Computers use static IPs C. The AP hands out addresses D. The core switch forwards requests using ip helper-address 10.10.0.10 Answer: D. The core holds the VLAN gateways and relays DHCP to the server. (Key point: DHCP across VLANs) ## Related in Internet and Intranet - [BoQ for IOE ICT building six labs](https://www.bctengineer.com/question.md?key=semester_8%2Finternet_and_intranet%3Aboq-ioe-ict-six-labs) โ€” Internet and Intranet, chapter 5 - [BoQ for NITC four research labs](https://www.bctengineer.com/question.md?key=semester_8%2Finternet_and_intranet%3Aboq-nitc-research-labs) โ€” Internet and Intranet, chapter 5 - [Redundant network for IOE Pulchowk campus](https://www.bctengineer.com/question.md?key=semester_8%2Finternet_and_intranet%3Aredundant-network-ioe-pulchowk) โ€” Internet and Intranet, chapter 5