How does polling-based network receive work? How is it different than interrupt based network? Describe both functions and list as many differences as you can identify. What factors determine the relative performance between a programmed I/O and a DMA?

In a polling-based network receives, the operating system keeps checking the status register after a fixed time interval to see whether a device needs hardware attention or not. CPU gradually checks the status of receive bit after a defined time interval of a clock and if the value of receive bit is true then data is moved from the RX register to the memory. In interrupt-based networks, whenever a device needs hardware processing then an interrupt is raised to inform the CPU for device attention. The data of the device is transferred to memory from the network card.

Polling is a protocol, not a hardware mechanism where the device gets attention by CPU. An interrupt is a hardware mechanism, not a protocol where the device gets attention by the interrupt handler. In polling, during receiving, parts of received data need to transfer to memory but in case of interrupts, whole data is transferred at once after receiving is done. An interrupt is a heavy operation as compared to pooling because hardware involves in it. For large amounts of data, Polling becomes an inefficient method. In polling, CPU gradually checks devices at regular intervals whereas the Interrupt handler can send interrupt at any time. Each device has its own Command ready bit which indicates whether the device requires servicing or not. In polling, CPU wastes numerous clock cycles by repetitively checking the command-ready little bit of each device. In Interrupt based network, the interrupt request line indicates whether the device requires servicing or not. In interrupt-based network, CPU cycles not wasted but CPU is disturbed only when device send interrupt signals.  

Relative performance between programmed I/O and DMA:

For determining the relative performance between programmed I/O and DMA, important factors are:

1.Size of data  

2.Frequency of operations

pH = 5.85

Explanation:

The reaction of methylamine with HCl is:

CH₃NH₂ + HCl → CH₃NH₃⁺ + Cl⁻

As the titration was made with two solutions with the same molarity, the final concentration of the CH₃NH₃⁺ ion is 0.100M ([CH₃NH₃⁺] = 0.100M)

Then, the conjugate acid will be in equilibrium with water as follows:

CH₃NH₃⁺(aq) + H₂O(l) ⇄ CH₃NH₂(aq) + H₃O⁺(aq)

And the equilibrium constant, Ka, of this reaction is:

Kw/Kb = Ka =  1.0x10⁻¹⁴ / 5.0x10⁻⁴= 2x10⁻¹¹

2x10⁻¹¹ =  [CH₃NH₂] [H₃O⁺] / [CH₃NH₃⁺]

As both CH₃NH₂ and H₃O⁺ comes from the same equilibrium we can say:

[CH₃NH₂] = [H₃O⁺] = X

And as we know [CH₃NH₃⁺] = 0.100M:

2x10⁻¹¹ =  [X] [X] / [0.100M]

2x10⁻¹² = X²

X = 1.414x10⁻⁶M = [H₃O⁺]

And as pH= -log [H₃O⁺]