EBK OM
EBK OM
6th Edition
ISBN: 9781305888210
Author: Collier
Publisher: YUZU
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Chapter 7, Problem 13PA

(a)

Summary Introduction

Interpretation: A process flow diagram is to be constructed along withlabeling and the times and percentages.

Concept Introduction:

Any flow unit, when flows through a process, follows L’s Law which states that the Work in Progress and given by,

  WorkInProgress(I)=FlowTime(T)×Throughput(R)

(a)

Expert Solution
Check Mark

Explanation of Solution

Process flow diagram is as follows.

  EBK OM, Chapter 7, Problem 13PA , additional homework tip  1

(b)

Summary Introduction

Interpretation: The throughout in patients is to be determined per hour of each stage in the process.

Concept Introduction:

The throughput of any process is the rate at which outputs are coming out of the process. When the capacity (i.e. the service rate) of a process step is more than or equal to the arrival rate or the demand rate, the throughput will be equal to the demand rate.

On the contrary, if the capacity is less than the demand rate, it means that the process step is a bottleneck and the throughput will be equal to the capacity.

(b)

Expert Solution
Check Mark

Explanation of Solution

The throughout in patients per hour of each stage in the process is shown in table below.

    Resource Demand or arrival rateCapacity or Service rateCondition Throughout rate
    RC10 per hour60/3=20perhourCapacity>Demand10 per hour
    RRC10 per hour60/6=10perhourCapacity=Demand10 per hour
    NP10 per hour60/5=12perhourCapacity>Demand10 per hour
    PA10×0.4=4perhour60/6=10perhourCapacity>Demand4 per hour
    MD10×0.6=6perhour60/15=4perhourCapacity<Demand4 per hour
    BC8 per hour60/5=12perhourCapacity>Demand8 per hour

(c)

Summary Introduction

Interpretation: The labor utilization rates are to be calculated and whether these values appropriate. In case they are not appropriate the way the process can be redesigned is to be determined along with then bottlenecks.

Concept Introduction:

The labor utilization is defined as the fraction of time a resource is busy producing output in the long run. It can be calculated by using the following formula.

  Utilization(U)=DemandrateServiceRate×NumberofServers

(c)

Expert Solution
Check Mark

Explanation of Solution

However, note that the labor utilization cannot be more than 100% in long run. However, in short run, we can write it as more than 100% using the above formula.

    Resource Demand or arrival rateCapacity or Service rateNumber of servers (N)Utilization (U) U=D/(SR×N)
    RC10 per hour60/3=20perhour1 50.00%
    RRC10 per hour60/6=10perhour1100.00%
    NP10 per hour60/5=12perhour183.33%
    PA10×0.4=4perhour60/6=10perhour140.00%
    MD10×0.6=6perhour60/15=4perhour1150.00%
    BC8 per hour60/5=12perhour166.67%

The resource RRC and MD are overloaded as the utilization figures suggest. Similarly, the resource RC and PA are under loaded. Possible improvement strategy is to share the loads of work (as they are similar in nature) between RC and RRC. Also, it should be checked whether the 60/40 ratio of diversion to MD and PA can be modified in order to balance their load

At present, the resource MD, having the capacity less than the arrival rate and thus the flow is constrained at MD. So, it is the bottleneck.

(d)

Summary Introduction

Interpretation: The way by which the given change affect your answer to the preceding questions.

Concept Introduction:

Any flow unit, when flows through a process, follows L’s Law which states that the Work in Progress and given by,

  WorkInProgress(I)=FlowTime(T)×Throughput(R)

(d)

Expert Solution
Check Mark

Explanation of Solution

Due to the change, the process layout will be as follows.

  EBK OM, Chapter 7, Problem 13PA , additional homework tip  2

The throughput rates for each resource will be as follows.

    Resource Demand or arrival rate (D)Capacity or Service rateCondition Throughout rate
    RC10 per hour60/3=20perhourCapacity>Demand10 per hour
    RRC10 per hour60/6=10perhourCapacity=Demand10 per hour
    NP10 per hour60/5=12perhourCapacity>Demand10 per hour
    PA10×0.4=4perhour60/6=10perhourCapacity>Demand4 per hour
    MD10×0.6+4×0.5=8perhour60/15=4perhourCapacity<Demand4 per hour
    BC4×0.5+4=6perhour60/5=12perhourCapacity>Demand6 per hour

The labor utilization of the resources are as follows.

    Resource Demand or arrival rateCapacity or Service rateNumber of servers (N)Utilization (U)

      U=D/(SR×N)

    RC10 per hour60/3=20perhour1 50.00%
    RRC10 per hour60/6=10perhour1100.00%
    NP10 per hour60/5=12perhour183.33%
    PA10×0.4=4perhour60/6=10perhour140.00%
    MD10×0.6+4×0.5=8perhour60/15=4perhour1200.00%
    BC4×0.5+4=6perhour60/5=12perhour150.00%

Note that the utilization of MD has just jumped to 200%.

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