2.2-08. Download delays for 100 objects (HTTP 1.1 with local web caching). Consider an HTTP 1.1 client and server. The RTT delay between the client and server is 2 seconds. Suppose the time a server needs to transmit an object into its outgoing link is 3 seconds.There is also a local web cache, as shown in the figure below, with negligible (zero) propagation delay and object transmission time. The client makes 100 requests one after the other, waiting for a reply before sending the next request. All requests first go to the cache (which also has a 2.0 sec. RTT delay to the server but zero RTT to the client).How much time elapses between the client transmitting the first request, and the receipt of the last requested object, assuming no use of the IF-MODIFIED-SINCE header line anywhere, and assuming that 50% of the objects requested are "hits" (found) in the local cache?Group of answer choices352 secs203 secs252 secs150 secs350 secs
Question
2.2-08. Download delays for 100 objects (HTTP 1.1 with local web caching). Consider an HTTP 1.1 client and server. The RTT delay between the client and server is 2 seconds. Suppose the time a server needs to transmit an object into its outgoing link is 3 seconds.There is also a local web cache, as shown in the figure below, with negligible (zero) propagation delay and object transmission time. The client makes 100 requests one after the other, waiting for a reply before sending the next request. All requests first go to the cache (which also has a 2.0 sec. RTT delay to the server but zero RTT to the client).How much time elapses between the client transmitting the first request, and the receipt of the last requested object, assuming no use of the IF-MODIFIED-SINCE header line anywhere, and assuming that 50% of the objects requested are "hits" (found) in the local cache?Group of answer choices352 secs203 secs252 secs150 secs350 secs
Solution 1
Para resolver este problema, sigamos los siguientes pasos:
-
Entender los tiempos involucrados:
- RTT entre el cliente y el servidor: 2 segundos.
- Tiempo de transmisión del objeto desde el servidor: 3 segundos.
- RTT entre el cliente y la caché local: 0 segundos (negligible).
- Tiempo de transmisión del objeto desde la caché local: 0 segundos (negligible).
-
Dividir los objetos en "hits" y "misses":
- 50% de los objetos son "hits" en la caché local.
- 50% de los objetos son "misses" y deben ser obtenidos del servidor.
-
Calcular el tiempo para los "hits":
- Para los "hits", el tiempo total es solo el tiempo de ida y vuelta (RTT) a la caché local, que es 0 segundos.
- Por lo tanto, el tiempo para cada "hit" es 0 segundos.
-
Calcular el tiempo para los "misses":
- Para los "misses", el tiempo total incluye el RTT al servidor y el tiempo de transmisión del objeto.
- Tiempo para cada "miss": RTT (2 segundos) + tiempo de transmisión (3 segundos) = 5 segundos.
-
Calcular el tiempo total para todos los objetos:
- Hay 100 objetos en total.
- 50 objetos son "hits" y 50 objetos son "misses".
- Tiempo total para los "hits": 50 * 0 segundos = 0 segundos.
- Tiempo total para los "misses": 50 * 5 segundos = 250 segundos.
-
Considerar el tiempo de espera entre solicitudes:
- El cliente espera la respuesta antes de enviar la siguiente solicitud.
- Para cada solicitud, hay un RTT de 2 segundos (ya sea a la caché o al servidor).
-
Calcular el tiempo total incluyendo el tiempo de espera:
- Para los "hits", el tiempo de espera es 2 segundos por cada solicitud.
- Para los "misses", el tiempo de espera ya está incluido en los 5 segundos calculados anteriormente.
- Tiempo total de espera para los "hits": 50 * 2 segundos = 100 segundos.
- Tiempo total de espera para los "misses": 0 segundos (ya incluido).
-
Sumar todos los tiempos:
- Tiempo total = tiempo para los "hits" + tiempo para los "misses" + tiempo de espera para los "hits".
- Tiempo total = 0 segundos + 250 segundos + 100 segundos = 350 segundos.
Por lo tanto, la respuesta correcta es 350 segundos.
Solution 2
Para resolver este problema, sigamos los siguientes pasos:
-
Entender los tiempos de respuesta:
- RTT entre el cliente y el servidor: 2 segundos.
- Tiempo de transmisión de un objeto desde el servidor: 3 segundos.
- RTT entre el cliente y la caché local: 0 segundos (negligible).
-
Dividir las solicitudes en "hits" y "misses":
- 50% de las solicitudes son "hits" en la caché local.
- 50% de las solicitudes son "misses" y deben ser obtenidas del servidor.
-
Calcular el tiempo para un "hit":
- Para un "hit" en la caché local, el tiempo es 0 segundos (negligible).
-
Calcular el tiempo para un "miss":
- Para un "miss", el cliente envía la solicitud a la caché, la caché la reenvía al servidor, y el servidor responde.
- Tiempo total para un "miss" = RTT (cliente-servidor) + tiempo de transmisión del objeto = 2 segundos + 3 segundos = 5 segundos.
-
Calcular el tiempo total para todas las solicitudes:
- Número de "hits": 50% de 100 = 50.
- Número de "misses": 50% de 100 = 50.
- Tiempo total para "hits": 50 * 0 segundos = 0 segundos.
- Tiempo total para "misses": 50 * 5 segundos = 250 segundos.
-
Considerar el tiempo de espera entre solicitudes:
- El cliente espera la respuesta antes de enviar la siguiente solicitud.
- Para cada solicitud, hay un RTT de 2 segundos (cliente-caché-servidor-cliente).
-
Calcular el tiempo total incluyendo el RTT:
- Tiempo total de RTT para 100 solicitudes: 100 * 2 segundos = 200 segundos.
- Tiempo total de transmisión de objetos para "misses": 250 segundos.
-
Sumar todos los tiempos:
- Tiempo total = Tiempo de RTT + Tiempo de transmisión de objetos = 200 segundos + 250 segundos = 450 segundos.
Sin embargo, dado que el tiempo de transmisión de objetos ya incluye el RTT para los "misses", debemos ajustar el cálculo:
- Tiempo total correcto = Tiempo de transmisión de objetos (250 segundos) + Tiempo de RTT para "hits" (50 * 2 segundos) = 250 segundos + 100 segundos = 350 segundos.
Por lo tanto, la respuesta correcta es:
350 segundos.
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