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BoQ for IOE ICT building six labs

Short answer

Assume 360 cable runs (180 dual-port PCs) plus 6 APs, with average lengths of 45 m (2nd floor) and 55 m (3rd floor) and 10% wastage. Key quantities: 69 boxes of CAT6 cable, 180 dual-port faceplates plus 6 single-port, 366 RJ45 jacks, 16 patch panels, 8 48-port access switches plus 1 PoE switch, 2 core L3 switches, and 732 patch cords.

IOE ICT building has six computer labs each having 30 computers. Two labs are located at 2nd floor and four labs are located at 3rd floor. Each computer has redundant network connection via dual ports CAT6 information outlet connected directly from Network Control Room (NCR) located at ground floor. Prepare the 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 Panels and 48-ports Switches.

Assumptions

  • 6 labs x 30 computers = 180 computers. Dual-port outlets, so 2 cables per computer = 360 cables.
  • Labs 1-2 on the 2nd floor, labs 3-6 on the 3rd floor. NCR on the ground floor.
  • One wireless AP per lab = 6 APs, single CAT6 drop each (PoE).
  • Average run (horizontal plus vertical riser and slack): 2nd-floor labs 45 m, 3rd-floor labs 55 m. All are under 90 m.
  • Wastage and slack 10%. Cable box 305 m. A computer's two ports go to two different switches (dual-homing).

Layout

 3rd floor: Lab3 Lab4 Lab5 Lab6  (30 PC each, dual outlet)
 2nd floor: Lab1 Lab2            (30 PC each, dual outlet)
              |  CAT6 riser (55 m / 45 m)
 Ground  :  [Patch panels x16]
            [Switch x8 (48p) + 1 PoE]
                  |
            [Core L3 x2] -- Router/FW -- Internet

Quantity estimation

  1. Dual-port faceplates: 180. Single-port faceplates for APs: 6. Total faceplates = 186. RJ45 jacks = 360 + 6 = 366.
  2. Cable length:
    • Labs 1-2: 2 labs x 61 runs (60 + 1 AP) x 45 m = 5,490 m.
    • Labs 3-6: 4 labs x 61 runs x 55 m = 13,420 m.
    • Total = 18,910 m. With 10%: 20,801 m. Boxes = 20,801 / 305 = 68.2, so 69 boxes.
  3. Patch panels: 366 / 24 = 15.25, so 16 panels.
  4. Switches (48-port): PC ports 360 / 48 = 7.5, so 8 switches for PCs; plus 1 PoE 48-port switch for 6 APs = 9 switches. Two core L3 switches (24-port, SFP+) are added for redundant uplinks, routing and server links = 2.
  5. Patch cords: user side 2 m = 360 + 6 = 366; panel to switch 1 m = 366; access to core uplinks (fibre or CAT6) = 18. Total CAT6 cords = 732 plus 18 uplink cords.
  6. Rack space: 16 panels (16U) + 11 switches (11U) + managers (8U) + router/FW (2U) + UPS (4U) = 41U. Use 2 racks of 42U.

Bill of Quantity

S.N.ItemSpecificationQty
1CAT6 UTP cable305 m box69 boxes
2Dual-port faceplateWith back box180
3Single-port faceplateFor APs6
4RJ45 CAT6 jackKeystone366
5Patch panel24-port CAT616
6Patch cord 2 mCAT6366
7Patch cord 1 mCAT6366
8Access switch48-port GE, 4 SFP+8
9PoE switch48-port PoE+1
10Core L3 switch24-port, 4 SFP+2
11Wireless APDual-band, PoE6
12Router / firewall1
13Rack42U2
14UPS5 kVA online1
15Conduit, trunking, cable trayslot
16Labels, ties, testerlot

IP plan (10.20.0.0/16)

A lab has 30 PCs plus a gateway, so a /27 (30 hosts) is too small. A /26 (62 hosts) is used.

NetworkSubnetUsableGateway
Lab 110.20.1.0/2662.1
Lab 210.20.1.64/2662.65
Lab 310.20.1.128/2662.129
Lab 410.20.1.192/2662.193
Lab 510.20.2.0/2662.1
Lab 610.20.2.64/2662.65
Wi-Fi (6 APs)10.20.3.0/2730.1
Management10.20.4.0/2730.1
Servers10.20.5.0/2730.1

The two NICs of a PC are teamed (active-backup) so the PC keeps one address. DHCP is served by the core switch or a server using relay.

✓ Answer checked by Claude Opus 5.5 (high effort)

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