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KIM [24]
3 years ago
11

Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running

times of 6, 4, 1, 3, and 7 minutes. Their ( externally determined ) priorities are 3, 5, 2, 1, and 4, respectively, with 5 being the highest priority. For each of the following scheduling algorithms, determine the mean process waiting time. Ignore process switching overhead.
a) Round Robin ( assume quantum = 1 )
b) Priority scheduling
c) First-come first-served
d) Shortest job first
For a), assume that the system is multitasking, and that each job gets its fair share of the CPU. For b) through d) assume that only one job at a time runs, until it finishes. All jobs are completely CPU bound.
Engineering
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1   2   3   4   5   6   7   8   9   10   11   12   13   14   15   16   17   18   19   20   21   22   23   24   25   26   27   28   29    30

A  B  C   D  E  A  B   C* D  E    A    B    D   E    A   B   D*   E     A   B     E     A    B* E   A     E  A    E*  A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6       7-14        15-24      25-26        27-30

   

 B           E             A             C            D

     AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10      11-16      17-18      19-22      23-30

 

   A            B              C            D              E

   

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2        3-6         7-12         13-20      21-30

C           D            B               E                A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

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Alla [95]

Answer:

Explanation:

?

5 0
3 years ago
Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg
Dennis_Churaev [7]

Answer:

h = 1429.74\,\frac{kJ}{kg}

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

\nu_{f} = 0.001080\,\frac{m^{3}}{kg}

\nu_{g} = 0.50850\,\frac{m^{3}}{kg}

Since \nu_{f} < \nu < \nu_{g}, it is a liquid-vapor mixture. The quality of the mixture is:

x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}

x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }

x = 0.392

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

h_{f} = 589.16\,\frac{kJ}{kg}

h_{g} = 2733.5\,\frac{kJ}{kg}

The specific enthalpy is:

h = h_{f}+x\cdot (h_{g}-h_{f})

h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)

h = 1429.74\,\frac{kJ}{kg}

6 0
4 years ago
Read 2 more answers
PLS HELP
RSB [31]

Answer:

T=kg·m^2/s^2

Explanation:

T = kg (m^2/s^2) m^3 /m^3

Here I wrote down the unit for every dimension.

T=kg m^2 / s^2

m^3 is divided between m^3, this is equal to 1.

Result: T=kg·m^2/s^2

PD: I'm not sure if this is what you ask for. I hope it helps

4 0
3 years ago
Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of
Savatey [412]

Answer:

option C

option A

Explanation:

Enthalpy gained by air= 1023-290

                                       = 733 kJ/kg

Rate of energy gain= mass flow rate × Enthalpy gained by air

                               = 0.1 × 733

                            = 73.3 kJ/s

rate of heat transfer between compressor and air= 77kW

Heat loss by air to surroundings= 77-73.3

                                                     =3.7kW

Enthalpy lost by steam in turbine= 1407.6-1236.4

                                                     = 171.2 Btu/lb

Rate of energy transfer to turbine= Enthalpy lost by steam× mass flow rate

                                                    = 171.2×5

                                                     = 856 Btu/s

Net rate of energy transfer to turbine=rate of  Energy transfer to turbine- rate of heat transfer to turbine

                          = 856-40

                         = 816 Btu/s

8 0
4 years ago
The pattern in a simple____ truss is a zigzag
liberstina [14]

Answer:

Bridge

Explanation:

A common, simply bridge truss is the zigzag.

6 0
3 years ago
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