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GOS numerical: 1400 calls, 14 lost

Short answer

GoS is the fraction of calls lost in the busy hour. The busy hour is the peak 60‑minute period used for dimensioning. For 1400 offered calls, 14 lost, mean holding 3 min, traffic offered = (1400×3)/60 = 70 E, GoS = 14/1400 = 0.01, traffic carried = (1386×3)/60 = 69.3 E.

Define Grade of Service (GOS) and Busy hour. During a busy hour, 1400 calls were offered to a group of trunks and 14 calls were lost. The average call duration has 3 minutes. Find (i) Traffic offered (ii) GOS (iii) Traffic carried.

Grade of Service (GoS)

GoS is the proportion of calls lost (or delayed) in the busy hour because all servers are busy: GoS=calls lostcalls offeredGoS = \dfrac{\text{calls lost}}{\text{calls offered}}. Smaller value = better service.

Busy hour

The busy hour is the continuous 60-minute period of the day during which the traffic is maximum. Exchanges and trunks are dimensioned for busy-hour traffic so that GoS is met even at peak.

Numerical

Given: calls offered = 1400, calls lost = 14, h = 3 min = 1/20 h, T = 1 h.

(i) Traffic offered

Ao=Co hT=1400×360=70 E\begin{aligned} A_o &= \frac{C_o\,h}{T} = \frac{1400 \times 3}{60} = 70\ \text{E} \end{aligned}

(ii) Grade of service

GoS=141400=0.01\begin{aligned} GoS &= \frac{14}{1400} = 0.01 \end{aligned}

(iii) Traffic carried Calls carried = 1400 − 14 = 1386

Ac=1386×360=69.3 E\begin{aligned} A_c &= \frac{1386 \times 3}{60} = 69.3\ \text{E} \end{aligned}

(Check: Ac=Ao(1−GoS)=70×0.99=69.3A_c = A_o(1 - GoS) = 70 \times 0.99 = 69.3 E.)

Answer: traffic offered = 70 E, GoS = 0.01 (1%), traffic carried = 69.3 E.

✓ Answer checked by Claude Opus 5.5 (high effort)

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